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How to show that this sequence is monotonic

We are given that $(s_{n})$ is a bounded sequence. My goal is to show that $(s_{n})$ is monotonic, and thus it converges. It has the following property $$s_{n+1} \geq s_{n} - \frac{1}{2^n}$$ I have tried induction, because intuitively the sequence monotonically increasing, but it fails at the basis step but works on the induction step. Any ideas?

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That doesn't look like a definition but rather like a property of the sequence... – DonAntonio Mar 3 '14 at 11:02
True, edited now. – user29163 Mar 3 '14 at 11:05
Do you need to prove monotonicity or you need only to prove the sequence converges? – DonAntonio Mar 3 '14 at 11:14
Thanks, I only need to prove that the sequence converges. I since it is bounded I assumed (big mistake) that the right move is to try to prove monotonicity. I will try other things. Learned much from this case. Thanks – user29163 Mar 3 '14 at 11:22

If you need to prove convergence:

$$s_n-s_{n+1}\le\frac1{2^n}\implies \;\forall\,m>n\;:$$

$$s_n-s_m=(s_n-s_{n+1})+(s_{n+1}-s_{n+2})+\ldots+(s_{m-1}-s_m)\le$$

$$\le\frac1{2^n}+\frac1{2^{n+1}}+\ldots+\frac1{2^{m-1}}\le\frac{m-n-1}{2^n}$$

So our sequence is a Cauchy sequence and thus it converges (one can see this without the last step above as the left expression in the last line is the difference $\;S_n-S_m\;$ of the sequence of partial sums of the convergent series $\;\sum\frac1{2^n}\;$...)

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Counter-example. Bounded but non-monotonic. $s_{n+1}$ = \begin{cases} 0, & n+1 \mbox{ is even, }\\ \frac{-1}{2^n}, & n+1 \mbox{ is odd.} \end{cases}

E.g. $s_1 = 0, s_2 = -0.5, s_3 = 0, ....$

$s_1 > s_2$ but $s_2 < s_3$.

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$s_n$ is not necessarily monotonic given the information provided.

Example. Let $s_n = \frac{2}{3} \left(\frac{-1}{2}\right)^n$. Then $s_n$ is bounded, and \begin{align*} s_{n+1} - s_n &= \frac{2}{3}\left(\frac{-1}{2}\right)^{n+1} - \frac{2}{3}\left(\frac{-1}{2}\right)^n \\ &= \frac{2}{3}\left(\frac{-1}{2}\right)^n \left( \frac{-1}{2} - 1\right) \\ &= -\left(\frac{-1}{2}\right)^n \\ &= (-1)^{n+1} \left(\frac{1}{2}\right)^n \ge \frac{-1}{2^n}. \\ \end{align*} However, $s_n$ is clearly not monotonic.

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text: To cover the answer again, click "Refresh" ("Reload"). We explain Adding Radical Expressions with Unlike Radicands with video tutorials and quizzes, using our Many Ways(TM) approach from multiple teachers. Now that the radicands have been multiplied, look again for powers of 4, and pull them out. â¦ Since only the radicals in a are like, we can only combine (add and subtract) the radicals in a. Rule #3 Radicals - Adding Radicals Objective: Add like radicals by ï¬rst simplifying each radical. Last edited: Jul 23, 2013. topsquark. âxy â â6 cannot be subtracted, different radicands. The indices are different. 1 Answer Jim H Mar 22, 2015 Make the indices the same (find a common index). And so then we are all done. The same rule applies for adding two radicals! Before the terms can be multiplied together, we change the exponents so they have a common denominator. $$2\sqrt[5]{1000q}$$ ... (-4\sqrt[4]{1000q}\) are not like radicals. Come to Polymathlove.com and master a line, equations in two variables and plenty additional algebra subject areas radicals with different radicands cannot be added or subtracted. Since the radicals are like, we subtract the coefficients. Always check to see whether you can simplify the radicals. Subtract Radicals Subtraction of radicals follows the same set of rules and approaches as additionâthe radicands and the indices must be the same for two (or more) radicals to be subtracted. Further, get to intensify your skills by performing both the operations in a single question. Forums. Examples: a. Subtraction of radicals follows the same set of rules and approaches as additionâthe radicands and the indices (plural of index) must be the same for two (or more) radicals to be subtracted. hhsnb_alg1_pe_0901.indd 484snb_alg1_pe_0901.indd 484 22/5/15 8:57 AM/5/15 8:57 AM The trick is to get rid of the exponents, we need to take radicals of both sides, and to get rid of radicals, we need to raise both sides of the equation to that power. In the three examples that follow, subtraction has been rewritten as addition of the opposite. 1. EXAMPLE 2: Add and subtract the pairs of radical expressions given in EXAMPLE 1 above. 3âx + 5ây + 2â6 are three radicals that cannot be added together, each radicand is different. The questions in these pdfs contain radical expressions with two or three terms. In some cases, the radicals will become like radicals. Adding and Subtracting Radical Expressions You could probably still remember when your algebra teacher taught you how to combine like terms. Forum Staff. Identify and pull out powers of 4, using the fact that . To see the answer, pass your mouse over the colored area. The above expressions are simplified by first transforming the unlike radicals to like radicals and then adding/subtracting When it is not obvious to obtain a common radicand from 2 different radicands, decompose them into prime numbers. How do you multiply radical expressions with different indices? Do you want to learn how to multiply and divide radicals? Gear up for an intense practice with this set of adding and subtracting radicals worksheets. So that the domain over here, what has to be under these radicals, has to be positive, actually, in every one of these cases. d. Ë 57 6Ë Ë 54 e. Ë4 6Ë !Ë 54 Ë4 6Ë Ë 54 4 6Ë 54 Ë Otherwise, we just have to keep them unchanged. Note : When adding or subtracting radicals, the index and radicand do not change. Therefore, radicals cannot be added and subtracted with different index . After seeing how to add and subtract radicals, itâs up to the multiplication and division of radicals. First we provide a formal definition ... {125y}\) are not like radicals. 6Ë Ë c. 4 6 !! Algebra Radicals and Geometry Connections Multiplication and Division of Radicals. Adding and subtracting radical expressions is similar to adding and subtracting like terms. Iâll explain it to you below with step-by-step exercises. $$-5 \sqrt{2}$$ b. Next Iâll also teach you how to multiply and divide radicals with different indexes. To add and , one adds the numbers on the outside only to get .-----The Rules for Adding and Subtracting Radicals. They incorporate both like and unlike radicands. $$9 \sqrt[3]{y}$$ c. $$7 \sqrt[4]{x}-2 \sqrt[4]{y}$$ The indices are the same but the radicals are different. It is valid for a and b greater than or equal to 0.. Adding radicals is very simple action. Right from dividing and simplifying radicals with different indexes to division, we have every part covered. Adding and Subtracting Radicals â Practice Problems Move your mouse over the "Answer" to reveal the answer or click on the "Complete Solution" link to reveal all of the steps required for adding and subtracting radicals. Now this problem is ready to be simplified because I have 3 different terms that they all have the same radicals. Consider the following example. In this tutorial, you will learn how to factor unlike radicands before you can add two radicals together. Nov 2012 744 2 Hawaii Jul 23, 2013 #1 Did I do it right? Square root of 9 I know is regular 3 multiplied by -3, thatâll give me 9 square roots of 5x. Radicals may be added or subtracted when they have the same index and the same radicand (just like combining like terms). 4 Ë5Ë Ë5 Ë b. 55.4 KB Views: 8. Break down the given radicals and simplify each term. When we have two terms that contain the same type of root (the radical in both terms is a square root, the radical in both terms is a cube root, etc.) To multiply radicals, first verify that the radicals have the same index, which is the small number to the left of the top line in the radical symbol. By doing this, the bases now have the same roots and their terms can be multiplied together. Improve your math knowledge with free questions in "Add and subtract radical expressions" and thousands of other math skills. SOLUTIONS: Since only the radicals in a are like, we can only combine (add or subtract) the radicals in a. a. Rule #1 - When adding or subtracting two radicals, you must simplify the radicands first. image.jpg. Adding and Subtracting Radical Expressions. In order to add or subtract radicals, we must have "like radicals" that is the radicands and the index must be the same for each term. Problem 1. To simplify two radicals with different roots, we first rewrite the roots as rational exponents. âx 2 + 2âx We cannot add or subtract the radicands to combine or simplify them, they are different. Rewrite as the product of radicals. Rule #2 - In order to add or subtract two radicals, they must have the same radicand. Radicals that are "like radicals" can be added or subtracted by adding or subtracting the coefficients. Adding and Subtracting Radicals Worksheets. Since all the radicals are fourth roots, you can use the rule to multiply the radicands. Adding and Subtracting Higher Roots We can add and subtract higher roots like we added and subtracted square roots. This means that when we are dealing with radicals with different radicands, like 5 \sqrt{5} 5 and 7 \sqrt{7} 7 , there is really no way to combine or simplify them. A. asilvester635. The following video shows more examples of adding radicals that require simplification. In this lesson, we are only going to deal with square roots only which is a specific type of radical expression with an index of \color{red}2.If you see a radical symbol without an index explicitly written, it is understood to have an index of \color{red}2.. Below are the basic rules in multiplying radical expressions. It is the symmetrical version of the rule for simplifying radicals. When learning how to add fractions with unlike denominators, you learned how to find a common denominator before adding. And if you make the assumption that this is defined for real numbers. Adding and subtracting radicals is very similar to adding and subtracting with variables. You canât add radicals that have different index or radicand. If these were the same root, then maybe we could simplify this a little bit more. Add Radicals. Pre-University Math Help. 5x +3x â 2x Combineliketerms 6x OurSolution 5 11 â +3 11 â â 2 11 â Combineliketerms 6 11 â OurSolution Simplify the radicands first before subtracting as we did above. 5 plus 8 is 13 13 minus 9 is 4, so my final answer will be 4 square roots of 5x. Crack the questions one by one, and add and subtract radicals like a pro! 5â20 + 4â5 they can't be added because their radicands are different. Factorize the radicands and express the radicals in the simplest form. 2. and identical radicands (the expressions under the radical signs in the two terms are the same), they are like terms, and adding and subtracting is â¦ Attachments. Multiplying Radical Expressions. There is only one thing you have to worry about, which is a very standard thing in math. adding radicals subtracting; Home. Just keep in mind that if the radical is a square root, it doesnât have an index. These are not like radicals. Adding and Subtracting Radicals with Fractions. The goal is to add or subtract variables as long as they âlookâ the same. However, when dealing with radicals that share a base, we can simplify them by combining like terms. And we have fully simplified it. The only thing you can do is match the radicals with the same index and radicands and add them together. Algebra. Solution: 5â20 = 10â5 Therefore, 10â5 + 4â5 = 14â5 *Answer Do the same thing if the problem is subtraction. But if you simplify the first term they will be able to be added. Example 1. They can only be added and subtracted if they have the same index. How to add and subtract radicals. The radicands are different. different radicands. Use prime factorization method to obtain expressions with like radicands and add or subtract them as indicated. Multiply. Here the radicands differ and are already simplified, so this expression cannot be simplified. 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Triangles

