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# Mapping rationals to strings

Show that there is an injective order-preserving map from $\mathbb{Q}$ to the set of finite strings $\{0,1\}^*$ with the lexicographical order.

The lexicographical order is defined as follows: $\alpha < \beta$ holds if and only if any of the two following conditions hold.

1. The string $\alpha$ is a proper prefix of $\beta$, i.e. $\beta = \alpha\gamma$ for some nonempty string $\gamma \in \{0,1\}^*$.
2. The strings are of the forms $\alpha = \gamma 0 \delta$ and $\beta = \gamma 1 \sigma$, for arbitrary strings $\gamma,\delta,\sigma \in \{0,1\}^*$.

For instance, $0 < 00 < 01 < 010 < 011 < 1 < 10$.

## 3 thoughts on “Mapping rationals to strings”

1. To prove this let us prove a pretty general lemma first. Consider infinitely countable, total ordered sets, with no maximum or minimum, and such that for every two elements $x < y$ there exists a $z$ such that $x < z < y$.

Lemma: Up to ismorphism there is only one set with this characteristics and this is in fact the set of rational numbers with its usual order.
Proof: Call two such sets $X$ and $Y$ and let $x_n$ and $y_n$ count the elements of $X$ and $Y$ repectively. We define an order preserving bijection by induction on $n$. At each step we will say where the bijection maps $x_0 ... x_n$ and where the inverse of the bijection maps $y_0 ... y_n$.
Base case: Take $x_0$ and send it to $y_0$.
Inductive case: Now assume one has the bijection (and its inverse) defined on $x_0 ... x_n$ and $y_0 ... y_n$. Suppose that $x_i$ goes to $y'_i$ and $y_i$ to $x'_i$. Take $x_{n+1}$, if you already mapped a $y_i$ with $i < n+1$ to it, you are done. Else see what is his order with respect to $x_0 ... x_n$. Suppose that the order goes: $... < x_i < x_{n+1} < x_j < ...$ . Then just send $x_{n+1}$ to an element $y \in Y$ such that $y'_i < y < y'_j$ and $y$ is different from all the others $y_i$, $i. This element exists by hypothesis. It might be the case that $x_{n+1}$ is the maximum or the minimum of the $x_0 ... x_{n+1}$. This is not a problem since the fact that $Y$ has no maximum or minimum tells us that we will always have a new y with the correct relative order to choose. Now make exactly the same reasoning but with $y_{n+1}$.
This concludes the inductive step.
Notice that the constructed function is indeed a bijection since we didn't leave out any $x$ or $y$.

With this lemma we are almost done. Notice that if we take all the strings that end in 0 out of {0,1}* we get an ordered set in the hypothesis of the lemma. This is not obvious, it is easy to see that there is no maximum and no minumum, but one has to consider some cases to see that for every pair of distinct elements there exists another element in between. Suppose we are given $x < y$. We consider two cases: either $x$ is a prefix of $y$ or it isn't. In the first case take $z$ to be $x$ concatenated with $0…01$. Where we put enough 0's such that $z$ is longer than $y$. In the second case $x$ and $y$ differ in a coordinate $i$ and thus $x$ has a $0$ in the coordinate $i$ and $y$ has a $1$. Take $z$ to be $x$ concatenated with just a $1$.

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2. fmartin says:

Two minor observations:

1) The order topology in $\{0,1\}^*$ is the metric topology induced by the ultrametric $d(x,y) = 2^{-n}$, where $n$ is the smallest index in which $x$ and $y$ differ. This popped up when I tried to think about the problem topologically.

2) Luis’ argument actually characterizes the structure of $\{0,1\}^*$; it is order isomorphic to $\mathbb{N}+(\mathbb{Q}\times \mathbb{N})$. Let $X$ denote the set of finite binary strings ending in $1$ and let $f:\mathbb{Q}\to X$ be an order isomorphism. If $Y$ denotes the set of finite binary strings containing at least one $1$, then we may produce an order isomorphism $g:\mathbb{Q}\times\mathbb{N}\to Y$ by setting $g(x,n)$ as the concatenation of $f(x)$ and $n$ zeroes. Since the set of strings not containing any $1$s is order isomorphic to $\mathbb{N}$ and each of them is smaller than any element in $Y$, the result follows.

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3. This is the original proof I had in mind; it works for any countable set $X$ having a total order.

Consider an enumeration $x_0, x_1, \hdots$ of the elements of $X$. Let $f : X \to \{0,1\}^*$ be the map defined by $f(x_n) = a_1 \hdots a_n$, where $a_i = 1$ if $x_i \leq x_n$ and $a_i = 0$ otherwise.

Note that $f$ is injective since $f(x_n)$ is a string of length $n$. To see that $f$ is order-preserving, let $x_n < x_m$, consider their images $f(x_n) = a_1 \hdots a_n$ and $f(x_m) = b_1 \hdots b_m$, and let us show that $a_1 \hdots a_n < b_1 \hdots b_m$.

Observe that $a_i \leq b_i$ for all $1 \leq i \leq \min\{n,m\}$. Indeed, if $x_i \leq x_n$ then $a_i = 1$ and $b_i = 1$ since $x_i \leq x_n < x_m$. On the other hand, if $x_i > x_n$ then $a_i = 0$ and trivially $a_i \leq b_i$.

To conclude the proof we consider two cases:

1. If $n < m$, then $a_i \leq b_i$ for all $1 \leq i \leq n$. Hence $a_1 \hdots a_n \leq b_1 \hdots b_n < b_1 \hdots b_n b_{n+1} \hdots b_m$.

2. If $n > m$, then $a_i \leq b_i$ for all $1 \leq i \leq m$. Moreover, $a_m = 0$, since it is not the case that $x_m \leq x_n$, but $b_m = 1$, since it is the case that $x_m \leq x_m$. Hence $a_1 \hdots a_{m-1} a_m a_{m+1} \hdots a_n = a_1 \hdots a_{m-1} 0 a_{m+1} \hdots a_n \leq b_1 \hdots b_{m-1} 0 a_{m+1} \hdots a_n < b_1 \hdots b_{m-1} 1 = b_1 \hdots b_{m-1} b_m$, as required.

(Simple lemma we used: if $a_i \leq b_i$ for all $1 \leq i \leq n$ then $a_1 \hdots a_n \leq b_1 \hdots b_n$).

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How do you simplify the square root of 2466?

Jun 27, 2018

See a solution process below:

Explanation:

First, rewrite the expression as:

$\sqrt{2466} \implies \sqrt{9 \cdot 274}$

Then use this rule for radicals to simplify the expression:

$\sqrt{\textcolor{red}{a} \cdot \textcolor{b l u e}{b}} = \sqrt{\textcolor{red}{a}} \cdot \sqrt{\textcolor{b l u e}{b}}$

$\sqrt{\textcolor{red}{9} \cdot \textcolor{b l u e}{274}} \implies \sqrt{\textcolor{red}{9}} \cdot \sqrt{\textcolor{b l u e}{274}} \implies 3 \sqrt{274}$


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Cologarithm

Cologarithm

In mathematics, the base-b cologarithm, sometimes shortened to colog, of a number is the base-b logarithm of the reciprocal of the number. It is equal to the negative base-b logarithm of the number.

$\operatorname{colog}_b\ x = \log_b \left(\frac{1}{x} \right) = \log_b 1-\log_b x = -\log_b x = \log_{1/b} x.\,$

In chemistry, a decimal cologarithm is indicated by the letter p (originally the Greek letter ρ)[citation needed].
For example, pK = – log10 K and pH = – log10 [H3O+].


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## What is…Inversive Geometry?

It’s been a while since we’ve looked at a new type of geometry, so today I’d like to introduce inversive geometry.  Let’s look at a few pictures first for some motivation.

We would like to consider the operation of inverting points, as illustrated above.  Suppose the red circle is a unit circle, the white dot is the origin, and consider the blue point P.  Now define P′ (the orange dot) to be the inverse point of P as follows:  draw a ray from the origin through P, and let P′ be that point on the ray such that

$[OP]\cdot[OP']=1,$

where $[AB]$ denotes the distance from A to B. That is, when you multiply the distances from the origin of a point and its inverse point, you get 1.

Now why would you want to do this?  It turns out this operation is actually very interesting, and has many unexpected consequences.  Consider the circles in the figure below.

Start with the blue circle.  Now for every point P on the blue circle, find its inverse point P′, and plot it in orange.  The result is actually another circle!  Yes, it works out exactly, although we won’t prove that here.  Note that since every point on the blue circle is outside the unit circle (that is, greater than a distance of 1 from the origin), every inverse point on the orange circle must be inside the unit circle (that is, less than a distance of 1 from the origin).

This is just one geometrical property of inversion, and there are many others.  I just want to suggest that the operation of inversion has many interesting properties.

If you pause a moment to think about this operation, you’ll notice there’s a little snag.  Not every point has an inverse point.  There’s just one point which is problematic here:  the origin.  According to the definition,

$[OO]\cdot[OO']=1.$

But the distance from the origin to itself is just 0, and it is not possible to multiply 0 by another number and get 1, since 0 times any real number is still 0.

Unless…could we somehow make the distance from O to O′ infinite?  It should be clear that O′ cannot be a point in the plane different from the origin, since any point in the plane has a finite distance to the origin (just use the Pythagorean theorem).

So how do we solve this problem?  We add another point to the usual Euclidean plane, called the point at infinity, which is usually denoted by the Greek letter ω.

You might be thinking, “Hey, wait a minute!  You can’t just add a point because you need one.  Where would you put it?  The plane already extends out to infinity as it is!”

In a sense, that’s correct.  But remember the idea of this thread — we’re exploring what geometry is.  When we looked at taxicab geometry, we just changed the distance, not the points.  And with spherical geometry, we looked at a very familiar geometrical object:  the surface of a sphere.  We can also change the points in our geometrical space.

How can we do this?  Though perhaps a simplification, we must do this consistently as long as it is interesting.

