Hello, so in this video we're going to go
over sig figs, so basically how to find the
number of sig figs in a number, how to deal
with sig figs when adding, subtracting, multiplying,
dividing using exponents, using logarithms,
and also how to determine the number of sig
figs after square rooting. Anyways, so firstly
we're going to go over how to determine the
number of sig figs in a number. In order to
do that, the first step would be to determine
if a decimal point in that number is present
or absent. What you do basically is if the
decimal point is present, you start on the
left side and then you start looking for a
non-zero digit. So in this case you start
on the left side, the first number is 2 and
since 2 is non-zero, it's significant, and
every number to the right, so every digit
to the right of 2, is significant, so two
is significant, three is significant, 5 and
7--they're all significant, so that gives
you four. It doesn't matter what the numbers
are to the right if the decimal point is present
after the first non-zero digit counting from
the left everything is significant. If you
look here we start on the left side and we
start looking for the first non-zero digit.
The first digit we look at is this. It's zero,
right? So that's not significant. The next
one is also zero, which is also not significant.
Likewise, the 0 next to that is not significant.
The number next to that is 3, and now we have
found our first non-zero digit, and as I said,
every number to the right of three is significant.
So two is also significant, which gives us
2 sig figs.The same thing here: we start from
the left the first digit we're going to look
at is 2, and since that’s not zero it's
significant, and every digit to the right
of this number, 5, 6 and 0 are significant
so that gives us 4 sig figs. So even though
this number is zero, since the decimal point
is present, it’s going to be significant,
so that would be four. So let's try doing
the last one; try doing it like pause the
video or something on your own. So you start
counting from the left, right? The first non-zero
digit you run into is the first digit. It’s
5. So every number to the right of this 5,
6 and all these zeros, are significant. That
gives us 5 sig figs, alright? So that's how
you determine the number of sig figs when
a decimal point is present.
When it's absent though, you can't start counting
from the left side. What you do is you start
counting on the right side. Start counting
from the right and look for your first non-zero
digit like we did when the decimal point was
present. So the first non-zero digit would
be seven and every number to the left of that
digit is significant so in this case is 7,
5, 2 and 3. They're all significant so that
gives us 4 sig figs. Likewise, for 5,000,
we start counting on the right so we look
for our first non-zero digit. That's zero,
so that won't work. That's also zero--won't
work. That's another zero, and we run into
her first non-zero digit counting from the
right, and it's 5 and since there aren't any
numbers to the left of that the number of
sig figs is 1. So 6707, we start counting
from the right we look for the first non-zero
digit which happens to be the first digit
we look at: 7. And it doesn't matter that
this number next to it is zero since it's
on the left it’s significant. These are
all significant, which gives us 4 sig figs,
right? Now let’s try doing the last one.
So you start counting from the right. You
count to the left from the right, right? So
hopefully you got 2 sig figs because you look
at your first digit--it's zero, next is also
zero. That’s zero, so you look at this number
and everything to the left, since 3 is non-zero,
would be significant, so that gives us two.
Anyways that's how you determine the number
of sig figs in a number and let's say they
give you the number in scientific notation
like 2.57 times 10 to the 5th power. So basically
what you do is you just ignore this part and
you just determine the number of sig figs
in this part of the number. So since the decimal
point is present you start counting from the
right that gives you one two three sig figs.
So this has 3 sig figs and I mean the same
thing if it's like 1.57 8 times 10 to the
negative 7th you ignore this part so you just
count the number of sig figs here. The decimal
point is present, and we get four sig figs
cuz if you count 1, 2, 3, and 4. Anyways so
that's how you determine the number of sig
figs. So the next thing would be determining
the number of sig figs when adding and subtracting.
Let's just start with an example, maybe 4.357
+ 2.111. What you do basically is you just
add it like you would add normal numbers,
so 6.4681. And you look not at the sig figs
of the two numbers that you're adding but
the number of decimals, so the number of digits
to the right of the decima, or the number
of decimal places. So that is three decimal
places and that is 4 decimal places so what
you do is you take the least number so it'll
be 3 in this case so the smaller one and round
off to three decimal places so since one is
less than 5, it doesn't round up. We just
drop it and get 6.468. So let’s do another
example, maybe like 5.823 + 2.57 actually
let's change that to an 8. If we do that,
what we do is we would add it up like a normal
number right so if we added we get 8.398,
right? Then we look at the number of decimal
places; its 3 and 2 and since two is smaller
we use this number to determine the number
of decimal places in our answer so it should
be just two and since there is that eight
and eight is greater than five we have to
round up so basically what we get would be
8.40. So we're still keeping the two decimal
places in our answer from this number right
here. Anyways, let's do an example with subtracting
so 5.83 - 2.375. So try doing this on your
own the rules are exactly the same as addition.