 Mind Maps

Class X - Maths: Similar Triangles
Q) The similar triangles are called?

Q) What are Similar Figures?

Q) PS/SQ = PT/TR and ∠ PST = ∠ PRQ.
Prove that PQR is an isosceles triangle?

Q) If PQ ||RS, prove that Δ POQ ~ ΔSOR?

Q) Full formula of SSS?

Q) Give two different examples of pair of : a) Similar Figures b) Non- Similar Figures.

Q) What can you say about the similarity of two triangles?

Q) If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio. Explain

Q) If a line intersects sides AB and AC of a Δ ABC at D and E respectively
and is parallel to BC, prove that AD/AB = AE/AC.

Q) ABCD is a trapezium with AB ||DC. E and F are points on non-parallel sides AD and BC respectively such that EF is parallel to AB (see Fig.). Show that AE/ED = BF/FC.

Q) Draw two triangles ABC and DEF such that AB = 2 cm, ∠ A = 50° AC = 4 cm, DE = 3 cm, ∠ D = 50° and DF = 6 cm.

Q) In Fig.CM and RN are respectively the medians of ΔABC and
ΔPQR. If Δ ABC ~ ΔPQR, prove that:
(i) ΔAMC ~ Δ PNR
(ii)CM/RN = AB/ PQ
(iii) ΔCMB ~ Δ RNQ?

Q) In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. Explain.

Q) In a triangle, if square of one side is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle. Explain.

Q) O is any point inside a rectangle ABCD (see Fig.). Prove that OB2 + OD2 = OA2+ OC2.

Contents
• 6.1 Introduction
• 6.2 Similar Figures
• Activity 1
• 6.3 Similarity of Triangles
• EXERCISE 6.1
• Activity 2
• Theorem 6.1
• Activity 3
• Theorem 6.2
• EXERCISE 6.2
• 6.4 Criteria for Similarity of Triangles
• Activity 4
• Theorem 6.3
• Activity 5
• Theorem 6.4
• Activity 6
• Theorem 6.5
• 6.5 Areas of Similar Triangles
• Theorem 6.6
• 6.6 Pythagoras Theorem
• Theorem 6.7
• Theorem 6.8
• Theorem 6.9
• EXERCISE 6.5
• 6.7 Summary

6.1 Introduction
You are familiar with triangles and many of their properties from your earlier classes. In Class IX, you have studied congruence of triangles in detail.
Recall that two figures are said to be congruent, if they have the same shape and the same size.
In this chapter, we shall study about those figures which have the same shape but not necessarily the same size.
Two figures having the same shape (and not necessarily the same size) are called similar figures.
In particular, we shall discuss the similarity of triangles and apply this knowledge in giving a simple proof of Pythagoras Theorem learnt earlier.
Can you guess how heights of mountains (say Mount Everest) or distances of some long distant objects (say moon) have been found out? Do you think these have

been measured directly with the help of a measuring tape? In fact, all these heights and distances have been found out using the idea of indirect measurements.
which is based on the principle of similarity of figures (see Example 7, Q.15 of Exercise 6.3 and also Chapters 8 and 9 of this book).

6.2 Similar Figures
In Class IX, you have seen that all circles with the same radii are congruent, all squares with the same side lengths are congruent and all equilateral triangles with the same side lengths are congruent.
Now consider any two (or more) circles [see Fig. 6.1 (i)]. Are they congruent?
Since all of them do not have the same radius, they are not congruent to each other. Note that some are congruent and some are not, but all of them have the same shape.
So they all are, what we call, similar. Two similar figures have the same shape but not necessarily the same size.
Therefore, all circles are similar. What about two (or more) squares or two (or more) equilateral triangles [see Fig. 6.1 (ii) and (iii)]? As observed in the case of circles, here also all squares are similar and all equilateral triangles are similar.
From the above, we can say that all congruent figures are similar but the similar figures need not be congruent.
Can a circle and a square be similar? Can a triangle and a square be similar? These questions can be answered by just looking at the figures (see Fig. 6.1). Evidently these figures are not similar. (Why?)

What can you say about the two quadrilaterals ABCD and PQRS (see Fig 6.2)?Are they similar?
These figures appear to be similar but we cannot be certain about it.Therefore, we must have some definition of similarity of figures and based on this definition some rules to decide whether the two given figures are similar or not.
For this, let us look at the photographs given in Fig. 6.3:

You will at once say that they are the photographs of the same monument (Taj Mahal) but are in different sizes.
Would you say that the three photographs are similar? Yes,they are.
What can you say about the two photographs of the same size of the same person one at the age of 10 years and the other at the age of 40 years? Are these photographs similar?
These photographs are of the same size but certainly they are not of the same shape. So, they are not similar.
What does the photographer do when she prints photographs of different sizes from the same negative?
You must have heard about the stamp size, passport size and postcard size photographs.
She generally takes a photograph on a small size film, say of 35mm size and then enlarges it into a bigger size, say 45mm (or 55mm).
Thus, if we consider any line segment in the smaller photograph (figure), its corresponding line segment in the bigger photograph (figure) will be 45/35(or 55/35) of that of the line segment.
This really means that every line segment of the smaller photograph is enlarged (increased) in the ratio 35:45 (or 35:55).
It can also be said that every line segment of the bigger photograph is reduced (decreased) in the ratio 45:35 (or 55:35).
Further, if you consider inclinations (or angles) between any pair of corresponding line segments in the two photographs of different sizes, you shall see that these inclinations(or angles) are always equal.
This is the essence of the similarity of two figures and in particular of two polygons.
We say that:
Two polygons of the same number of sides are similar, if (i) their corresponding angles are equal and (ii) their corresponding sides are in the same ratio (or proportion).

Note that the same ratio of the corresponding sides is referred to as the scale factor (or the Representative Fraction) for the polygons.
You must have heard that world maps (i.e., global maps) and blue prints for the construction of a building are prepared using a suitable scale factor and observing certain conventions.
In order to understand similarity of figures more clearly, let us perform the following activity:

Activity 1 :
Place a lighted bulb at a point O on the ceiling and directly below it a table in your classroom.
Let us cut a polygon, say a quadrilateral ABCD, from a plane cardboard and place this cardboard parallel to the ground between the lighted bulb and the table.
Then a shadow of ABCD is cast on the table.
Mark the outline of this shadow as A′B′C′D′ (see Fig.6.4).
Note that the quadrilateral A′B′C′D′ is an enlargement (or magnification) of the quadrilateral ABCD.
This is because of the property of light that light propagates in a straight line. You may also note that A′ lies on ray OA, B′ lies on ray OB, C′ lies on OC and D′ lies on OD.
Thus, quadrilaterals A′B′C′D′ and ABCD are of the same shape but of different sizes.
We can also say that quadrilateral ABCD is similar to the quadrilateral A′B′C′D′. Here, you can also note that vertex A′ corresponds to vertex A, vertex B′ corresponds to vertex B, vertex C′ corresponds to vertex C and vertex D′ corresponds to vertex D.
Symbolically, these correspondences are represented as A′ ↔ A, B′ ↔ B, C′ ↔ C and D′ ↔ D.
By actually measuring the angles and the sides of the two quadrilaterals, you may verify that
(i) ∠ A = ∠ A′, ∠ B = ∠ B′, ∠ C = ∠ C′, ∠ D = ∠ D′ and
(ii) AB/A′B′ = BC/B′C′ = CD/C′D′ = DA/D′A′
This again emphasises that two polygons of the same number of sides are
similar, if (i) all the corresponding angles are equal and
(ii) all the corresponding sides are in the same ratio (or proportion).
From the above, you can easily say that quadrilaterals ABCD and PQRS of Fig. 6.5 are similar.