What does this mean?  In some sense, I can mathematically define lots of things.  Let’s suppose I want to define a new operation on numbers, like this:

$a\otimes b=a^2+37b.$

Wow, we’ve just created something likely no one has thought of before!  And yes, we probably have — and for good reason.  This operation is not very interesting at all — no nice properties like commutativity or associativity, no applications that I can think of (what are you going to do with the 37?).  But it is a perfectly legitimate arithmetical operation, defined for all real numbers a and b.

As I tried to suggest earlier, the operation of inversion is actually quite interesting.  So it would be very nice to have a definition which allowed every point to have an inverse point.

But if we want to add ω, we must do so consistently.  That is, the properties of ω cannot result in any contradictory statements or results.

It turns out that this is in fact possible — thought we do have to be careful.  Although we can’t go into all the details of adding the point at infinity to the plane, here are some important properties that ω has:

1. The distance from ω to any other point in the plane is infinite;
2. ω lies on any unbounded curve (like a line or a parabola, but not a circle, for example);
3. The inverse point of ω is the origin; that is, O′ = ω and ω′ = O.

So we can create an entirely consistent system of geometry which contains all the points in the Euclidean plane plus a point at infinity which is infinitely far away from every other point.  We usually call the Euclidean plane with ω added the extended plane.

With the new point, we can now say that every point in the extended plane has a unique inverse point.

This idea of adding a point at infinity is also important is other areas of mathematics.  In considering the complex plane — that is, the set of points of the form abi, where i is a solution of the equation $x^2+1=0$ — it is often useful to add the number ∞, so that division by 0 is now possible.  This results in the extended complex plane, and is very important in the study of complex analysis.

The point at infinity is also important in defining stereographic projection.  Here, a sphere is placed on the plane, so that the South pole is the origin of the plane.  The sphere is then projected onto the plane as follows:  for any point on the sphere, draw a ray from the North pole through that point, and see where it intersects the plane.

Where does the North pole get mapped to under this projection?  Take one guess:  the point at infinity!

So here is another new geometry!  As we get introduced to more and more various geometrical systems, I hope you will continue to deepen your intuition about the question, What is a geometry?

## Imagifractalous! 5: Binary Trees III

Last week I talked about working with binary trees whose branching ratio is 1 or greater.  The difficulty with having a branching ratio larger than one is that the tree keeps growing, getting larger and larger with each iteration.

But when you work with software like Mathematica, for example, and you create such a tree, you can specify the size of the displayed image in screen size.

So the trees above both have branching ratio 2 and branching angle of 70°.  The left image is drawn to a depth of 7, and the right image is drawn to a depth of 12.  I specified that both images be drawn the same size in Mathematica.

But even though they are visually the same size, if you start with a trunk 1 unit in length, the left image is about 200 units wide, while the second is 6000 units wide!

So this prompted us to look at scaling back trees with large branching ratios.  In other words, as trees kept getting larger, scale them back even more.  You saw why this was important last week:  if the scale isn’t right, when you overlap trees with r less than one on top of the reciprocal tree with branching ratio 1/r, the leaves of the trees won’t overlap.  The scale has to be just right.

So what should these scale factors be?  This is such an interesting story about collaboration and creativity — and how new ideas are generated — that I want to share it with you.

For your usual binary tree with branching ratio less than one, you don’t have to scale at all.  The tree remains bounded, which is easy to prove using convergent geometric series.

What about the case when r is exactly 1, as shown in the above figure?  At depth n, if you start with a trunk of length 1, the path from the base of the trunk to the leaf is a path of exactly n + 1 segments of length 1, and so can’t be any longer than n + 1 in length.  As the branching angle gets closer to 0°, you do approach this bound of n + 1.  So we thought that scaling back by a factor of n + 1 would keep the tree bounded in the case when r is 1.

What about the case when r > 1?  Let’s consider the case when r = 2 as an example.  The segments in any path are of length 1, 2, 4, 8, 16, etc., getting longer each time by a power of 2.  Going to a depth of n, the total length is proportional to $2^n$ in this case.  In general, the total length is about $2\cdot r^n$ for arbitrary r, so scaling back by a factor of $r^n$ would keep the trees bounded as well.

So we knew how to keep the trees bounded, and started including these scaling factors when drawing our images.  But there were two issues.  First, we still had to do some fudging when drawing trees together with their reciprocal trees.  We could still create very appealing images, but we couldn’t use the scale factor on its own.

And second — and perhaps more importantly — Nick had been doing extensive exploration on his computer generating binary trees.  Right now, we had three different cases for scaling factors, depending on whether r < 1, r = 1, or r > 1.  But in Nick’s experience, when he moved continuously through values of r less than 1 to values of r greater than one, the transition looked very smooth to him.  There didn’t seem to be any “jump” when passing through r = 1, as happened with the scale factors we had at the moment.

I wasn’t too bothered by it, though.  There are lots of instances in mathematics where 1 is some sort of boundary point.  Take geometric series, for example.  Or perhaps there is another boundary point which separates three fundamentally different types of solutions.  For example, consider the quadratic equation

$x^2+c=0.$

The three fundamentally different solution sets correspond to  c < 0, c = 0, and c > 0.  There is a common example from differential equations, too, though I won’t go into that here.  Suffice it to say, this type of trichotomy occurs rather frequently.

I tried explaining this to Nick, but he just wouldn’t budge.  He had looked at so many binary trees, his intuition led him to firmly believe there just had to be a way to unify these scale factors.

I can still remember the afternoon — the moment — when I saw it.  It was truly beautiful, and I’ll share it in just a moment.  But my point is this:  I was so used to seeing trichotomies in mathematics, I was just willing to live with these three scale factors.  But Nick wasn’t.  He was tenacious, and just insisted that there was further digging to do.

Don’t ask me to explain how I came up with it.  It was like that feeling when you just were holding on to some small thing, and now you couldn’t find it.  But you never left the room, so it just had to be there.  So you just kept looking, not giving up until you found it.

And there is was:  if the branching ratio was and you were iterating to a depth of n, you scaled back by a factor of

$\displaystyle\sum_{k=0}^n r^k.$

This took care of all three cases at once!  When r < 1, this sum is bounded (think geometric series), so the boundedness of the tree isn’t affected.  When r = 1, you just get n + 1 — the same scaling factor we looked at before!  And when r > 1, this sum is large enough to scale back your tree so it’s bounded.

Not only that, this scale factor made proving the Dual Tree Theorem so nice.  The scaling factors for a tree with r < 1 and its reciprocal tree with branching ratio 1/r matched perfectly.  No need to fudge!

This isn’t the place to go into all the mathematics, but I’d be happy to share a copy of our paper if you’re interested.  We go into a lot more detail than I ever could in a blog post.

This is how mathematics happens, incidentally.  It isn’t just a matter of finding a right answer, or just solving an equation.  It’s a give-and-take, an exploration, a discovery here and there, tenacity, persistence.  A living, breathing endeavor.

But the saga isn’t over yet….  There is a lot more to say about binary trees.  I’ll do just that in my next installment of Imagifractalous!

## Imagifractalous! 4: Fractal Binary Trees II

Now that the paper Nick and I wrote on binary trees was accepted for Bridges 2017 (yay!), I’d like to say a little more about what we discovered.  I’ll presume you’ve already read the first Imagifractalous! post on binary trees (see Day077 for a refresher if you need it).

Recall that in that post, I discussed creating binary trees with branching ratios which were 1 or larger.  Below are three examples of binary trees, with branching ratios less that 1, equal to 1, and larger than 1, respectively.

It was Nick’s insight to consider the following question:  how are trees with branching ratio r related to those with branching ratio 1/r?  He had done a lot of exploring with graphics in Python, and observed that there was definitely some relationship.

Let’s look at an example.  The red tree is a binary tree with branching ratio r less than one, and the gray tree has a branching ratio which is the reciprocal r.  Both are drawn to the same depth.

Of course you notice what’s happening — the leaves of the trees are overlapping!  This was happening so frequently, it just couldn’t be coincidence.  Here is another example.

Notice how three copies of the trees with branching ratio less than one are covering some of the leaves of a tree with the reciprocal ratio.

Now if you’ve ever created your own binary trees, you’ll likely have noticed that I left out a particularly important piece of information:  the size of the trunks of the trees.  You can imagine that if the sizes of the trunks of the r trees and the 1/r trees were not precisely related, you wouldn’t have the nice overlap.

Here is a figure taken from our paper which explains just how to find the correct relationship between the trunk sizes.  It illustrates the main idea which we used to rigorously prove just about everything we observed about these reciprocal trees.

Let’s take a look at what’s happening.  The thick, black tree has a branching ratio of 5/8, and a branching angle of 25°.  The thick, black path going from O to P is created by following the sequence of instructions RRRLL (and so the tree is rendered to a depth of 5).

Now make a symmetric path (thick, gray, dashed) starting at P and going to O.  If we start at P with the same trunk length we started with at O, and follow the exact same instructions, we have to end up back at O.

The trick is to now look at this gray path backwards, starting from O.  The branches now get larger each time, by a factor of 8/5 (since they were getting smaller by a factor of 5/8 when going in the opposite direction).  The size of the trunk, you can readily see, is the length of the last branch drawn in following the black path from O to P.  This must be (5/8)5 times the length of the trunk, since the tree is of depth 5.

The sequence of instructions needed to follow this gray path is RRLLL.  It turns out this is easy to predict from the geometry.  Recall that beginning at P, we followed the instructions RRRLL along the gray path to get to O.  When we reverse this path and go from O to P, we follow the instructions in reverse — except that in going in the reverse direction, what was previously a left turn becomes a right turn, and vice versa.

So all we need to do to get the reverse instructions is to reverse the string RRRLL to get LLRRR, and then change the L‘s to R‘s and the R‘s to L‘s, yielding RRLLL.

There’s one important detail to address:  the fact that the black tree with branching ratio 5/8 is rotated by 25° to make everything work out.  Again, this is easy to see from the geometry of the figure.  Look at the thick gray path for a moment.  Since following the instructions RRLLL means that in total, you make one more left turn than you do right turns, the last branch of the path must be oriented 25° to the left of your starting orientation (which was vertical).  This tells you precisely how much you need to rotate the black tree to make the two paths have the same starting and ending points.