Remark :
You can verify that if one polygon is similar to another polygon and this second polygon is similar to a third polygon, then the first polygon is similar to the third polygon.
You may note that in the two quadrilaterals (a square and a rectangle) of Fig. 6.6, corresponding angles are equal, but their corresponding sides are not in the same ratio.

So, the two quadrilaterals are not similar.
Similarly, you may note that in the two quadrilaterals (a square and a rhombus) of Fig. 6.7, corresponding sides are in the same ratio, but their corresponding angles are not equal.
Again, the two polygons (quadrilaterals) are not similar.

Thus, either of the above two conditions (i) and (ii) of similarity of two polygons is not sufficient for them to be similar.

EXERCISE 6.1

1. Fill in the blanks using the correct word given in brackets :
(i) All circles are . (congruent, similar)
(ii) All squares are . (similar, congruent)
(iii) All triangles are similar. (isosceles, equilateral)
(iv) Two polygons of the same number of sides are similar, if (a) their corresponding angles are and (b) their corresponding sides are .
(equal, proportional)
2. Give two different examples of pair of
(i) similar figures. (ii) non-similar figures.
3. State whether the following quadrilaterals are similar or not:

6.3 Similarity of Triangles

What can you say about the similarity of two triangles?
You may recall that triangle is also a polygon.
So, we can state the same conditions for the similarity of two triangles. That is:
Two triangles are similar, if
(i) their corresponding angles are equal and
(ii) their corresponding sides are in the same ratio (or proportion).
Note that if corresponding angles of two triangles are equal, then they are known as equiangular triangles.
A famous Greek mathematician Thales gave an important truth relating to two equiangular triangles which is as follows: The ratio of any two corresponding sides in two equiangular triangles is always the same.
It is believed that he had used a result called the Basic Proportionality Theorem (now known as the Thales Theorem) for the same.
To understand the Basic Proportionality Theorem, let us perform the following activity:

Activity 2 :
Draw any angle XAY and on its one arm AX, mark points (say five points) P, Q, D, R and B such that AP = PQ = QD = DR = RB.
Now, through B, draw any line intersecting arm AY at C (see Fig. 6.9).
Also, through the point D, draw a line parallel to BC to intersect AC at E.

Measure AE and EC. What about AE/EC ? Observe that AE/EC is also equal to 3/2 .
Thus, you can see that in Δ ABC, DE ||BC and AD/DB = AE/EC.
Is it a coincidence? No, it is due to the following theorem (known as the Basic Proportionality Theorem):

Theorem 6.1 :
If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.
Proof :
We are given a triangle ABC in which a line parallel to side BC intersects other two sides AB and AC at D and E respectively (see Fig. 6.10).

We need to prove that
(AD)/(DB) = (AE)/(EC).

Let us join BE and CD and then draw DM _|_ AC and EN _|_ AB.
Now, area of Δ ADE = 1/2 (base**height) = 1/2 (AD ** EN).
Recall from Class IX, that area of Δ ADE is denoted as ar(ADE).
So, area (ADE) =1/2 ( AD **EN )
Similarly, ar(BDE) =1/2 ( DB **EN )
ar(ADE) =1/2 ( AE** DM ) and ar(DEC) =1/2 ( EC** DM )
Therefore, ar(ADE)/ar(BDE) =[1/2(AD** EN)]/[1/2 (DB** EN)]
ar(ADE)/ar(DEC) = [1/2 ( AE**DM )]/[1/2(EC **DM)]...........................(2)
Note that Δ BDE and DEC are on the same base DE and between the same parallels BC and DE.
So, ar(BDE) = ar(DEC)....................................................(3)
Therefore, from (1), (2) and (3), we have :
(AD)/(DB)= (AE)/(EC)
Is the converse of this theorem also true (For the meaning of converse, see Appendix 1)? To examine this, let us perform the following activity:

Activity 3 :
Draw an angle XAY on your notebook and on ray AX,
mark points B1, B2,B3, B4 and B such that AB1 = B1 B2 = B2 B3 = B3 B4 = B4 B. Similarly, on ray AY, mark points C1 , C2 , C3 , C4 and C such that AC1 = C1 C2 = C2 C3 = C3 C4 = C4 C. Then join B1 C1 and BC(see Fig. 6.11).
Note that [AB_1]/(B_1 B )= [AC_1]/[C_1C] (Each equal to 1/4 )
You can also see that lines B1C1 and BC are parallel to each other, i.e.,
B1C1 ||BC.................. (1)
Similarly, by joining B2C2, B3C3 and B4C4, you can see that:
[AB_2]/[B_2B]= [AC_2]/[C_2C]=(2/3) and B2C2||BC.........(2)
[AB_3]/[B_3B]= [AC_3]/[C_3 C]=(3/2)and B3C3||BC........ (3)
[AB_4]/[B_4B ]= [AC_4]/[C_4C]=(4/1)and B4C4 ||BC..........(4)

From (1), (2), (3) and (4), it can be observed that if a line divides two sides of a triangle in the same ratio, then the line is parallel to the third side.
You can repeat this activity by drawing any angle XAY of different measure and taking any number of equal parts on arms AX and AY .
Each time, you will arrive at the same result.
Thus, we obtain the following theorem, which is the converse of Theorem 6.1:

Theorem 6.2 : If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.
This theorem can be proved by taking a line DE such that
AD/DB = AE/EC and assuming that DE is not parallel to BC (see Fig. 6.12).
If DE is not parallel to BC, draw a line DE′parallel to BC.So,AD/DB = AE'/E'C (Why ?)
Therefore,AE/EC = AE'/E'C (Why ?)
Adding 1 to both sides of above, you can see that E and E' must coincide.
(Why ?)
Let us take some examples to illustrate the use of the above theorems.

Example 1 :
If a line intersects sides AB and AC of a Δ ABC at D and E respectively
and is parallel to BC, prove that AD/AB = AE/AC (see Fig. 6.13).

Solution :
DE ||BC
So,[AD]/[DB] = [AE]/[EC]
or
[DB]/[AD] = [EC]/[AE]
or
[DB]/[AD] + 1 = [EC]/[AE] + 1

or,
[AB]/[AD] = [AC]/[AE]

So,(AD)/(AB) = (AE)/(AC)

2: ABCD is a trapezium with AB ||DC.
E and F are points on non-parallel sides AD and BC respectively such that EF is parallel to AB
(see Fig. 6.14). Show that AE/ED = BF/FC.

Solution :
Let us join AC to intersect EF at G (see Fig. 6.15).
AB ||DC and EF ||AB (Given)
So, EF ||DC (Lines parallel to the same line are parallel to each other)
EG ||DC (As EF ||DC)
So,
AE/ED = AG/GC (Theorem 6.1)........... (1)
Similarly, from Δ CAB,
CG/AG = CF/BF

AG/GC = BF/FC............................................................(2)
Therefore, from (1) and (2),
AE/ED = BF/FC

3: In Fig. 6.16, [PS]/[SQ] = [PT]/[TR] and ∠ PST = ∠ PRQ. Prove that PQR is an isosceles triangle.
Solution :
It is given that [PS]/[SQ] = [PT]/[TR].
So, ST ||QR (Theorem 6.2)

Therefore, ∠ PST = ∠ PQR (Corresponding angles)........ (1)
Also, it is given that
∠ PST = ∠ PRQ..................... (2)
So, ∠ PRQ = ∠ PQR [From (1) and (2)]
Therefore, PQ = PR (Sides opposite the equal angles)
i.e., PQR is an isosceles triangle

EXERCISE 6.2

1. In Fig. 6.17, (i) and (ii), DE ||BC. Find EC in (i) and AD in (ii).

2. E and F are points on the sides PQ and PR respectively of a Δ PQR.
For each of the following cases, state whether EF ||QR :
(i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm
(ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm
(iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm

3. In Fig. 6.18, if LM ||CB and LN ||CD, prove that
[AM]/[AB] = [AN]/[AD].

4. In Fig. 6.19, DE ||AC and DF ||AE. Prove that
BF/FE = BE/ EC.

5. In Fig. 6.20, DE ||OQ and DF ||OR. Show that EF ||QR.

6. In Fig. 6.21, A, B and C are points on OP, OQ and OR respectively such that AB ||PQ and AC ||PR.
Show that BC ||QR.

7. Using Theorem 6.1, prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side.
(Recall that you have proved it in Class IX).

8. Using Theorem 6.2, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side.
(Recall that you have done it in Class IX).

9.ABCD is a trapezium in which AB ||DC and its diagonals intersect each other at the point O. Show that AO/BO = CO/DO

10. The diagonals of a quadrilateral ABCD intersect each other at the point O.
such that AO/BO = CO/DO.
Show that ABCD is a trapezium.

6.4 Criteria for Similarity of Triangles

In the previous section, we stated that two triangles are similar,
if (i) their corresponding angles are equal and
(ii) their corresponding sides are in the same ratio (or proportion).
That is, in Δ ABC and Δ DEF, if
(i) ∠ A = ∠ D, ∠ B = ∠ E, ∠ C = ∠ F and
(ii)[AB]/[DE] = [BC]/[EF] = [CA]/[FD], then the two triangles are similar (see Fig. 6.22).