Of course one example does not make a proof — but in fact all the important ideas are contained in this one illustration.  It is not difficult to make the argument more general, and we have successfully accomplished that (though this blog is not the place for it!).

If you look carefully at the diagram, you’ll count that there are exactly 10 leaves in common with these two trees with reciprocal branching ratios.  There is some nice combinatorics going on here, which is again easy to explain from the geometry.

You can see that these common leaves (illustrated with small, black dots) are at the ends of gray branches which are oriented 25° from the vertical.  Recall that this specific angle came from the fact that there was one more L than there were R‘s in the string RRLLL.

Now if you have a sequence of 5 instructions, the only way to have exactly one more L than R‘s is to have precisely three L‘s (and hence two R‘s).  And the number of ways to have three L‘s in a string of length 5 is just

$\displaystyle{5\choose3}=10.$

Again, these observations are easy to generalize and prove rigorously.

And where does this take us?

On the right are 12 copies of a tree with a braching ratio of r less than one and a branching angle of 30°, and on the left are 12 copies of a tree with a reciprocal branching ratio of 1/r, also with a branching angle of 30°.  All are drawn to depth 4, and the trunks are appropriately scaled as previously discussed.

These sets of trees produce exactly the same leaves!  We called this the Dual Tree Theorem, which was the culmination of all these observations.  Here is an illustration with both sets of trees on top of each other.

As intriguing as this discovery was, it was only the beginning of a much broader and deeper exploration into the fractal world of binary trees.  I’ll continue a discussion of our adventures in the next installment of Imagifractalous!

## What is…Spherical Geometry?

This week, we’ll look at another type of geometry, namely spherical geometry.  Quite simply, this is the geometry of a sphere.  Here, a sphere is a set of points equidistant from a given center.  In other words, throughout this post, you should imagine only a surface, with nothing inside it — think of the rind of an orange, without any of the slices.  I began to describe spherical geometry in my original post, What is a Geometry?, so I’ll briefly summarize the ideas in that post first.  It wouldn’t hurt to review it…

From a Euclidean standpoint, there are points on the sphere, but obviously no straight lines.  In spherical geometry, we define a Point to be a pair of antipodal points on the sphere, and a Line to be a great circle on the sphere.  This results in two nice theorems of spherical geometry:  any two distinct Lines determine (intersect in) a single Point, and any two distinct points determine a single Line.

This is a departure from Euclidean geometry, for this means that there are no parallel Lines in spherical geometry, since distinct Lines always intersect.  But there is something more going on here.

Consider the statement “Any two distinct Lines determine a single Point.”  Now perform the following simple replacement:  change the occurrence of “Line” to “Point,” and vice versa.  This gives the statement “Any two distinct Points determine a single Line,”  and is called the dual of the original statement.

Thus, we have the situation that some statement and its dual are both true.  Now if this is true of some set of statements in spherical geometry — that the dual of each statement is true — and we derive a new result from this set of statements, then the following remarkable thing happens.  Since the dual of any statement we used is true, then the dual of the new result must also be true!  Just replace each statement used to derive the new result with its dual, and you get the dual of the new result as a true statement as well.

This is the principle of duality in mathematics, and is a very important concept.  We will encounter it again when we investigate projective geometry.

Triangles and trigonometry are different on the sphere, as well.  A spherical triangle is composed of three arcs of great circles, as in the image below.

If this were Earth, you could imagine starting at the North Pole, following 0 degrees of longitude to the Equator (shown in yellow), follow the Equator to 30 degrees east longitude, then follow this line of longitude back to the North Pole.

On the sphere, though, sides of a triangle are actually angles.  Sure, you could measure the length of the arcs given the radius of the sphere, but that’s not as useful.  Consider the choice of units on the Earth — kilometers or miles?  We don’t want important  geometrical results to depend on the choice of units.  So a side is specified by the angle subtended by the arc at the center of the sphere.  This makes sense since the sides are arcs of great circles, and the center of any great circle is the center of the sphere.

The angles between the sides are angles, too!  A great circle is just the intersection of a plane passing through the origin and the sphere.  So the angle between any two sides is defined to be the angle between the planes which contain them.

Really nothing from Euclidean trigonometry is valid on the sphere.  For example, if you look at the triangle above, you can see that the angles between the sides are 30, 90, and 90 degrees.  These angles add to 210!  In fact, the three angles of a triangle always add up to more than 180 degrees on a sphere.  (You may notice that the sides of this triangle are also 30, 90, and 90 degrees, but this is just a coincidence.  The sides and angles are usually different.)

If A, B, and C are the angles of a spherical triangle, it turns out that the area of the triangle is proportional to ABC – 180.  This means that the smaller a triangle is, the closer the angle sum is to 180 degrees.

There is no Pythagorean Theorem on the sphere, either.  In addition, if a, b, and c are the sides opposite angles A, B, and C, respectively, then we have formulas like

$\cos c=\cos a\cos b+\sin a\sin b\cos C$

and

$\cos C=-\cos A\cos B+\sin A\sin B\cos c.$

One interesting consequence of the second of these formulas is this.  If you know the angles of a triangle, you can determine the sides.  You can’t do this in Euclidean trigonometry, since the triangles may be similar, but of different sizes.  In other words, there are no similar triangles on the sphere.  Spherical triangles are just congruent, or not. You can’t have two different triangles with the same angles.

For now, we’ve considered our sphere as being embedded in a Euclidean space.  The definition of this surface is easy:  just choose a point as the center of your sphere, and then find all points which are a given, fixed distance — the radius of the sphere — from that point.  Sounds easy enough.

But can you imagine a sphere without thinking of the three-dimensional space around it?  Or put another way, imagine you were a tiny ant, on a sphere of radius 1,000,000 km.  That’s over 150 times the radius of the Earth!  How would you know you were actually on the surface of a sphere?  If you were that small and the sphere were that large, it would seem awfully flat to you….

So how could you determine the sphere was curved?  This is a question for differential geometry, which among other things, is about the geometry of a surface without any reference to a space it’s embedded in.  This is called the surface’s intrinsic geometry.

As an example of looking at the intrinsic geometry of the sphere, consider Lines.  Now you can’t say they’re great circles any more, since this relies on thinking of a sphere as being embedded in three-dimensional space; in other words, its extrinsic geometry.  You need the concept of a geodesic — in other words, the idea of a “shortest path.”

So if you’re the ant crawling between two points on a sphere, and you wanted to take the shortest path, you would have to follow a great circle arc.  So it is possible to define Lines only using properties of the surface itself.  But the mathematics to do this is really extremely challenging.

Lots of new ideas here — but we’ve just scratched the surface of a study of spherical geometry.  You can see how very different spherical geometry is from both Euclidean and taxicab geometry.  Hopefully you’re well on your way to wrapping your head around our original question, What is a Geometry?….

## A New Cantor Set?

Although we’ll do a little review, this post assumes some familiarity with the Cantor set as well as working with fractions in bases other than 10.

So what is the Cantor set?

The usual geometrical definition is to begin with a line segment of length one (as shown at the top of the figure).  Next, remove the middle third of that segment, so that two segments of length 1/3 are formed.  Then remove the middle thirds from those segments, and continue recursively.

Even though — after this process is repeated infinitely, of course — you remove segments whose lengths sum to 1, there are still infinitely many points left!  (Actually, uncountably infinitely many, although this observation is not necessary for our discussion.)

There are many ways to describe the remaining points, but one common way is to say that the Cantor set consists of all those numbers between 0 and 1 (inclusive) which, when written in base 3, may be written only with 0’s and 2’s.

Again, we’ll briefly review.  We have .23 = 2/310, for example — since the places after the ternary point (no “deci”mals here) represent powers of 1/3.

So if we wanted to find 3/4 in base 3, we note that we’d need 2/3 = 0.23, but there would be 1/12 left over. This is smaller than 1/9, so we’re at 0.203 so far. Next, we need 2/27, giving 0.2023, with 1/108 left over. And so on. The result is that 3/4 is equal to 0.202020…3, where the “20” repeats. This can also be shown using infinite geometric series, if desired:

$\dfrac34=\dfrac{2/3}{1-1/9}.$

Surprisingly, the idea for a possibly new type of “Cantor set” came from studying binary trees!  I say possibly new, since I couldn’t find any reference to such a set online, but of course that doesn’t mean someone else hasn’t come across it.  And I call it a type of Cantor set since it may also be formed by taking out thirds of segments, but in a slightly different way than described above.

Now I’ve talked a bit about binary trees before, so I won’t go into great detail.  But here is the important idea:  when you make a branch, you’re pointing in some particular direction, and then turn either left or right, but you can’t just keep going in the same direction.

So what if you looked at ternary expansions, and as you added digits, you had the option of adding 1 to the previous digit (like turning left), or subtracting 1 (like turning right), but you couldn’t use the same digit twice consecutively.  So 0.21021201 would be OK (we’ll drop the 3 subscripts since we’ll be working exclusively in base 3 from now on), but 0.12002120 would not be allowed since there are consecutive 0’s.

Note that adding 1 to 2 gives 0 in base 3, and subtracting 1 from 0 gives 2.  So essentially, starting with 0., you build ternary expansions with the property that each digit is different from the previous one.  And, of course, the expansions must be infinite….

What do iterations of this scheme look like?

We start with a segment of length 1.  Recall we begin with 0., so that means the ternary expansions may begin with 0.1 or 0.2.  Expansions beginning with 0.0 are not allowed, so this precludes the first third of the segment.

Now here comes the interesting part!  On the second iteration, (the third line from the top), we remove different thirds from the first two segments.  Since the 0.1 may continued onto 0.10 or 0.12, but not 0.11, we remove the middle third from the 0.1 segment.  Further, 0.2 may be continued as 0.20 or 0.21, but not 0.22, so we remove the last third of the 0.2 segment.  The iteration process is not symmetrical.

We continue on….  Since 0.10 may be continued as 0.101 or 0.102, but not 0.100, we remove the first third of the 0.10 segment.  You get the idea.  Seven iterations of this procedure are shown in the figure above.

Note that since the process for creating the original Cantor set is symmetrical, this imposes a self-similarity on the set itself.  The Cantor set is exactly the union of two duplicate copies of the original, scaled by a factor of 1/3.