Here, you can see that A corresponds to D, B corresponds to E and C corresponds to F. Symbolically, we write the similarity of these two triangles as ‘Δ ABC ~ Δ DEF’ and read it as ‘triangle ABC is similar to triangle DEF’.
The symbol ‘~’ stands for ‘is similar to’.
Recall that you have used the symbol ‘≅’ for ‘is congruent to’ in Class IX.

It must be noted that as done in the case of congruency of two triangles,
the similarity of two triangles should also be expressed symbolically,
using correct correspondence of their vertices.
For example, for the triangles ABC and DEF of Fig. 6.22, we cannot write Δ ABC ~ Δ EDF or Δ ABC ~ Δ FED.
However, we can write Δ BAC ~ Δ EDF.

Now a natural question arises :
For checking the similarity of two triangles, say ABC and DEF, should we always look for all the equality relations of their
corresponding angles (∠ A = ∠ D, ∠ B = ∠ E, ∠ C = ∠ F) and all the equality relations of the ratios
of their corresponding sides
[AB]/[DE] = [BC]/[EF] = [CA]/[FD]?Let us examine.
You may recall that in Class IX, you have obtained some criteria for congruency of two triangles involving only three pairs of corresponding parts (or elements) of the two triangles.
Here also, let us make an attempt to arrive at certain criteria for similarity of two triangles involving relationship between less number of pairs of corresponding parts of the two triangles,
instead of all the six pairs of corresponding parts.
For this, let us perform the following activity:

Activity 4 :
Draw two line segments BC and EF of two different lengths, say 3 cm and 5 cm respectively.
Then, at the points B and C respectively, construct angles PBC and QCB of some measures, say, 60° and 40°.
Also, at the points E and F, construct angles REF and SFE of 60° and 40° respectively (see Fig. 6.23).

Let rays BP and CQ intersect each other at A and rays ER and FS intersect each other at D.
In the two triangles ABC and DEF, you can see that ∠ B = ∠ E, ∠ C = ∠ F and ∠ A = ∠ D.
That is, corresponding angles of these two triangles are equal.
What can you say about their corresponding sides ?
Note that [BC]/[EF] = 3/5 = 0.6
What about AB/DE and CA/FD ? On measuring AB, DE, CA and FD, you will find that [AB]/[DE] and [CA]/[FD] are also equal to 0.6 (or nearly equal to 0.6, if there is some error in the measurement).
Thus, [AB]/[DE] =[BC]/[EF] = [CA]/[FD].
You can repeat this activity by constructing several pairs of triangles having their corresponding angles equal. Every time,
you will find that their corresponding sides are in the same ratio (or proportion).
This activity leads us to the following criterion for similarity of two triangles.

Theorem 6.3 :
If in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio (or proportion) and hence the two triangles are similar.
This criterion is referred to as the AAA (Angle–Angle–Angle) criterion of similarity of two triangles.
This theorem can be proved by taking two triangles ABC and DEF such that ∠ A = ∠ D, ∠ B = ∠ E and ∠ C = ∠ F (see Fig. 6.24) Cut DP = AB and DQ = AC and join PQ.
So, Δ ABC ≅ Δ DPQ (Why ?) This gives ∠ B = ∠ P = ∠ E and PQ ||EF (How?)

Therefore,[DP]/[PE] = [DQ]/[QF] (Why?)
i.e.,[AB]/[DE] = [AC]/[DF] (Why?)
Similarly,[AB]/[DE] = [BC]/[EF] and so
[AB]/[DE] = [BC]/[EF] = [AC]/[DF].
Remark :
If two angles of a triangle are respectively equal to two angles of another triangle, then by the angle sum property of a triangle their third angles will also be equal.
Therefore, AAA similarity criterion can also be stated as follows:
If two angles of one triangle are respectively equal to two angles of another triangle,
then the two triangles are similar. This may be referred to as the AA similarity criterion for two triangles.
You have seen above that if the three angles of one triangle are respectively equal to the three angles of another triangle,
then their corresponding sides are proportional (i.e., in the same ratio).
What about the converse of this statement? Is the converse true?
In other words, if the sides of a triangle are respectively proportional to the sides of another triangle, is it true that their corresponding angles are equal?
Let us examine it through an activity :

Activity 5 :
Draw two triangles ABC and DEF such that AB = 3 cm, BC = 6 cm, CA = 8 cm, DE = 4.5 cm, EF = 9 cm and FD = 12 cm (see Fig. 6.25).

AB/DE = BC/EF = CA/FD (each equal to 2/3 )
Now measure ∠ A, ∠ B, ∠ C, ∠ D, ∠ E and ∠ F.
You will observe that ∠ A = ∠ D, ∠ B = ∠ E and ∠ C = ∠ F, i.e., the corresponding angles of the two triangles are equal.
You can repeat this activity by drawing several such triangles (having their sides in the same ratio).
Everytime you shall see that their corresponding angles are equal.
It is due to the following criterion of similarity of two triangles:

Theorem 6.4 :
If in two triangles, sides of one triangle are proportional to (i.e., in the same ratio of ) the sides of the other triangle, then their corresponding angles are equal and hence the two triangles are similiar.
This criterion is referred to as the SSS (Side–Side–Side) similarity criterion for two triangles.
This theorem can be proved by taking two triangles ABC and DEF such that
[AB]/[DE] = [BC]/[EF] = [CA]/[FD](< 1) (see Fig. 6.26):

Cut DP = AB and DQ = AC and join PQ.
It can be seen that
[DP]/[PE] =[DQ]/[QF] and PQ ||EF (How?)
So, ∠ P = ∠ E and ∠ Q = ∠ F.
[DP]/[PE] = [DQ]/[QF] = [PQ]/[EF] (How?)
[DP]/[PE] = [DQ]/[QF] = [BC]/[EF] (Why?)
So, BC = PQ (Why?)
Thus, Δ ABC Δ DPQ (Why ?)
So, ∠ A = ∠ D, ∠ B = ∠ E and ∠ C = ∠ F (How) ?
Remark :
You may recall that either of the two conditions namely,
(i) corresponding angles are equal and
(ii) corresponding sides are in the same ratio is not sufficient for two polygons to be similar.
However, on the basis of Theorems 6.3 and 6.4, you can now say that in case of similarity of the two triangles, it is not necessary to check both the conditions as one condition implies the other.
Let us now recall the various criteria for congruency of two triangles learnt in Class IX.
You may observe that SSS similarity criterion can be compared with the SSS congruency criterion.
This suggests us to look for a similarity criterion comparable to SAS congruency criterion of triangles.
For this, let us perform an activity.

Activity 6 :
Draw two triangles ABC and DEF such that AB = 2 cm, ∠ A = 50°, AC = 4 cm, DE = 3 cm, ∠ D = 50° and DF = 6 cm (see Fig.6.27).

Here, you may observe that AB/DE = AC/DF (each equal to 2/3 ) and ∠ A (included between the sides AB and AC) = ∠ D (included between the sides DE and DF). That is, one angle of a triangle is equal to one angle of another triangle and sides including these angles are in the same ratio (i.e., proportion).
Now let us measure ∠ B, ∠ C, ∠ E and ∠ F.
You will find that ∠ B = ∠ E and ∠ C = ∠ F.
That is, ∠ A = ∠ D, ∠ B = ∠ E and ∠ C = ∠ F. So, by AAA similarity criterion, Δ ABC ~ Δ DEF.
You may repeat this activity by drawing several pairs of such triangles with one angle of a triangle equal to one angle of another triangle and the sides including these angles are proportional.
Everytime, you will find that the triangles are similar.
It is due to the following criterion of similarity of triangles:

Theorem 6.5 :
If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar.
This criterion is referred to as the SAS (Side–Angle–Side) similarity criterion for two triangles.

As before, this theorem can be proved by taking two triangles ABC and DEF such that AB/AC = DE/DF (< 1) and ∠ A = ∠ D (see Fig. 6.28).
Cut DP = AB, DQ = AC and join PQ.
Now, PQ ||EF and Δ ABC Δ DPQ (How ?)
So, ∠ A = ∠ D, ∠ B = ∠ P and ∠ C = ∠ Q
Therefore, Δ ABC ~ Δ DEF (Why?)
We now take some examples to illustrate the use of these criteria.

Example 4 : In Fig. 6.29, if PQ ||RS, prove that Δ POQ ~ Δ SOR.

Solution :

PQ ||RS (Given)
So, ∠ P = ∠ S (Alternate angles)
and ∠ Q = ∠ R
Also, ∠ POQ = ∠ SOR (Vertically opposite angles)
Therefore, Δ POQ ~ Δ SOR (AAA similarity criterion)

Example 5 : Observe Fig. 6.30 and then find ∠ P.

Solution :
In Δ ABC and Δ PQR,
[AB]/[RQ] = [3.8]/[7.6] = 1/2 and  [CA]/[PR] = [3sqrt(3)]/[6(sqrt3)]=1/2.
That is,
[AB]/[RQ] = [BC]/[QP] = [CA]/[PR]
So, Δ ABC ~ Δ RQP (SSS similarity)
Therefore, ∠ C = ∠ P (Corresponding angles of similar triangles)
But ∠ C = 180° – ∠ A – ∠ B (Angle sum property)
= 180° – 80° – 60° = 40°
So, ∠ P = 40°

Example 6 :
In Fig. 6.31,
OA . OB = OC . OD.
Show that ∠ A = ∠ C and ∠ B = ∠ D.
Solution :
OA . OB = OC . OD (Given)
So,
OA/OC = OD/OB...... (1)

Also, we have ∠ AOD = ∠ COB (Vertically opposite angles)...... (2)
Therefore, from (1) and (2), Δ AOD ~ Δ COB (SAS similarity criterion)
So, ∠ A = ∠ C and ∠ D = ∠ B
(Corresponding angles of similar triangles)

Example 7 :
A girl of height 90 cm is walking away from the base of a lamp-post at a speed of 1.2 m/s.
If the lamp is 3.6 m above the ground, find the length of her shadow after 4 seconds.
Solution :
Let AB denote the lamp-post and CD the girl after walking for 4 seconds away from the lamp-post (see Fig. 6.32).
From the figure, you can see that DE is the shadow of the girl.
Let DE be x metres.