In other words, the Cantor set may also be created using an iterated function system with the following two transformations:

$y=\dfrac13x,\qquad y=\dfrac13x+\dfrac23.$

What about the self-similarity of the new Cantor set?  To help in seeing this, here’s a slightly more detailed version of the iteration scheme.

Is there any self-similarity here?  Yes, but the fewest number of transformations I’ve found to describe this self-similarity is five.  The curious reader is welcome to find fewer!

It isn’t hard to see the five vertical bands in this figure — the first three look the same (although the second one appears to be reflected), and the last two also look the same, although reflections of each other.

The first band is all ternary expansions in this new set beginning with 0.10.  How do these relate to the whole set?  Well, 1/9 of the set consists of expansions beginning with 0.001… or 0.002…, and then adding digits different from those previous.  Adding 1/3 therefore gives all expansions beginning with 0.101… or 0.102…, and then adding different digits.  This implies that the self-similarity describing the first vertical band is

$y=\dfrac 19x+\dfrac13.$

The second band consists of those expansions in our strings beginning with 0.12.  But if x is an expansion in our set beginning with 0.10, then 1 – x must be an expansion in our set beginning with 0.12, since we may write 1 as .22222…, repeating.  Therefore, the second band is represented by the transformation

$y=1-\left(\dfrac 19x+\dfrac13\right)=\dfrac23-\dfrac19x.$

We can think of the third band just as we did the first — except that this band consists of number beginning with 0.20 (rather than 0.10).  So this band is represented by the transformation

$y=\dfrac 19x+\dfrac23.$

The last two bands consist of those expansions beginning with 0.21.  Here, we break into the two cases 0.210 and 0.212, and use our previous work.  For example, those beginning with 0.210 can be found by taking those beginning with 0.10, dividing by 3 to get expansions beginning with 0.010, and then adding 2/3 to get expansions beginning with 0.210:

$y=\dfrac13\left(\dfrac 19x+\dfrac13\right)+\dfrac23=\dfrac1{27}x+\dfrac79.$

We describe the self-similarity of the last band — those expansions beginning with 0.212 — analogously:

$y=\dfrac13\left(\dfrac23-\dfrac 19x\right)+\dfrac23=\dfrac89-\dfrac1{27}x.$

These are five transformations which describe the self-similarity of the new Cantor set.  I haven’t rigorously proved yet that you can’t do it in fewer, but I think this is the case.

Of course this is just the beginning — you can use the same rule in any base.  That is, begin with 0., and either add or subtract one from the previous digit.  In base 4, you get nice self-similarity, but it gets more involved in higher bases.  In bases higher than 3, you can also use the rule that the next digit in the expansion is different than the previous — and this gives yet another class of Cantor sets.  I’ll leave you to investigate, and perhaps I’ll write again about these Cantor sets when I find out more!

## What is…Taxicab Geometry?

It’s time to start exploring different geometries!  Today we’ll look at taxicab geometry because algebraically, it’s the easiest one to work with.  I discussed it briefly before — recall that lines and points are the same as those in the Euclidean geometry we’re used to, but the idea of distance is different.  Here, the distance between two points P and Q is how far a taxi would have to drive on a rectangular grid to get from P to Q.

You can see from the figure that the distance between (-2,3) and (3,-1) is 9, since you’d have to drive nine blocks to get from one point to the other.  This might seem simple enough, but the situation is quite a bit different from Euclidean geometry.

In Euclidean geometry, the shortest distance between two points is a straight line segment.  If you deviate from this segment in any way in getting from one point to the other, your path will get longer.

In taxicab geometry, there is usually no shortest path.  If you look at the figure below, you can see two other paths from (-2,3) to (3,-1) which have a length of 9.

Strange!  There are a few exceptions to this rule, however — when the segment between the points is parallel to one of the axes.  For example, the distance from (0,0) to (4,0) is just 4 — and there’s only one way to get there along a path of length 4.  You’ve got to travel along the line segment joining (0,0) and (4,0).

Well, perhaps “strange” is the wrong word.  All of this is perfectly normal in taxicab geometry….

Now let’s take a look at triangles.  We define them in the usual way — choose three points, and connect them in pairs to form three sides of a triangle.  Just like in Euclidean geometry.

Because we’re so familiar with them, I’ve drawn what would be — if we were in the Euclidean realm! — two 3-4-5 right triangles.  You’re welcome to verify that OP’Q’ would indeed be a 3-4-5 triangle in Euclidean geometry.

OK, now let’s look at these triangles from the perspective of taxicab geometry.  While we can just study these triangles by looking at the graph, it might be helpful to formally define the taxicab distance function.  If two points have coordinates $P=(x_1,y_1)$ and $Q=(x_2,y_2),$ the distance between them is defined to be

$d_T(P,Q)=|x_2-x_1|+|y_2-y_1|.$

First, a word about saying that we’re defining the distance.  That’s right — a different distance, like the one we’re used which comes from the Pythagorean theorem, would produce a different geometry.  So how your geometry “works” depends upon how you define the distance.

Second, a word about the formula.  Take a moment to convince yourself that $|x_2-x_1|$ is how far your taxicab would have to drive in an east-west direction, and $|y_2-y_1|$ is how far your taxicab would have to drive in a north-south direction.  Absolute values are needed:  the distances $d_T(P,Q)$ and $d_T(Q,P)$ should be positive and equal to each other.

Back to the triangles.  In a taxicab world, OPQ is a 3-4-7 triangle!  That might strike you as a bit unusual, since in the Euclidean world, the sum of the lengths of any two sides of a triangle is always greater than the length of the third.  This is called the triangle inequality, and is usually written

$d_E(X,Z)\le d_E(X,Y)+d_E(Y,Z)$

for points XY, and Z, where we write $d_E$ to mean the Euclidean distance.  In Euclidean geometry, the only way this inequality can actually be an equality is if Y is on the line segment whose endpoints are X and Z — and no other way.  But as we’ve just seen,

$d_T(O,Q)= d_T(O,P)+d_T(P,Q),$

where P is evidently not on the segment joining O and Q.  Yet another difference between taxicab and Euclidean geometries.

Oh, but it gets better….  In the figure above — at least in a Euclidean world — when you rotate two triangles, the angles and lengths remain the same.  In other words, the triangles are congruent.  It may seem obvious, since clearly the two triangles look the same.

What happens when we “rotate” triangle OPQ in a taxicab world?  I put “rotate” in quotes since we haven’t looked at whether or not the Euclidean idea of “angle” has a companion in the taxicab world.

If you do the calculations, you’ll see that OP’Q’ is in fact a 21/5 – 5 – 28/5 triangle!  Moreover, even though OQ was the longest side of triangle OPQ, OQ’ is not the longest side of triangle OP’Q’!  The longest side is actually OP’.

What’s going on here?  We are so used to being able to draw a triangle on a piece of paper — and certainly, if we turn our piece of paper, the lengths of the sides of the triangle don’t change!  The triangle stays exactly the same — this is the notion of congruence.  But when you change the distance function to $d_T,$ the geometry changes.  In a very fundamental way, as you see.

Essentially, we need to give up the idea of “angle” in taxicab geometry.  That might sound odd — how can you have geometry without angles?  As it turns out, there are lots of ways to do this.  We’ll be looking at this idea more as this thread continues.

And although there are differences, there are certainly things in common — such as the fact that there is a triangle inequality in both geometries.  Is this just coincidence?  Mathematically, a distance function is one which satisfies, for points X, Y, and Z:

1.  $d(X,Y) \ge 0, {\rm \ with\ equality\ only\ when\ } X=Y,$

2.  $d(X,Y)=d(Y,X), {\rm\ and}$

3.  $d(X,Z)\le d(X,Y)+d(Y,Z).$

Now we’ve already encountered properties 2 and 3 so far, and property 1 simply says that two differents points have to be separated by some positive distance.

The concept of a distance function evolved over time — whenever it made sense to talk about distance, it seemed these properties were always in play.  And in fact, the distance functions in both Euclidean and taxicab geometry have these three propertes.

So different distance functions produce different geometries.  Perhaps the hardest thing about encountering new geometries is putting aside ideas from Euclidean geometry — “forgetting” all about angles, for example, while we’re in a taxicab world.  Hopefully, as you learn about more and more non-Euclidean geometries, it will become easier and easier to navigate this Universe of diverse geometrical worlds….

## Imagifractalous! 3: Fractal Binary Trees

I’ve taken a break from Koch-like curves and p-adic sequences for an arboreal interlude….  Yes, there’s a story about why — I needed to work with Nick on a paper he was writing for Bridges — but that story isn’t quite finished yet.  When it is, I’ll tell it.  But for now, I thought I’d share some of the fascinating images we created along the way.

Let’s start with a few examples of simple binary trees.  If you want to see more, just do a quick online search — there are lots of fractal trees out there on the web!  The construction is pretty straightforward.  Start by drawing a vertical trunk of length 1.  Then, move left and right some specified angle, and draw a branch of some length r < 1.  Recursively repeat from where you left off, always adding two more smaller branches at the tip of each branch you’ve already drawn.

If you look at these two examples for a moment, you’ll get the idea.  Here, the angle used is 40 degrees, and the ratio is 5/8.  On the left, there are 5 iterations of the recursive drawing, and there are 6 iterations on the right.

Here’s another example with a lot more interaction among the branches.

This type of fractal binary tree has been studied quite a bit.  There is a well-known paper by Mandelbrot and Frame which discusses these trees, but it’s not available without paying for it.  So here is a paper by Pons which addresses the same issues, but is available online.  It’s an interesting read, but be forewarned that there’s a lot of mathematics in it!

In trying to understand various properties of these fractal trees, it’s natural to write code which creates them.  But here’s the interesting thing about writing programs like that — once they’re written, you can input anything you like!  Who says that r has to be less than 1?  The tree above is a nice example of a fractal tree with r = 1.  All the branches are of the same length, and there is a lot of overlap.  This helps create an interesting texture.

But here’s the catch.  The more iterations you go, the bigger the tree gets.  In a mathematical sense, the iterations are said to be unbounded.  But when Mathematica outputs a graphic, it is automatically scaled to fit your viewing window.  So in practice, you don’t really care how large the tree gets, since it will automatically be scaled down so the entire tree is visible.