Now, BD = 1.2 m xx 4 = 4.8 m.
Note that in Δ ABE and Δ CDE,
∠ B = ∠ D (Each is of 90° because lamp-post
as well as the girl are standing
vertical to the ground)
and ∠ E = ∠ E (Same angle)
So, Δ ABE ~ Δ CDE (AA similarity criterion)
Therefore,
[BE]/[DE] = [AB]/[CD]
i.e.,(4.8 + x) / x = (3.6)/(0.9) (90 cm =90/100 m = 0.9 m)
i.e., 4.8 + x =4x
i.e., 3x = 4.8
i.e., x = 1.6
So, the shadow of the girl after walking for 4 seconds is 1.6 m long.

Example 8 :
In Fig. 6.33, CM and RN are respectively the medians of Δ ABC and
Δ PQR. If Δ ABC ~ Δ PQR, prove that :
(i) Δ AMC ~ Δ PNR
(ii)[CM]/[RN] = [AB]/ [PQ]
(iii) Δ CMB ~ Δ RNQ
Solution :
(i) Δ ABC ~ Δ PQR

So,
[AB]/[PQ] = [BC]/[QR] = [CA]/[RP]
and ∠ A = ∠ P, ∠ B = ∠ Q and ∠ C = ∠ R........ (2)
But AB = 2 AM and PQ = 2 PN
(As CM and RN are medians)
So, from (1),[2AM]/[2PN] = [CA]/[RP]
i.e.,[AM]/[PN] = [CA]/[RP] (3)
Also, ∠ MAC = ∠ NPR [From (2)]........ (4)
So, from (3) and (4),
Δ AMC ~ Δ PNR (SAS similarity) (5)
(ii) From (5),
[CM]/[RN] = [CA]/[RP]...... (6) But [CA]/[RP] = [AB]/[PQ] [From (1)]...... (7)
Therefore,[CM]/[RN] = [AB]/[PQ] [From (6) and (7)].......... (8)
(iii) Again,[AB]/[PQ] = [BC]/[QR] [From ...........(1)]
Therefore,[CM]/[RN] = [BC]/[QR] [From.... (8)]......... (9)
Also, [CM]/[RN] = [AB]/[PQ] = [2BM]/[2QN]
i.e.,[CM]/[RN] = [BM]/[QN] .......(10)
i.e.,[CM]/[RN] = [BC]/[QR] = [BM]/[QN]....... [From (9) and (10)]
Therefore, Δ CMB ~ Δ RNQ (SSS similarity)
[Note : You can also prove part (iii) by following the same method as used for proving part (i).]

6.5 Areas of Similar Triangles

You have learnt that in two similar triangles, the ratio of their corresponding sides is the same.
Do you think there is any relationship between the ratio of their areas and the ratio of the corresponding sides? You know that area is measured in square units.
So, you may expect that this ratio is the square of the ratio of their corresponding sides.
This is indeed true and we shall prove it in the next theorem.

Theorem 6.6 :
The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
Proof :
We are given two triangles ABC and PQR such that Δ ABC ~ Δ PQR (see Fig. 6.42)

We need to prove that
ar(ABC)/ar(PQR)={ AB/PQ }2= {BC/QR}2 = {CA/RP}2.
For finding the areas of the two triangles, we draw altitudes AM and PN of the triangles.
Now, ar(ABC) =1/2{BC xx AM}
and ar (PQR) =1/2{QR xx PN}
So, ar (ABC)/ar (PQR) =1/2{BC xx AM}/1/2{QR xx PN} = {BC xx AM}/{QR xx PN}.......(1)
Now, in Δ ABM and Δ PQN,
∠ B = ∠ Q (As Δ ABC ~ Δ PQR)
and ∠ M = ∠ N (Each is of 90°)
So, Δ ABM ~ Δ PQN (AA similarity criterion)
Therefore,
AM/PN =AB/PQ......... (2)
Also, Δ ABC ~ Δ PQR (Given)
So,AB/PQ =BC/QR=CA/RP..............(3)
Therefore, ar (ABC)/ar (PQR) = AB/PQ = AM/PN ...[From (1) and (3)]
= AB/PQ × AB/ PQ........[From (2)]
= [ AB/PQ ]2
Now using (3), we get
ar (ABC)/ar (PQR) = AB/PQ = BC/QR = CA/RP
Let us take an example to illustrate the use of this theorem

Example 9 : In Fig. 6.43, the line segment XY is parallel to side AC of Δ ABC and it divides the triangle into two parts of equal areas.
Find the ratio AX/AB
Solution :

We have XY ||AC (Given)
So, ∠ BXY = ∠ A and ∠ BYX = ∠ C (Corresponding angles)
Therefore, Δ ABC ~ Δ XBY (AA similarity criterion)
So,
ar (ABC)/ar (XBY) = [AB/XB]2
(Theorem 6.6)......(1)
Also, ar (ABC) = 2 ar (XBY) (Given)
So, ar (ABC)/ar (XBY) = 2/1........ (2)
Therefore, from (1) and (2),
[ AB/XB ]2 = 2 / 1
i.e., AB/XB = sqrt(2)/1
or,XB/AB = 1/sqrt(2)or,
1 – (XB)/(AB) =1 – 1/sqrt(2)
or,(AB – XB )/ (AB) = (sqrt(2)-1)/sqrt(2)
(AX)/(AB) = (sqrt(2)-1) / sqrt(2) ,
i.e., (AX)/(AB) = (sqrt(2)-1)/ sqrt(2) = (2-sqrt(2))/2

6.6 Pythagoras Theorem

You are already familiar with the Pythagoras Theorem from your earlier classes.
You had verified this theorem through some activities and made use of it in solving certain problems. You have also seen a proof of this theorem in Class IX.
Now, we shall prove this theorem using the concept of similarity of triangles.
In proving this, we shall make use of a result related to similarity of two triangles formed by the perpendicular to the hypotenuse from the opposite vertex of the right triangle.
Now, let us take a right triangle ABC, right angled at B.
Let BD be the perpendicular to the hypotenuse AC (see Fig. 6.45).
You may note that in Δ ADB and Δ ABC ∠ A = ∠ A and ∠ ADB = ∠ ABC (Why?)
So, Δ ADB ~ Δ ABC (How?).......... (1)
Similarly, Δ BDC ~ Δ ABC (How?) (2)
Fig. 6.44 Fig. 6.45

So, from (1) and (2), triangles on both sides of the perpendicular BD are similar to the whole triangle ABC.
Also, since Δ ADB ~ Δ ABC
and Δ BDC ~ Δ ABC
So, Δ ADB ~ Δ BDC (From Remark in Section 6.2)
The above discussion leads to the following theorem :

Theorem 6.7 :
If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse then triangles on both sides of the perpendicular are similar to the whole triangle and to each other.
Let us now apply this theorem in proving the Pythagoras Theorem:

Theorem 6.8 :
In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
Proof :
We are given a right triangle ABC right angled at B. We need to prove that AC2 = AB2+ BC2
Let us draw BD ⊥ AC (see Fig. 6.46).
Now, Δ ADB ~ Δ ABC (Theorem 6.7)
or, AD . AC = AB2............... (1)
Also, Δ BDC ~ Δ ABC (Theorem 6.7)
So,CD/BC =BC/AC
or, CD . AC = BC........... (2)
Pythagoras(569 – 479 B.C.E.)
AD . AC + CD . AC = AB2 + BC2 or, AC (AD + CD) = AB2 + BC2
or, AC . AC = AB2 + BC2
or, AC2 = AB2 + BC2
The above theorem was earlier given by an ancient Indian mathematician Baudhayan (about 800 B.C.E.) in the following form :

The diagonal of a rectangle produces by itself the same area as produced by its both sides (i.e., length and breadth).
For this reason, this theorem is sometimes also referred to as the Baudhayan Theorem.
What about the converse of the Pythagoras Theorem? You have already verified, in the earlier classes, that this is also true. We now prove it in the form of a theorem.

Theorem 6.9 :
In a triangle, if square of one side is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle.
Proof : Here, we are given a triangle ABC in which AC2 = AB2+ BC2.
We need to prove that ∠ B = 90°.
To start with, we construct a Δ PQR right angled at Q such that PQ = AB and QR = BC (see Fig. 6.47).

Now, from Δ PQR, we have :
PR2 = PQ2 + QR2 (Pythagoras Theorem,as ∠ Q = 90°)
or, PR2 = AB2 + BC2 (By construction)......... (1)
TRIANGLES 147
But AC2 = AB2 + BC2 (Given).......... (2)
So, AC = PR [From (1) and (2)]....... (3)
Now, in Δ ABC and Δ PQR,
AB = PQ (By construction)
BC = QR (By construction)
AC = PR [Proved in (3) above]
So, Δ ABC Δ PQR (SSS congruence)
Therefore, ∠ B = ∠ Q (CPCT)
But ∠ Q = 90° (By construction)
So, ∠ B = 90°
Note :
Also see Appendix 1 for another proof of this theorem.
Let us now take some examples to illustrate the use of these theorems.