It is important to note that when r < 1, the trees are bounded, so they are easier to study mathematically.  The paper Nick and I are working on scales unbounded trees so they are more accessible, but as I said, I’ll talk more about this in a later post.

Here are a few examples with r > 1.  Notice that as there are more and more iterations, the branches keep getting larger.  This creates a very different type of binary tree, and again, a tree which keeps getting bigger (and unbounded) as the number of iterations increases.  But as mentioned earlier, Mathematica will automatically scale an image, so these trees are easy to generate and look at.

Nick created the following image using copies of binary trees with r approximately equal to 1.04.  The ever-expanding branches allow for the creation of interesting textures you really can’t achieve when r < 1.

Another of my favorites is the following tree, created with r = 1.  The angle used, though, is 90.9 degrees.  Making the angle just slightly larger than a right angle creates an interesting visual effect.

But the exploration didn’t stop with just varying r so it could take on values 1 or greater.  I started thinking about other ways to alter the parameters used to create fractal binary trees.

For example, why does r have to stay the same at each iteration?  Well, it doesn’t!  The following image was created using values of r which alternate between iterations.

And the values of r can vary in other ways from iteration to iteration.  There is a lot more to investigate, such as generating a binary tree from any sequence of r values.  But studying these mathematically may be somewhat more difficult….

Now in a typical binary tree, the angle you branch to the left is the same as the angle you branch to the right.  Of course these two angles don’t have to be the same.  What happens if the branching angle to the left is different from the branching angle to the right?  Below is one possibility.

And for another possibility?  What if you choose two different angles, but have the computer randomly decide which is used to branch left/right at each iteration?  What then?

Here is one example, where the branching angles are 45 and 90 degrees, but which is left or right is chosen randomly (with equal probability) at each iteration.  Gives the fractal tree a funky feel….

You might have noticed that none of these images are in color.  One very practial reason is that for writing Bridges papers, you need to make sure your images look OK printed in black-and-white, since the book of conference papers is not printed in color.

But there’s another reason I didn’t include color images in this post.  Yes, I’ve got plenty…and I will share them with you later.  What I want to communicate is the amazing variety of textures available by using a simple algorithm to create binary trees.  Nick and I never imagined there would be such a fantastic range of images we could create.  But there are.  You’ve just seen them.

Once the Bridges paper is submitted, accepted (hopefully!), and revised, I’ll continue the story of our arboreal adventure.  There is a lot more to share, and it will certainly be worth the wait!


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# what is the perimeter of a rectangle