Example 10 :
In Fig. 6.48, ∠ ACB = 90°
and CD ⊥ AB. Prove that BC2/AC2 = BD/AD
Solution : Δ ACD ~ Δ ABC(Theorem 6.7)
Similarly, Δ BCD ~ Δ BAC (Theorem 6.7)
So,BC/BA =BD/BC or, BC2 = BA . BD .........(2)
Therefore, from (1) and (2),
(BC^2)/(AC^2)=(BAxxBD)/(ABxxAD)=(BD)/(AD)

Example 11 :
A ladder is placed against a wall such that its foot is at a distance of 2.5 m from the wall and its top reaches a window 6 m above the ground.
Find the length of the ladder.
Fig. 6.48
148 MATHEMATICS
Solution :
Let AB be the ladder and CA be the wall with the window at A (see Fig. 6.49).
Also, BC = 2.5 m and CA = 6 m From Pythagoras Theorem, we have:
AB2 = BC2 + CA2 = (2.5)2 + (6)2
= 42.25
So, AB = 6.5

Thus, length of the ladder is 6.5 m.

Example 12 : In Fig. 6.50, if AD ⊥ BC, prove that
AB2 + CD2 = BD2 + AC2
. Solution :

AC2 = AD2 + CD2(Pythagoras Theorem).......... (1)
AB2 = AD2 + BD2(Pythagoras Theorem) (2)Subtracting (1) from (2), we have
AB2 – AC2 = BD2 – CD2 or, AB2 + CD2 = BD2 + AC2

Example 13 : BL and CM are medians of a triangle ABC right angled at A. Prove that 4 (BL2 + CM2) = 5 BC2.
Solution :

BL and CM are medians of the Δ ABC in which ∠ A = 90° (see Fig. 6.51).
From Δ ABC,
BC2 = AB2 + AC2 (Pythagoras Theorem)................ (1)
From Δ ABL,BL2 = AL2 + AB2
or, BL2 = (AC/2)2+AB2(L is the mid-point of AC)
or, BL2 = AC2/4+AB2
or, 4 BL2 = AC2 + 4 AB2.............. (2)
From Δ CMA,
CM2 = AC2 + AM2
or, CM2= AC2 +2 AB2(M is the mid-point of AB)
or, CM2 = AC2 + AB2/4
or 4 CM2 = 4 AC2 + AB2..............(3)
Adding (2) and (3), we have
4 (BL2 + CM2) = 5 (AC2 + AB2)
i.e., 4 (BL2 + CM2) = 5 BC2........... [From (1)]

Example 14 : O is any point inside a rectangle ABCD (see Fig. 6.52). Prove that
OB2 + OD2 = OA2 + OC2.
Solution :

Through O, draw PQ ||BC so that P lies on
AB and Q lies on DC.
Now, PQ ||BC
Therefore, PQ ⊥ AB and PQ ⊥ DC (∠ B = 90° and ∠ C = 90°)
So, ∠ BPQ = 90° and ∠ CQP = 90°
Therefore, BPQC and APQD are both rectangles.
Now, from Δ OPB,
OB2 = BP2 + OP2.......... (1)
Similarly, from Δ OQD,
OD2 = OQ2 + DQ2...........(2)
From Δ OQC,we have
OC2 = OQ2 + CQ2........ (3)
and from Δ OAP, we have
OA2= AP2 + OP2........... (4)
OB2 + OD2 = BP2 + OP2 + OQ2 + DQ2
= CQ2 + OP2 + OQ2 + AP2(As BP = CQ and DQ = AP)
= CQ2 + OQ2 + OP2 + AP2
= OC2 + OA2 [From (3) and (4)]

EXERCISE 6.5
1. Sides of triangles are given below.
Determine which of them are right triangles.
In case of a right triangle, write the length of its hypotenuse.
(i) 7 cm, 24 cm, 25 cm
(ii) 3 cm, 8 cm, 6 cm
(iii) 50 cm, 80 cm, 100 cm
(iv) 13 cm, 12 cm, 5 cm
2. PQR is a triangle right angled at P and M is a
point on QR such that PM ⊥ QR. Show that
PM2 = QM . MR.
3. In Fig. 6.53, ABD is a triangle right angled at A
and AC ⊥ BD. Show that
(i) AB2 = BC . BD
(ii) AC2 = BC . DC
(iii) AD2 = BD . CD
4. ABC is an isosceles triangle right angled at C. Prove that AB^2 = 2AC^2.
5. ABC is an isosceles triangle with AC = BC. If AB2= 2 AC2, prove that ABC is a right triangle.
6. ABC is an equilateral triangle of side 2a. Find each of its altitudes.
7. Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.
8. In Fig. 6.54, O is a point in the interior of a triangle
ABC, OD ⊥ BC, OE ⊥ AC and OF ⊥ AB. Show that
(i) OA2 + OB2 + OC2 – OD2 – OE2 – OF2 = AF2 + BD2 + CE2,
(ii) AF2 + BD2 + CE2 = AE2 + CD2 + BF2
. 9. A ladder 10 m long reaches a window 8 m above the ground.
Find the distance of the foot of the ladder from base of the wall.
10. A guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end.
How far from the base of the pole should the stake be driven so that the wire will be taut?
11. An aeroplane leaves an airport and flies due north at a speed of 1000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1200 km per hour.
How far apart will be the two planes after1(1/2) hours?
12. Two poles of heights 6 m and 11 m stand on a plane ground.
If the distance between the feet of the poles is 12 m, find the distance between their tops.
13. D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that AE2 + BD2 = AB2 + DE2.
14. The perpendicular from A on side BC of a
Δ ABC intersects BC at D such that DB = 3 CD (see Fig. 6.55). Prove that 2 AB2 = 2 AC2 + BC2.
15. In an equilateral triangle ABC, D is a point on side BC such that BD =13 BC. Prove that 9 AD2 = 7 AB2.
16. In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes.
17. Tick the correct answer and justify : In Δ ABC, AB = 6 3 cm, AC = 12 cm and BC = 6 cm.
The angle B is :
(A) 120° (B) 60°
(C) 90° (D) 45°

6.7 Summary

In this chapter you have studied the following points :
1. Two figures having the same shape but not necessarily the same size are called similar figures.
2. All the congruent figures are similar but the converse is not true.
3. Two polygons of the same number of sides are similar, if (i) their corresponding angles are equal and (ii) their corresponding sides are in the same ratio (i.e., proportion).
4. If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio.
5. If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.
6. If in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio and hence the two triangles are similar (AAA similarity criterion).
7. If in two triangles, two angles of one triangle are respectively equal to the two angles of the other triangle, then the two triangles are similar (AA similarity criterion).
8. If in two triangles, corresponding sides are in the same ratio, then their corresponding angles are equal and hence the triangles are similar (SSS similarity criterion).
9. If one angle of a triangle is equal to one angle of another triangle and the sides including these angles are in the same ratio (proportional), then the triangles are similar (SAS similarity criterion).
10. The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
11. If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse,
then the triangles on both sides of the perpendicular are similar to the whole triangle and also to each other.
12. In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides (Pythagoras Theorem).
13. If in a triangle, square of one side is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle.

If in two right triangles, hypotenuse and one side of one triangle are proportional to the hypotenuse and one side of the other triangle,
then the two triangles are similar.
This may be referred to as the RHS Similarity Criterion.
If you use this criterion in Example 2, Chapter 8, the proof will become simpler

• Corresponding angles are equal
• Corresponding sides are in the same ratio (or proportion).
• The ratio of any two corresponding sides in two equiangular triangles is always the same.
• If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.
• In two triangles, sides of one triangle are proportional to (i.e., in the same ratio of ) the sides of the other triangle
• This criterion is referred to as the SSS (Side–Side–Side) similarity criterion for two triangles.
• Two polygons to be similar.
• The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
• The diagonal of a rectangle produces by itself the same area as produced by its both sides (i.e., length and breadth).
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text:
Curl, H, B, A problem

1. Nov 22, 2007

magleviee

N is the normal vector of a surface Sb and N dot B=0, div B=0 on Sb. Therefore, let B=curl A where A is a vector field. How can you prove that N cross A=0? Thanks.

Last edited: Nov 22, 2007
2. Nov 27, 2007

siddharth

What have you done with this question? You need to show some work before you get help.


text:
# Kunh-Tucker-Langrange - minimum points in $\text{int} (S)$

Let $$f(x,y,z) = x^3 - y^3 + z$$ and the constraint set $$S = \left\{ (x,y,z) \in \mathbb{R}^3 : x^2 + 2y^2 \leq 1, \, y = 2z \right\}$$ and assume that we want to find a $$\bar{x}$$ such that $$\min_\limits{x \in S} f(x) = f(\bar{x})$$.

Kuhn-Tucker-Lagrange:

Let $$f_0 = x^3 - y^3 + z, f_1 = x^2 + 2y^2 - 1, f_2 = y - 2z$$.

$$\begin{cases} \nabla f_0 + \lambda_1 \nabla f_1 + \lambda_2 \nabla f_2 = 0 \\ \lambda_1f_1 = 0 \\ f_2 = 0 \\ \lambda_1 \geq 0, \, \lambda_2 \in \mathbb{R} \end{cases}$$

• Case $$\lambda_1 = 0$$ :

we yield the points : $$x_1 = \left( 0, \frac{1}{\sqrt{6}}, \frac{1}{2 \sqrt{6}}\right)$$ and $$x_2 = \left( 0, -\frac{1}{\sqrt{6}}, -\frac{1}{2 \sqrt{6}}\right)$$

which are both in $$\text{int}(S)$$.

Question: Does the fact that we found these candidates in $$\text{int}(S)$$ ensures that every other point we find on the boundary of $$S$$ (Case $$\lambda_1 > 0$$ ) cannot be a minimum point (and thus $$x_2$$ is the minimum) and why?

• Since $$\frac{\partial f(x,y,z)}{\partial z}=1\neq 0$$ so there are no extrempoints. – Dr. Sonnhard Graubner Jun 21 '19 at 12:01
• @Dr.SonnhardGraubner Can you elaborate on that? – Paris Jun 21 '19 at 13:21

Making the substitution $$y\to 2z$$ the problem assumes the reduced form

$$\min f=x^3-8z^3+z\ \ \ \text{s. t.}\ \ \ g = x^2+8z^2-1 \le 0$$

and the lagrangian with the contribution of a slack vatiable $$s$$ is

$$L(x,z,\lambda,s) = f+\lambda(g+s^2)$$

The stationary conditions are

$$\nabla L = 0 = \left\{ \begin{array}{l} 3 x^2+2 \lambda x \\ -24 z^2+16 \lambda z+1 \\ s^2+x^2+8 z^2-1 \\ 2 \lambda s \\ \end{array} \right.$$

with solutions

$$\left( \begin{array}{ccccc} f & x & z & \lambda & s^2 \\ 0. & 0. & -0.353553 & -0.353553 & 0. \\ 0. & 0. & 0.353553 & 0.353553 & 0. \\ -0.136083 & 0. & -0.204124 & 0. & 0.816497 \\ 0.136083 & 0. & 0.204124 & 0. & 0.816497 \\ -1.02035 & -0.993481 & -0.0403049 & 1.49022 & 0. \\ 1.02035 & 0.993481 & 0.0403049 & -1.49022 & 0. \\ -0.00604972 & 0.22368 & -0.344595 & -0.335521 & 0. \\ 0.00604972 & -0.22368 & 0.344595 & 0.335521 & 0. \\ \end{array} \right)$$

Attached a graphic showing the feasible region with the objective function level curves, with the stationary points in red dots. The red vectors represent at the stationary points, the restriction gradient, as well as the black vectors, the objective function gradient. With those elements we can qualify the stationary points, choosing the solution set.


text:
• This event has passed.

## May 15, 2018

The Department of Energy’s Office of Science has announced that the Office of Science Graduate Student Research (SCGSR) program is now accepting applications for the 2018 Solicitation 1.

The SCGSR program supports supplemental awards to outstanding U.S. graduate students to conduct part of their graduate thesis research at a DOE national laboratory/facility in collaboration with a DOE laboratory scientist for a period of three to 12 consecutive months, with the goal of preparing graduate students for scientific and technical careers critically important to the DOE Office of Science mission.


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# Analytic investigation of rotating holographic superconductors

Srivastav, Ankur (Aff1, 0000 0001 2188 427X, grid.452759.8, Department of Theoretical Sciences, S. N. Bose National Centre for Basic Sciences, Block-JD, Sector-III, Salt Lake City, 700106, Kolkata, India) ; Gangopadhyay, Sunandan (Aff1, 0000 0001 2188 427X, grid.452759.8, Department of Theoretical Sciences, S. N. Bose National Centre for Basic Sciences, Block-JD, Sector-III, Salt Lake City, 700106, Kolkata, India)

16 April 2019

Abstract: In this paper we have investigated, in the probe limit, s -wave holographic superconductors in rotating AdS3+1 \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$AdS_{3+1}$\end{document} spacetime using the matching method as well as the Stürm–Liouville eigenvalue approach. We have calculated the critical temperature using the matching technique in such a setting and our results are in agreement with previously reported results obtained using the Stürm–Liouville approach. We have then obtained the condensation operators using both analytical methods. The results obtained by both these techniques share the same features as found numerically. We observe that the rotation parameter of the black hole affects the critical temperature and the condensation operator in a non-trivial way.

Published in: EPJC 79 (2019) 340 DOI: 10.1140/epjc/s10052-019-6834-0


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# AA Similarity

## Two triangles are similar if two pairs of angles are congruent.

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If two right triangles contain corresponding sides with equal ratios, then they are said to be similar by $\bf{SAS\ Similarity}$.