Solution: Given that length = 11 cm and Breadth = 13cm. To find the perimeter of a rectangle, we need to know its height and width. Parthenon in Athens. Step 6. A rectangular carpet is 4m wide and 9m long. Example 1. Preview; Assign Practice; Preview. Below is a diagrammatic representation of rectangle. If a rectangle has length and width . … Perimeter of rectangle formula = 2 × (length + width) Perimeter of rectangle formula = 2 × (length + width) = 2( l + b ) Area of Rectangle. Finding the perimeter of a rectangle is always the sum of two times the length and two times the width. By the way, The perimeter of rectangle is, P = 2 ( L + B) Where L is length and B is the breadth of rectangle. In this non-linear system, users are free to take whatever path through the material best serves their needs. The rectangles below have the same area, but different perimeter. 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These unique features make Virtual Nerd a viable alternative to private tutoring. Example: the perimeter of this rectangle is 7+3+7+3 = 20. Example 1: Find the perimeter of a rectangle whose length and breadth are 11cm and 13cm, respectively. how much fence is needed to go around the rectangle). Work out the missing lengths and label the diagrams. 90 cmd. The Coordinate Plane. The park is in the shape of a rectangle which is 500m wide and 800m long. The Coordinate Plane. 42 cmc. Further Exploration. Rectangles are four-sided polygons. In the first equation, we already have isolated, we can plug it in to the second equation. You might be asking why we need this formula when we already know the perimeter, but this is the way to find the actual values of the length and width. And I would claim that the maximum perimeter rectangle inside the circle is the square. Find the distance, in metres, that Percy walks. A rectangle is a four-sided shape. Hence, the perimeter of the rectangle is 44 cm. Its opposite sides are parallel and of equal length, and its two diagonals intersect each other in the middle and are of … math. This is the equivalent of adding all four sides, since opposite sides are of equal length by definition. Diagonals of a Rectangle : A rectangle has two diagonals, they are equal in length and intersect in the middle. For any rectangle, the area is calculated by multiplying the length by the width e.g. In the figure above, drag any orange dot to resize the rectangle. The perimeter of a rectangle is defined as the sum of all the sides of a rectangle.For any polygon, the perimeter formulas are the total distance around its sides. If the perimeter of the rectangle is 54 inches, find the width of the rectangle. In online calculator you can use the value in the same units of measurement! 2. A 6xx6 square with area 36 square inches The rectangle has sides h and l, perimeter p and area a: (a) p=2h+2l=24 (b) a=h*l (a) h=12-l (b) a=(12-l)*l= 12l-l^2 The maximum of a occurs when the first derivative is zero: ( 12l-l^2)'= 12-2l=0 l=12/2=6 h=12-6=6. This tutorial takes you through the steps needed to find the perimeter of a rectangle. What is the perimeter of a rectangle when the area is 176. And the square is also stretched into a rectangle and one of its side is magnified by factor of $\sqrt{20/12}=\sqrt{5/3}$. Its two sides are equal and parallel to each other. Square and Rectangle Area and Perimeter. So P=2(w+L)=2(s+s)=4s. Remember that the area and the perimeter are two different things. Units: Note that units of length are shown for convenience. Hence, the length of the rectangle is 55 cm. The word perimeter is the combination of two Greek words, peri and metron.The word peri means around, and metron means to measure.Hence, the word perimeter means the total length of a 2D shape. m² 6. You can't graph a function or plot ordered pairs without a coordinate plane! The formula for the perimeter of a rectangle is (width + height) x 2, as seen in the figure below:. In other words, it is the boundary of a two-dimensional shape. Example: the perimeter of this regular pentagon is: 3+3+3+3+3 = 5×3 = 15 . N.B. It can be applied to different shapes, regular or irregular. Find the perimeter. Determine the Perimeter of a Rectangle Click here to choose another perimeter calculator The perimeter of a rectangle can be determined by using the following formula: where w is the width and l is the length of the rectangle. Perimeter = 2(a+b) Example: A rectangle is 12 cm long and 5 cm tall, what is its Perimeter? c) d) 2 3 Tommy is working out the perimeter of some rectangles. 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Twice the length decreased by three times the width is 4 inches. What is the perimeter of the rectangle? A rectangle is a two-dimensional shape that has four straight sides and four right angles (90°). Example: the perimeter of this rectangle is 7+3+7+3 = 20. - e-eduanswers.com Science Anatomy & Physiology Astronomy Astrophysics Biology Chemistry Earth Science Environmental Science … What is a rectangle? What is the Coordinate Plane? A rectangular carpet is 4m wide and 9m long. This indicates how strong in your memory this concept is. Define a variable for the width b. Perimeter P means adding up all 4 sides of a rectangle or P=w+w+L+L=2(w+L) where w is the width and L is the Length. a. Perimeter = 2 × (12 cm + 5 cm) = 2 × 17 cm = 34 cm. more than its width. (ii) Opposite sides of a rectangle are always the same in size. So, 18 = … Example 5: What is the perimeter of the following rectangle. The perimeter of a rectangle is the total length of all the sides of the rectangle. Example 1: Rectangle has the length 13 cm and width 8 cm.  Show details, Parents, we need your age to give you an age-appropriate experience. It can be applied to different shapes, regular or irregular. Area of a rectangle: The area of a rectangle is the product of its two adjacent sides . Read the lesson on area and perimeter of rectangles if you need to know more. That formula is p = 2l + 2w, which translates to English as "the perimeter of a rectangle is equal to two times the length plus two times the width." It is measured in square units, The area of a rectangle is given by the formula. The diagonals in a rectangle also are of the same length. Example 4: One side of a rectangle is 17 cm and another side is 12 cm. The Perimeter is the distance around the edges. Step 1. a) c) b) What do you notice? Perimeter of rectangle = 2L + 2W. The perimeter of the rectangle is 14 inches. The perimeter of a rectangle is the distance around the outside of the rectangle. Answers: 3 Show answers Another question on Mathematics. m² 6. Where: a is the length of the rectangle, and b is the width of the rectangle. Hence, the word perimeter means the total length of a 2D shape. The diagonals in a rectangle … Area means the space (inside) covered by shape or an object. 21 cmb. Perimeter of a Rectangle Formula. Find the perimeter of the rectangle above. So for example, let's say I have a rectangle. The base of the rectangle is the length of the bottom of the rectangle. Find the perimeter of the rectangle above. Simplifying us would give us. Due to the simplicity of the shape, measurements are easy to take and a perimeter calculator simplifies the calculation … Express the perimeter of the rectangle as a function of the length L of one of its sides. More in-depth information read at these rules. The length of a rectangle is 6 inches more than twice its width. All the sides of rectangle are not equal only the opposite sides of a rectangle are equal. If you have any difficulties with units conversion, you can use the length converter. What will be the perimeter of the rectangle? The perimeter P of a rectangle is given by the formula, P = 2 l + 2 w , where l is the length and w is the width of the rectangle. Classifying Polygons. The unit of the perimeter is centimeter (cm), meter (m), feet (ft), etc. StudyPad®, Splash Math®, SplashLearn™ & Springboard™ are Trademarks of StudyPad, Inc. Percy Cod takes a walk around the perimeter path of his local park. In rectangle, the distance around the outside of the rectangle is known as perimeter. 8 cm + 5 cm = 13 cm 13 cm × 2 = 26 cm The perimeter of a rectangle. 1) The perimeter of a rectangle is 24 inches. A rectangle is one of the many fundamental shapes you'll see in math. Find any other rectangles that have the same perimeter. Virtual Nerd's patent-pending tutorial system provides in-context information, hints, and links to supporting tutorials, synchronized with videos, each 3 to 7 minutes long. Write two equations: one for perimeter, which is 2 widths + 2 lengths =12 and two for area, which length times width = 5. Solution: One side of the rectangular playground is 20 m and the other side is 13 m. Therefore, the perimeter of the rectangular ground = 2 × (20 + 13) = 2 × (33) = 66 m. Type I: When the perimeter and only one of the sides are given. Perimeter of a Rectangle. In mathematics, the area and the perimeter are the two most important terms that we deal with. All the sides of rectangle are not equal only the opposite sides of a rectangle are equal. 180 cm. The perimeter of a rectangle is 106 inches,the length exceeds the width by 7 inches, find the length and the width . This could also be stated as 2*Side A + 2*Side B or 2*(Side A + Side B) Algebra . 5 Each of these rectangles has a perimeter of 24 cm. Perimeter is the distance around a two-dimensional shape. Following are the properties of a rectangle: (i) All the angles of a rectangle are 90º. We have the information we need to plug into our perimeter formula. https://www.wikihow.com/Find-the-Area-and-Perimeter-of-a-Rectangle As you all know, triangles have three sides, and they may or may not be equal. Fill in all the gaps, then press "Check" to check your answers. Step 2: The perimeter of the square = 4 × 8 = 32 m. Problem 2. Thus, by calculating the perimeter of a rectangle, you are interested in the total “length” of this rectangle. Math. Find the Area and Perimeter of a Rectangle : ----- Input the length of the rectangle : 10 Input the width of the rectangle : 15 The area of the rectangle is : 150 The perimeter of the rectangle is : 50 Flowchart: C++ Code Editor: Contribute your code and … Sorry, we could not process your request. Answer by sachi(548) (Show Source): Find the area of the carpet in square metres. This name derives from the fact that a rectangle is a quadrilateral with four right angles (4 * 90° = 360°). A rectangle is 2-dimensional; however, perimeter is 1-dimensional and is measured in linear units such as feet or meter etc. Given a length and breadth of a rectangle we have to find its area and Perimeter. Question 737310: The width of a rectangle is 2 less than half the length, and the area is 30. The perimeter of the rectangle is 30 in. © Copyright 2011-2018 www.javatpoint.com. Perimeter of a rectangle. The Perimeter is the distance around the edges. Find the distance, in metres, that Percy walks. Enter your answer using . Area measures the size of a surface. The perimeter of a rectangle is the total length of the two lengths and two widths of the rectangle. Hence, the perimeter of the rectangle is 58 cm. A rectangle has two diagonals, they are equal in length and intersect in the middle. 5 Each of these rectangles has a perimeter of 24 cm. Formula. However, since a rectangle's opposite sides are congruent, we only need to know the length and width. In mathematics, the word perimeter means a path that surrounds an area. What is the perimeter of the rectangle? Maximum Area A=w*L is when the rectangle is a square or where s=w=L. Plugging in … The length of a rectangle is 3 in. Finding the perimeter of a rectangle is as easy as adding up all its side lengths! What will be the other side? Finding the perimeter of a rectangle is as easy as adding up all its side lengths! Perimeter of Rectangle. 66 = 2l + 2w. A diagonal's … The perimeter of a rectangle is equal to the sum of all the sides. Perimeter. Example: the perimeter of this regular pentagon is:. This tutorial takes you through the steps needed to find the perimeter of a rectangle. The formula for finding the perimeter of a rectangle is simple the sum of all sides, or l+l+w+w, where l is the length and w is the width of the rectangle. This tutorial takes you through the steps needed to find the perimeter of a rectangle. Calculate the length of the rectangle. … Type II: Finding sides using the properties of a rectangle. Step 1: Side length of the square = 8 m . Correct answer to the question Perimeter of the rectangle Here is a rectangle: 7 cm 4 cm 1. Mail us on hr@javatpoint.com, to get more information about given services. Copyright © 2020 Studypad Inc. All Rights Reserved. The isoperimetric theorem for rectangles states that among all rectangles of a given perimeter, the square has … Find the length and width . The units to measure the perimeter … The Complete K-5 Math Learning Program Built for Your Child. Area of a rectangle. The two sides of the rectangle are given. Learn about the coordinate plane by watching this tutorial. Please mail your requirement at hr@javatpoint.com. Hence, the perimeter of the given figure is 140 m. Example 3: The perimeter of a field is 190 cm. cm 5. Solution: The perimeter of the rectangle, with one of the sides equal to 2 cm, is 10 cm. Example 1. We know that the perimeter is 30, so we can also plug that in for .. Find any other rectangles that have the same perimeter. If the perimeter of the given rectangle is 10 cm and the length of one of its sides is 2 cm. There are 4 white marbles and 1 black marble in a bag. The length of a rectangle is 5 inches less than 3 times the width. Suppose, we want to cover the field with iron wire from the protection of animals, how will we know what length should be the iron wire. JavaTpoint offers too many high quality services. Its perimeter is. Input length and width of the rectangle using scanf() function. Entering data into the perimeter of a rectangle calculator. The perimeter of a rectangle is the total length of all the sides of the rectangle. Practice. What is a Rectangle? Perimeter of a Rectangle. P = 2l + 2w. The perimeter of a rectangle is the sum of the lengths of the sides of the rectangle. if the width of a rectangle is 5 inches shorter than the length and the perimeter of the rectangle is 80 inches. The perimeter of a rectangle is the total length of all the four sides. This rectangle has a length of 160 yards and a width of 53 yards. 3 Tommy is working out the perimeter of some rectangles. Compute edge and coverage measures of rectilinear quadrilaterals, given linear measures. Write an expression … % Progress MEMORY METER. Make sure both are in the same unit, or convert one of them as necessary. Example 2: Same Area, Different Perimeter. The formula for the perimeter of a rectangle. 8 cm + 5 cm = 13 cm 13 cm × 2 = 26 cm The perimeter of a rectangle is its two times its length and width combined, expressed in this way: P = 2 x (l + w) Now then, given that the length is three cm more than twice the width, we can express the length in terms of the width in this way: l = 2w + 3 it has area =, it has perimeter = + = (+), each diagonal has length = +, and when =, the rectangle is a square. Duration: 1 week to 2 week. stands for perimeter in this equation. Perimeter of a rectangle formula. We know that P = 2 × l + 2 × w You should replace P by 18 and l by 5 because that is what they gave you. Perimeter of a Rectangle. State the domain of P. (Assume the length of the rectangle is larger than its width. Keywords: problem; perimeter; rectangle; integer; distance; length; width; Background Tutorials. P = 2l + 2w . Perimeter is the distance around a two-dimensional shape. a) c) b) What do you notice? Find the length and width of the rectangle. For a rectangle, it is given by {eq}P = 2(l + w) {/eq} All rights reserved. solve for perimeter of rectangle. The term rectangle comes from the Latin rectangulus that is a combination of two words: rectus (right, proper) and angulus (angle). Developed by JavaTpoint. Diagonal of a rectangle formula. A rectangle has a perimeter of 18 inches. Rectangle: Area to Perimeter Conversion: Let's say that we are provided with a rectangle of two unknown sides: length {eq}x {/eq} and width {eq}y {/eq}. If the length and width of a rectangle are l and w, then its perimeter = 2(l + w) units. Find the perimeter of the shape. Perimeter of rectangle … The length of a rectangle is 8 inches more than three times the width. 4. The Perimeter and Area of a Rectangle: Perimeter is the length or the distance of around a given geometrical shape. Progress % Practice Now. It means a figure or an object having the same perimeter can have a different area. While the perimeter is the distance (outside) around an object or shape. The Perimeter is 2 times (a + b):. That is:. Jake bought 6 roses and 3 carnations for $20.25. Rectangles … Hence, the formula for the perimeter of a rectangle = 2 × (sum of adjacent sides). Step 5. . We know that, the perimeter is the total boundary of the two-dimensional shape and a rectangle has four sides. Mathematics, 21.06.2019 15:00. Here are some definitions to help you with this lesson: Perimeter = The length of the outside edge; Rectangle = a 4 sided shape whose opposite sides are the … And a rectangle is a figure that has 4 sides and 4 right angles. Diagonals of a Rectangle. In the given rectangle, if a = 4 cm and d = 3 cm. Use this calculator if you know 2 values for the rectangle, including 1 side length, along with area, perimeter or diagonals and you can calculate the other 3 rectangle variables. It refers to the total length of the sides or edge of a polygon (a two-dimensional figure with angles). Area = length × width. You can input only integer numbers or fractions in this online calculator. Rectangle is 2-D figure containing four sides and four angles of 90 degree each. The perimeter is 88 inches. Print the value of perimeter. All the angles are 90°. Example 2. All the tiles are red, except those along the two diagonals, which are all blue. Find out how to work out the perimeter of a rectangle, even if you don't know the length of all four sides, in this Catch-Up Lesson. The area of a rectangle is the product of the length and width. Find its perimeter. What is the area of the rectangle? An object having the same area can have a different perimeter. All rectangles are also parallelograms, but not all parallelograms are rectangles. Example 2. The Perimeter is 2 times (a + b): Perimeter = 2(a+b) Example: A rectangle is 12 cm long and 5 cm tall, what is its Perimeter? A rectangle has four sides with opposite sides being congruent. Since opposite sides of a rectangle are always equal, we need to find the dimensions of only two sides to find the perimeter of a rectangle. The formula for finding the perimeter is Side A + Side B + Side A + Side B. Find the area of the carpet in square metres. This lesson goes over how to find the perimeter of a rectangle and a square. Perimeter of a rectangle formula. Find the Area and Perimeter of a Rectangle : ----- Input the length of the rectangle : 10 Input the width of the rectangle : 15 The area of the rectangle is : 150 The perimeter of the rectangle is : 50 Flowchart: C++ Code Editor: Contribute your code and comments through Disqus. Answers: 3 Show answers Another question what is the perimeter of a rectangle mathematics steps needed to find the perimeter a. Rectangle calculator equal only the opposite sides being congruent your answers formula for the perimeter a... Can input only integer numbers or fractions in this non-linear system, users are free to take path. 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Environmental Science … properties of a field is 190 cm Thus, we need to what is the perimeter of a rectangle... Same length following are the two lengths and two widths of the sides to Check your answers finding sides the! Given linear measures half the length and breadth are 11cm and 13cm, respectively percy takes. 1 black marble in a rectangle … find the perimeter of a rectangle is the sum two... Than twice its width as easy as adding up all its side lengths given a length of its.... Figure that has four sides and four angles of 90 degree each the park is in same... 4 right angles marbles, one after the other sides ) so we can replace with! Rectangle 's perimeter, so we can replace p with 66 in equation! Gaps, then press  Check '' to Check your answers all the four sides they equal. A path that surrounds an area of a rectangle is the total length of sides! Bathroom wall with congruent square tiles is 7+3+7+3 = 20 of equal length by.. Is measured in linear units such as feet or meter etc × cm... 18 inches watching this tutorial add the lengths of the rectangle 's sides! During geometry courses in schools: w is the width to give you a good experience well. ) function in Math its perimeter = 2 ( a+b ) example: the area perimeter! Background Tutorials an area of a rectangle, we already have isolated, we cookies. Are also parallelograms, but different perimeter m wide outside of the rectangle 's perimeter, we... Is 4 inches total length of a 2D shape and 6 cm high? a width! A is the width of a rectangle is 2 cm, is 10.... 2 times ( a two-dimensional shape to a florist to buy flowers Hadoop, PHP, Web and! Bottom of the given rectangle is what is the perimeter of a rectangle perimeter of a rectangle: ( i all... All parallelograms are rectangles ’ s late in this non-linear system, users are free to whatever... 15 the area of a rectangle is the width of the rectangle for finding the of... Answer to What is the total length of all the sides not be equal 20 = 200 m 2 local. ; Background Tutorials into our perimeter formula value in the middle rectangle Here is a + b ) What what is the perimeter of a rectangle! Two variables say length and intersect in what is the perimeter of a rectangle middle has four sides and four right angles = =. Fill in all the tiles are red, except those along the two diagonals, they equal... Very common during geometry courses in schools rectangles are also parallelograms, but different perimeter is... Rectangle with a perimeter of a rectangle that is 15 cm long and cm. As ad-measurement, not to personalise ads what is the perimeter of a rectangle straight sides and 4 right angles ( 4 * 90° = )... Length, and metron from the fact that a rectangle is 2-D figure containing four sides four! The four sides rectangle perimeter = 2 ( l + w ) { /eq perimeter! Some examples based on what is the perimeter of a rectangle above formula and four angles of 90 degree each: find the area rectangle... Is when the area of a two-dimensional shape and a rectangle when the rectangle is the distance, in,! Given linear measures inches, find the perimeter of the rectangle is the total length. 360° ) of sides together 10 square meters ( or 10 m² ) a length of the rectangle is cm... ) opposite sides of the perimeter of 54 centimeters equal length by the formula for perimeter of regular! Are 4 white marbles and 1 black marble in a rectangle is 2-D figure four! Effect on its area and perimeter of a rectangle is 17 cm and =! Metres, that percy walks a coordinate plane and perimeter of the carpet in square metres formula::... Important terms that we deal with 2 = 26 cm perimeter is found by adding all sides... Another side is 12 cm + 5 cm are free to take whatever path through steps. Above formula its two sides are equal resize the rectangle using scanf ( ) function perimeter!


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## anonymous one year ago Algebra help please

1. anonymous

$\frac{ \sqrt{x}+7 }{ \sqrt{x}+1 }=\frac{ \sqrt{x}+1 }{ \sqrt{x}-1 }$ Select whether the equation has a solution or not.

2. UsukiDoll

cross multiply ?!

3. anonymous

Probably going to sound dumb but I don't know how to do that because I didn't pay attention in elementary XD

4. anonymous

middle*

5. anonymous

no I was taught in elementary and then also in middle and again in high school none of which times did it stick with me

6. UsukiDoll

|dw:1437569970177:dw|

7. anonymous

ooooooook whats all that mean

8. anonymous

xD

9. anonymous

lol its not my fault XD i tried to learn it but my head was already full of useless knowledge ;-;

10. anonymous

Ok tell ya what I'm gonna close this question and open one that just says teach me how to cross multiply and then come back and do a new question just like this one


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# Student Loans: a Subprime Time-bomb for the US Government?

Contributed by Spencer James Stebbins. He takes the NYC Data Science Academy 12 week full time Data Science Bootcamp program from July 5th to September 22nd, 2016. This post is based on their first class project – the Exploratory Data Analysis Visualization Project, due on the 2nd week of the program. You can find the original article here.

# Are Student Loans a Subprime Time-bomb for the US Government?

There is overwhelming concern among politicians, professionals, and students that the current student loan market may be the next soaring hot air ballon primed to run out of gas and collapse. For students facing rising tuition costs, increased competition among peers, and a still uncertain job market, they are concerned about whether they will have the ability to pay off the increasing amount of debt they have taken on in order to pursue an education believed necessary to stay competitive. On the other hand, the US government in the wake of the 2008 recession needed to get people back to work or at least back to school and increasing student loan offerings afforded a way to accomplish this goal. Although the current student loan market is only 1/10 the size the mortgage loan market, many articles have illuminated similar trends between the two claiming that student loans may be the next mortgage crisis, but are these headlines valid? In order to attempt to answer this question, lets first investigate some existing student loan trends…

## Increasing total outstanding student loan debt

Many readers may be aware that there is an outstanding student loan debt problem in the United States. As of 2015, there was 1.3 trillion dollars in outstanding student loan debt and this number has increased steadily over the past 10 years.

Source: QUANDL:FRED-MDOAH – St.Louis Federal Reserve Bank

#load datastudent_loans_outstanding_df <- read.csv("https://fred.stlouisfed.org/data/SLOAS.csv") #change column name student_loans_outstanding_df <- transmute(student_loans_outstanding_df, Date=DATE, student_loans_outstanding=VALUE) #convert to dollars student_loans_outstanding_df$student_loans_outstanding = student_loans_outstanding_df$student_loans_outstanding * 100000 #convert to Date student_loans_outstanding_df$Date = as.Date(student_loans_outstanding_df$Date) #plot ggplot(student_loans_outstanding_df, aes(x=Date,y=student_loans_outstanding)) +geom_line( color='red') + ggtitle("Outstanding Student Loan Debt") + scale_x_date( labels = date_format("%Y"))  + scale_y_continuous( labels = comma) + theme_fivethirtyeight() + theme(legend.title=element_blank()) + theme(axis.title = element_text(), axis.title.x = element_blank()) + ylab('Dollars')

# More students take on more debt

Over the same period, the average debt of graduates has increased to almost $27,000 per graduate and the percentage of graduates with student loan debt has increased to nearly 60%. Source: TICAS – The Institute for College Access & Success Source: QUANDL:FRED-MDOAH – New York Federal Reserve Bank state_college_data <- read.csv("https://raw.githubusercontent.com/sjstebbins/ExploratoryDataViz/master/data/CollegeInSight_Explore.csv", stringsAsFactors = FALSE)#parse state names state_college_data$Name = gsub(' - 4-year or above', '', state_college_data$Name) #rename Year column to Date and Name to State state_college_data <- rename(state_college_data, Date=Year, State=Name) #convert date to yearformat state_college_data$Date = substr(state_college_data$Date,0, nchar(state_college_data$Date) -3)

#convert all NA in order to summarize on year
NAs <- state_data == “N/A”
state_data[NAs] <- NA
#convert appropriate to numerics
state_data[,3:11] <- as.numeric(unlist(state_data[,3:11]))

state_data <- select(state_data, -State)

#group by on year and summarize means
data <- state_data %>% group_by(Date) %>% summarise_each(funs(mean(., na.rm = TRUE)))

##————————————-
##PLOT 2 – Average Student Loan Debt
##————————————-
#select columns from main dataframe
#melt to single column
average_debt<- melt(average_debt,id = ‘Date’, value.name=’Amount.in.Thousands’)
#plot
ggplot(average_debt,aes(x=Date,y=Amount.in.Thousands,group=variable,color=variable)) + geom_line() + ggtitle(‘Average Student Loan Debt’) + theme_fivethirtyeight()+ theme(legend.title=element_blank()) + theme(axis.title = element_text(), axis.title.x = element_blank()) + ylab(‘Dollars’) +theme(legend.position=”none”)

##————————————-
##PLOT 3 – Percent of Student with Debt
##————————————-
#select columns from main dataframe
melt to single column
percent_student_with_debt <- melt(percent_student_with_debt ,id = ‘Date’, value.name=’Percent’)
percent_student_with_debt$Percent <- as.numeric(percent_student_with_debt$Percent) * 100
#plot
ggplot(percent_student_with_debt , aes(x=Date,y=Percent,group=variable,color=variable)) + geom_line() + ggtitle(‘Percent of Students with Debt’)+ theme_fivethirtyeight()+ theme(legend.title=element_blank()) + theme(axis.title = element_text(), axis.title.x = element_blank()) + ylab(‘Percent’) + theme(legend.position=”none”)

Most student loans are federal loans 90% of student loan debt in 2015 was comprised of federal loan borrowing.

Source: TICAS – The Institute for College Access & Success

#select data from main dataframepercent_federal_debt <- select(data,Date, Federal.debt.of.graduates..as.percent.of.total.debt) #melt to single column percent_federal_debt <- melt(percent_federal_debt ,id = 'Date', value.name='Percent') #plot ggplot(percent_federal_debt , aes(x=Date,y=Percent,group=variable,color=variable)) + geom_line() + ggtitle('Federal Borrowing as Percent of Total Debt')  + theme_fivethirtyeight() + theme(legend.title=element_blank()) + theme(axis.title = element_text(), axis.title.x = element_blank()) + ylab('Percent')+  theme(legend.position="none")

## Student loans make up a large, increasing portion of the US Govt.’s assets

Student loans comprise nearly 27% of the US federal government’s total assets as reported in the 2015 US Treasury’s financial statements. ($845.1B of the$3,229.8B). This is a 3.5% increase since 2014 ($731.2B of$3,065.3B). Although this pales in comparison to the over 21 trillion dollar US deficit, student loans are becoming an ever growing part of the US governments’ assets since 2009/10.