qid: 100283


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Home OALib Journal OALib PrePrints Submit Ranking News My Lib FAQ About Us Follow Us+
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 Physics , 2015, Abstract: We study high-resolution hydrodynamic simulations of Milky Way type galaxies obtained within the "Evolution and Assembly of GaLaxies and their Environments" (EAGLE) project, and identify the those that best satisfy observational constraints on the Milky Way total stellar mass, rotation curve, and galaxy shape. Contrary to mock galaxies selected on the basis of their total virial mass, the Milky Way analogues so identified consistently exhibit very similar dark matter profiles inside the solar circle, therefore enabling more accurate predictions for indirect dark matter searches. We find in particular that high resolution simulated haloes satisfying observational constraints exhibit, within the inner few kiloparsecs, dark matter profiles shallower than those required to explain the so-called Fermi GeV excess via dark matter annihilation.
 Physics , 2013, DOI: 10.1093/mnras/stt691 Abstract: We investigate the chemical and kinematic properties of the diffuse stellar haloes of six simulated Milky Way-like galaxies from the Aquarius Project. Binding energy criteria are adopted to defined two dynamically distinct stellar populations: the diffuse inner and outer haloes, which comprise different stellar sub-populations with particular chemical and kinematic characteristics. Our simulated inner- and outer-halo stellar populations have received contributions from debris stars (formed in sub-galactic systems while they were outside the virial radius of the main progenitor galaxies) and endo-debris stars (those formed in gas-rich sub-galactic systems inside the dark matter haloes). The inner haloes possess an additional contribution from disc-heated stars in the range $\sim 3 - 30 %$, with a mean of $\sim 20%$. Disc-heated stars might exhibit signatures of kinematical support, in particular among the youngest ones. Endo-debris plus disc-heated stars define the so-called \insitu stellar populations. In both the inner- and outer-halo stellar populations, we detect contributions from stars with moderate to low [$\alpha$/Fe] ratios, mainly associated with the endo-debris or disc-heated sub-populations. The observed abundance gradients in the inner-halo regions are influenced by both the level of chemical enrichment and the relative contributions from each stellar sub-population. Steeper abundance gradients in the inner-halo regions are related to contributions from the disc-heated and endo-debris stars, which tend to be found at lower binding energies than debris stars. (Abridged).
 Physics , 2013, DOI: 10.1089/ast.2012.0842 Abstract: Every Galactic environment is characterized by a stellar density and a velocity dispersion. With this information from literature, we simulated flyby encounters for several Galactic regions, numerically calculating stellar trajectories as well as orbits for particles in disks; our aim was to understand the effect of typical stellar flybys on planetary (debris) disks in the Milky Way Galaxy. For the Solar neighborhood, we examined nearby stars with known distance, proper motions, and radial velocities. We found occurrence of a disturbing impact to the Solar planetary disk within the next 8 Myr to be highly unlikely; perturbations to the Oort cloud seem unlikely as well. Current knowledge of the full phase space of stars in the Solar neighborhood, however, is rather poor, and thus we cannot rule out the existence of a star that is more likely to approach than those for which we have complete kinematic information. We studied the effect of stellar encounters on planetary orbits within the habitable zones of stars in more crowded stellar environments, such as stellar clusters. We found that in open clusters habitable zones are not readily disrupted; this is true if they evaporate in less than 108 years. For older clusters the results may not be the same. We specifically studied the case of Messier 67, one of the oldest open clusters known, and show the effect of this environment on debris disks. We also considered the conditions in globular clusters, the Galactic nucleus, and the Galactic bulge-bar. We calculated the probability of whether Oort clouds exist in these Galactic environments.
 Physics , 2013, DOI: 10.1093/mnras/stt1600 Abstract: We analyse the structure and chemical enrichment of a Milky Way-like galaxy with a stellar mass of 2 10^{10} M_sun, formed in a cosmological hydrodynamical simulation. It is disk-dominated with a flat rotation curve, and has a disk scale length similar to the Milky Way's, but a velocity dispersion that is ~50% higher. Examining stars in narrow [Fe/H] and [\alpha/Fe] abundance ranges, we find remarkable qualitative agreement between this simulation and observations: a) The old stars lie in a thickened distribution with a short scale length, while the young stars form a thinner disk, with scale lengths decreasing, as [Fe/H] increases. b) Consequently, there is a distinct outward metallicity gradient. c) Mono-abundance populations exist with a continuous distribution of scale heights (from thin to thick). However, the simulated galaxy has a distinct and substantive very thick disk (h_z~1.5 kpc), not seen in the Milky Way. The broad agreement between simulations and observations allows us to test the validity of observational proxies used in the literature: we find in the simulation that mono-abundance populations are good proxies for single age populations (<1 Gyr) for most abundances.
 Physics , 2015, DOI: 10.1088/2041-8205/800/1/L4 Abstract: In this letter we present, for the first time, a study of star formation rate, gas fraction and galaxy morphology of a constrained simulation of the Milky Way (MW) and Andromeda (M31) galaxies, compared to other MW-mass galaxies. By combining with unconstrained simulations we cover a sufficient volume to compare these galaxies environmental densities ranging from the field to that of the Local Group (LG). This is particularly relevant as it has been shown that, quite generally, galaxy properties depend intimately upon their environment, most prominently when galaxies in clusters are compared to those in the field. For galaxies in loose groups such as the LG, however, environmental effects have been less clear. We consider the galaxy's environmental density in spheres of 1200 kpc (comoving) and find that whilst environment does not appear to directly affect morphology, there is a positive trend with star formation rates. This enhancement in star formation occurs systematically for galaxies in higher density environments, regardless whether they are part of the LG or in filaments. Our simulations suggest that the richer environment at Mpc-scales may help replenish the star-forming gas, allowing higher specific star formation rates in galaxies such as the MW.
 Physics , 2011, DOI: 10.1111/j.1365-2966.2011.19027.x Abstract: We study the stellar discs and spheroids in eight simulations of galaxy formation within Milky Way-mass haloes in a Lambda Cold Dark Matter cosmology. A first paper in this series concentrated on disc properties. Here, we extend this analysis to study how the formation history, structure and dynamics of discs and spheroids relate to the assembly history and structure of their haloes. We find that discs are generally young, with stars spanning a wide range in stellar age: the youngest stars define thin discs and have near-circular orbits, while the oldest stars form thicker discs which rotate ~2 times slower than the thin components, and have 2-3 times larger velocity dispersions. Unlike the discs, spheroids form early and on short time-scales, and are dominated by velocity dispersion. We find great variety in their structure. The inner regions are bar- or bulge-like, while the extended outer haloes are rich in complex non-equilibrium structures such as stellar streams, shells and clumps. Our discs have very high in-situ fractions, i.e. most of their stars formed in the disc itself. Nevertheless, there is a non-negligible contribution (~15 percent) from satellites that are accreted on nearly coplanar orbits. The inner regions of spheroids also have relatively high in-situ fractions, but 65-85 percent of their outer stellar population is accreted. We analyse the circular velocities, rotation velocities and velocity dispersions of our discs and spheroids, both for gas and stars, showing that the dynamical structure is complex as a result of the non-trivial interplay between cooling and SN heating.
 Andrew R. Casey Physics , 2014, Abstract: Hundreds of globular clusters and dwarf galaxies encircle the Milky Way. Many of these systems have undergone partial disruption due to tidal forces, littering the halo with stellar streams. These tidal tails are sensitive to the Galactic potential, facilitating an excellent laboratory to investigate Galaxy formation and evolution, as well as local chemical signatures of differing star formation environments. To better understand the emergence of the Milky Way, this thesis examines the dynamics and chemistry of a number of known stellar streams.
 Physics , 2011, DOI: 10.1111/j.1365-2966.2011.19971.x Abstract: We investigate the formation and evolution of satellite galaxies using smoothed particle hydrodynamics (SPH) simulations of a Milky Way(MW)-like system, focussing on the best resolved examples, analogous to the classical MW satellites. Comparing with a pure dark matter simulation, we find that the condensation of baryons has had a relatively minor effect on the structure of the satellites' dark matter halos. The stellar mass that forms in each satellite agrees relatively well over three levels of resolution (a factor of ~64 in particle mass) and scales with (sub)halo mass in a similar way in an independent semi-analytical model. Our model provides a relatively good match to the average luminosity function of the MW and M31. To establish whether the potential wells of our satellites are realistic, we measure their masses within observationally determined half-light radii, finding that the most massive examples have somewhat higher mass-to-light ratios than those derived for the MW dSphs from stellar kinematic data. A statistical test yields a ~9 percent probability that the simulated and observationally derived distributions of masses are consistent. Our results may suggest that either the MW halo is less massive than assumed in our simulations (~1.4e12 M_sun) or that there is substantial scatter in the satellite luminosity function or distribution of mass-to-light ratios at fixed host halo mass. Alternatively, feedback processes not properly captured by our simulations may have reduced the central densities of (sub)halos, or the subhalos may have initially formed with lower concentrations as would be the case, for example, if the dark matter were made of warm, rather than cold particles.
 EPJ Web of Conferences , 2012, DOI: 10.1051/epjconf/20121901005 Abstract: The formation of the Milky Way in the currently accepted concordance cosmology presents a couple of pitfalls. Mostly, these involve the much reported overcooling problem. If gas is allowed to cool, it will condense into massive central concentrations unlike any observed in real galaxies. These concentrations are characterized by a high central peak in the rotation curve and a rise in the surface brightness profile much steeper than the exponentially increasing disk. Such galaxies may contain no disk component and form many more stars than predicted by abundance matching studies.
 Physics , 2012, Abstract: We investigate the physical properties of the progenitors of today living Milky Way-like galaxies that are visible as Damped Lya Absorption systems and Lya Emitters at higher redshifts (z ~ 2.3,5.7). To this aim we use a statistical merger-tree approach that follows the formation of the Galaxy and its dwarf satellites in a cosmological context, tracing the chemical evolution and stellar population history of the progenitor halos. The model accounts for the properties of the most metal-poor stars and local dwarf galaxies, providing insights on the early cosmic star-formation. Fruitful links between Galactic Archaeology and more distant galaxies are presented.
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Contents

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# Contents

## Idea

In formal logic, a logical connective is a type formation rule which from given (sets and) propositions forms new propositions.

For example, the logical connective “AND” combines a pair of propositions $P_1$, $P_2$ to a new proposition $P_1 \wedge P_2$.

To the extent that propositions are regarded as functions from sets/types $S$ (whose elements/terms they are propositions about) to the set/type of truth values (the type of propositions)

$\frac{ \Gamma ,\;\; s \colon S \;\;\; \vdash \;\;\; P(s) \colon Prop }{ \Gamma \;\;\; \vdash \;\;\; P \,\colon\, S \to Prop }$

one may regard logical connectives as forming the resulting functions. This is what is expressed by the notorious truth tables of classical logic.

## Examples

$\phantom{-}$symbol$\phantom{-}$$\phantom{-}$in logic$\phantom{-}$
$\phantom{A}$$\in$$\phantom{A}$element relation
$\phantom{A}$$\,:$$\phantom{A}$typing relation
$\phantom{A}$$=$$\phantom{A}$equality
$\phantom{A}$$\vdash$$\phantom{A}$$\phantom{A}$entailment / sequent$\phantom{A}$
$\phantom{A}$$\top$$\phantom{A}$$\phantom{A}$true / top$\phantom{A}$
$\phantom{A}$$\bot$$\phantom{A}$$\phantom{A}$false / bottom$\phantom{A}$
$\phantom{A}$$\Rightarrow$$\phantom{A}$implication
$\phantom{A}$$\Leftrightarrow$$\phantom{A}$logical equivalence
$\phantom{A}$$\not$$\phantom{A}$negation
$\phantom{A}$$\neq$$\phantom{A}$negation of equality / apartness$\phantom{A}$
$\phantom{A}$$\notin$$\phantom{A}$negation of element relation $\phantom{A}$
$\phantom{A}$$\not \not$$\phantom{A}$negation of negation$\phantom{A}$
$\phantom{A}$$\exists$$\phantom{A}$existential quantification$\phantom{A}$
$\phantom{A}$$\forall$$\phantom{A}$universal quantification$\phantom{A}$
$\phantom{A}$$\wedge$$\phantom{A}$logical conjunction
$\phantom{A}$$\vee$$\phantom{A}$logical disjunction
symbolin type theory (propositions as types)
$\phantom{A}$$\to$$\phantom{A}$function type (implication)
$\phantom{A}$$\times$$\phantom{A}$product type (conjunction)
$\phantom{A}$$+$$\phantom{A}$sum type (disjunction)
$\phantom{A}$$0$$\phantom{A}$empty type (false)
$\phantom{A}$$1$$\phantom{A}$unit type (true)
$\phantom{A}$$=$$\phantom{A}$identity type (equality)
$\phantom{A}$$\simeq$$\phantom{A}$equivalence of types (logical equivalence)
$\phantom{A}$$\sum$$\phantom{A}$dependent sum type (existential quantifier)
$\phantom{A}$$\prod$$\phantom{A}$dependent product type (universal quantifier)
symbolin linear logic
$\phantom{A}$$\multimap$$\phantom{A}$$\phantom{A}$linear implication$\phantom{A}$
$\phantom{A}$$\otimes$$\phantom{A}$$\phantom{A}$multiplicative conjunction$\phantom{A}$
$\phantom{A}$$\oplus$$\phantom{A}$$\phantom{A}$additive disjunction$\phantom{A}$
$\phantom{A}$$\&$$\phantom{A}$$\phantom{A}$additive conjunction$\phantom{A}$
$\phantom{A}$$\invamp$$\phantom{A}$$\phantom{A}$multiplicative disjunction$\phantom{A}$
$\phantom{A}$$\;!$$\phantom{A}$$\phantom{A}$exponential conjunction$\phantom{A}$

Last revised on February 18, 2023 at 14:05:06. See the history of this page for a list of all contributions to it.