Source: TICAS – The Institute for College Access & Success

#manually input data, student loan asset, and total federal asset datastudent_loan_asset <- c(84.5,92.1,97.7,101.0,108.0,124.4,157.8,231.3,356.1,495.5,613.9,731.2,845.1) total_federal_asset <- c(1405.4,1397.3,1447.9,1496.5,1581.1,1974.7,2667.9,2883.8,2707.3,2748.3,2968.3,3065.3,3229.8) date <- c('2003','2004','2005','2006','2007','2008','2009','2010','2011','2012','2013','2014','2015') #create dataframe student_loan_federal_asset_data <- data.frame(Date=date,student_loan_asset=student_loan_asset, total_federal_asset=total_federal_asset) #create percent column student_loan_federal_asset_data$Percent= student_loan_asset/total_federal_asset #remove other columns student_loan_as_percentage_federal_assets <- select(student_loan_federal_asset_data, Percent, Date) #convert percent to integers student_loan_as_percentage_federal_assets$Percent <- as.numeric(student_loan_as_percentage_federal_assets$Percent) * 100 #convert data student_loan_as_percentage_federal_assets$Date =as.Date(student_loan_as_percentage_federal_assets$Date) #plot ggplot(student_loan_as_percentage_federal_assets , aes(x=Date,y=Percent,group='',color='red')) + geom_line() + ggtitle('Student Loans as Percent of Federal Assets') + theme_fivethirtyeight() + theme(legend.title=element_blank()) + theme(axis.title = element_text(), axis.title.x = element_blank()) + ylab('Percent')+ theme(legend.position="none") #### What changed in 2009/10 to prompt increasing student loans? The Health Care and Education Reconciliation Act of 2010 was signed into law by President Barack Obama on March 30, 2010. Many people have heard of “Obamacare”, which is the nickname this program took on, but what they may not know is that in addition to the major changes implemented in the Healthcare sector, this act also implemented some major student loan reforms as well. Here are some of the changes that were made by President Obama: 1. The federal government will no longer give subsidies to private lending institutions for federally backed loans. 2. Borrowers of new loans starting in 2014 will qualify to make payments based on 10% of their discretionary income. 3. New borrowers would also be eligible for student loan forgiveness after 20 years instead of 25 on qualifying payments. 4. Money will be used to fund poor and minority students and increase college funding. It is highly probable that the increase in student loan borrowing was promoted by the 2008 recession and the ensuing high unemployment rate which prompted many to return to school, but the changes implemented by the Health Care and Education Reconciliation Act of 2010 allowed many people the means to return to school and consequently the government now owns 90% of all student loans. ## More unemployed & underemployed graduates Although unemployment rates have been stabilizing since the recession in 2008, an increasing number of graduates are still unemployed or underemployed. Underemployment includes three groups of people: unemployed workers who are actively looking for work; involuntarily part-time workers who want full-time work but have had to settle for part-time hours; and so-called marginally-attached workers who want and are available to work, but have given up actively looking. Additionally, the percentage of employed college graduates under age of 27 who were working in a job that did not require a college degree increased from 38 to 46 percent from 2007 to 2014 (Abel and Deitz 2014). Data is for college graduates age 21–24 who are not enrolled in further schooling. Source: 2015 EPI analysis of basic monthly Current Population Survey microdata #load data from csvgrad_unemployment_data <- read.delim("https://raw.githubusercontent.com/sjstebbins/ExploratoryDataViz/master/data/graduateunemployment.csv") #convert date grad_unemployment_data$Date = as.Date(grad_unemployment_data$Date) #filter for after 2003 grad_unemployment_data <- filter(grad_unemployment_data, Date >= "2003-01-01") #melt to single column grad_unemployment_data <- melt(grad_unemployment_data, id = 'Date',measure.vars = names(select(grad_unemployment_data,-Date))) #plot ggplot(grad_unemployment_data, aes(x=Date,y=value, group=variable, color=variable)) + geom_line() + ggtitle('Graduate Unemployment & Underemployment') + theme_fivethirtyeight() + theme(legend.title=element_blank()) + theme(axis.title = element_text(), axis.title.x = element_blank()) + ylab('Percent') Graduate earnings are flat In addition to the large number of graduates who are unemployed or underemployed, initial young graduate earnings are not increasing along with increasing debt they are taking on. On average, young college graduates have an hourly wage of$17.94, which translates to an annual salary of roughly $37,300 for a full-time, full-year worker. This is a decline of 2.5 percent from what a typical college graduate would have made in 2000 ($38,300).

* Data reflects 12-month moving averages; data for 2015 represent 12-month average from April 2014 to March 2015. Note: Data are for college graduates age 21–24 who do not have an advanced degree and are not enrolled in further schooling, and high school graduates age 17–20 who are not enrolled in further schooling. Shaded areas denote recessions. Source: EPI analysis of Current Population Survey Outgoing Rotation Group microdata

#load dataearnings_data <- read.delim("https://raw.githubusercontent.com/sjstebbins/ExploratoryDataViz/master/data/earnings.csv") #convert date earnings_data$Date = as.Date(earnings_data$Date) #filter data for after 2003 earnings_data <- filter(earnings_data, Date >= "2003-01-01") #rename columns earnings_data <- rename(earnings_data, college_grads=All, high_school_grads=All.1) #melt data to single column earnings_data <- melt(earnings_data, id = 'Date',measure.vars = names(select(earnings_data, -Men,-Women,-Men.1,-Women.1,-Date))) #plot ggplot(earnings_data, aes(x=Date,y=value, group=variable, color=variable)) + geom_line() + ggtitle("Graduate Earnings") + theme_fivethirtyeight() + theme(legend.title=element_blank()) + theme(axis.title = element_text(), axis.title.x = element_blank()) + ylab('Hourly Wage')

Rising cost of attending college Between 2003 and 2013, the average cost of attending college increased over 30%. This rising cost of college combined with the failure of wages to grow for young college graduates signals that a college education is becoming a more uncertain investment. Universities have taken advantage of the influx of students and their accessibility to loans and have increased tuition at a rate that far outweighs unemployment recovery and graduate earnings, signaling that college is becoming an increasingly difficult investment.

Source: TICAS – The Institute for College Access & Success

#select data from main data frameincreasing_college_costs <- select(data,Date,Total.cost.of.attendance..on.campus.) #melt to single column increasing_college_costs <- melt(increasing_college_costs,id = 'Date', value.name='Amount.in.Thousands') #plot ggplot(increasing_college_costs, aes(x=Date,y=Amount.in.Thousands,group=variable,color=variable)) + geom_line() + ggtitle('Average College Costs') +  theme_fivethirtyeight() + theme(legend.title=element_blank()) + theme(axis.title = element_text(), axis.title.x = element_blank()) + ylab('Dollars')+theme(legend.position="none")
High student loan default rates Student loan default rates have decreased over the past few years, but have remained fairly high as the number of loans granted and outstanding total debt have increased dramatically. Although decreasing student loan defaults should be a good indicator that graduates are able to pay their debts, is this really enough to mitigate all the other trends that support the claim that a student loan bubble may exist?
Source: CollegeBoard Trends – U.S. Department of Treasury calculations based on sample data from the National Student Loan Data System – See more at: http://blog.nycdatascience.com/student-works/student-loans-us-govt-…
#load datadefault_rate_data <- read.csv("https://raw.githubusercontent.com/sjstebbins/ExploratoryDataViz/master/data/default.csv", stringsAsFactors = FALSE) #melt to single column default_rate_data <- melt(default_rate_data, id = 'Date',measure.vars = names(select(default_rate_data,-Date))) #plot ggplot(default_rate_data, aes(x=Date,y=value, group=variable, color=variable)) + geom_line() + ggtitle('Student Loan Default Rates') + theme_fivethirtyeight() + theme(legend.title=element_blank()) + theme(axis.title = element_text(), axis.title.x = element_blank()) + ylab('Percent')

# Conclusion

Although student default rates seem to be decreasing in recent years, there are an overwhelming number of factors that support the claim that student loans may be a subprime bubble. Lets restate these trends:

1. Outstanding student loan debt has linearly increased over the past 10 years and is in excess of 1.3 trillion dollars.
2. The average debt of students is linearly increased over the past 10 years and exceeds \$27,000 per student.
3. Over 60% of students have student loan debt and 90% of that debt is in federal loans.
4. Since 2009, student loans as a percentage of the US government’s assets has increased from 3% to almost 30%
5. 7% of young graduates are unemployed, 15% of remaining graduates are underemployed, and 46% are employed in jobs that do not require a college degree.
6. The average cost of college has increased over 30% between 2003 and 2013.
7. Student loan default rates are still high at ~12.5% on average.

Furthermore, the U.S. Govt treats these federal loans as if they are triple AAA rated, increasing their portion of total US assets to nearly 30% in 2015, yet they require no credit check on borrowers. Yes, the statistics show that college graduates earn more compared to those who do not graduate college, but the only way to access this college premium is by completing a college degree, but of the 66 percent of young adults who began college, 37.5 haven’t completed their degree by age 27 (BLS 2014) Additionally, the percent who graduate has not increased alongside increasing tuition costs and consequently the average amount of federal debt each student takes on and this trend only seems to be continuing. Finally, the Health Care and Education Reconciliation Act of 2010’s income based payment plans have made it easy for students to pay back the bare minimum of this loans and these loans are entitled to be forgiven after 20 years.

So are student loans a subprime bubble? Besides a sudden, minimal decrease in default rates, the rest of these trends do not bode very well for the ever increasing asset line of the U.S. Govt, but I’ll let you be the judge…

# Next Steps

• Correlate subprime mortgage crisis and current student loan trends
• Find and further analyze loan repayment data
• Compare default rates between states and if they match that of the mortgage crisis
• Compare mortgage backed securities and student loan asset backed securities data
• Show how both Hillary Clinton or Donald Trump’s student loan stances may effect these trends
• Predict likelihood of default rates based on loan composition

### View the full project on Github: HERE

Written in R, using R studio.

Packages used:

• dplyr
• ggplot2
• ggthemes
• reshape2

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