So what you should have got is 3.455, except
for the 5 rounds up, and you get 3.46 because
we have to limit the answer to 2 decimal places.
But now you won't see this often but sometimes
you might have to subtract or add two numbers
that don't have decimal places so like that
plus 5321 or something what you do is you
just do the same thing you just add first
and then since this and then you start like
looking for your first significant digit and
it's starting at the hundredth place, right?
So 5500 is only significant starting at the
hundredth so all these numbers here aren’t
significant, so you can't write these two
numbers as significant digits. What we have
to do is have to change these to zero, and
obviously you would round, but since 2 is
less than 5 you round down and get 10800 as
an answer. But you won't really see this often;
maybe let's do another example like 5300 +
2.578. now all these numbers are significant
but this is only significant starting from
the hundredth again. We can't write any digits
that are smaller than a hundred so we only
can write that, right? So like if you take
account of sig figs adding 2.578 to 5300 doesn't
really do anything to the number. Anyways
so that's basically How you do sig figs when
adding and subtracting the next thing would
be multiplying and dividing. So multiplying
and dividing. This is actually easier than
adding and subtracting because all you need
to do is basically let say we have 2.35 * 8.7.
This has 3 sig figs, right? This has two,
so our product will also have two sig figs.
So you basically just take the smaller one,
the smaller number of sig figs, and write
that as a product. So if we actually calculate
this, what you’re going to get is 2.35 * 8.7
equals 20.445. And we can only write 2 sig
figs so we just leave it 20.. And we have
to have the point because if you don't have
the point then it would look like one sig
fig, right? Maybe let's try another one, so
3.8 or like 13.8 times 105.7. This has three
sig figs, this has four, so this would have
three sig figs as well, right> try doing this
on your own and see what you get. So what
you should have got is 1460. And 60 because
the exact answer is that but what you have
to look at is your rounding again. So you
look there that's greater than 5 so you round
up, and it becomes 1460. all right so maybe
like let's do a division example. So if we
have 25 / 5.0, they both have 2 sig figs right?
So our answer would also have 2 sig figs.
If we do that that would just be 5.0 because
that's 2 and that's also 2. Maybe 365.8 / 27
so try doing this and see what you get. All
right, so as an answer you should have got
something like 13.5481481 now if you look
here that only has two significant digits
so we can only have two. So our answer is
13.54 blah blah blah but since it's only two
sig figs because that one has four, that one
has two, and always we take the smaller one.
We can only have two sig figs and since that’s
a 5 we round up. So we round to two sig figs
and get forteen as an answer.
All right, so the next thing I'm going to
quickly go over it is how to deal with sig
figs in powers. So basically sig figs when
raising something to power is like kind of
like an extension of multiplication. So I'm
only going to deal with powers to like whole
numbers, and we're assuming that these sowers,
so like let's say 3 to the fifth, the five
is an exact number and it has like an infinite
amount of sig figs. Anyways, so like basically
what you do is let's say we have 2.75 to the
26th power, our answer would be equal to 2.75
* 2.75 * 2.75... Like basically 2.75 multiplied
by itself 26 times and if you think about
it remember that's just multiplication, and
multiplication determining the number of sig
figs and its product what you do is you find
the number with the least sig figs and that
would be the number of sig figs in your product.
But in this case all these numbers have three
sig figs so our product will also have three
sig figs, and if we actually calculate this,
we're going to get something like 2.64636
blah blah times 10 to the 11th power. If you
look here remember we can only have three
sig figs and since that's a 6 we're going
to round up and get 2.65 * 10 to the 11th
power. So when we have an exponent and we
assume this is an exact number so like the
26 is exact, that means it has infinite number
of sig figs, we just say the product would
have the same number of sig figs as the base
so like let's say it's 2.7 to the 5th, if
you actually calculate that, you're going
to get something like a 143.4 since this only
has two sig figs, our answer can only have
two sig figs. So this is equal to eight, but
remember two sig figs, so we're going to round
to 140. Just another example, try leaving
this on your own: 3.02 to the 6th, so how
many sig figs should this have? It should
have three, right? I mean the product. The
answer should have 3 sig figs because that's
three and therefore our answer would be 759,
so hopefully that's what you got. All right,
so the next thing is a square root. So let's
say we're taking the square root of a number,
so let's say 8.9. This has two sig figs; 8.9
as 2 sig figs, and so our answer will have
three. So it'd be like 2.983, but since three
smaller than 5, we will just drop it, and
basically the answer is 2.98. So if we have
a number inside a square root so let's say
if the number of sig figs of n, sig figs of
m is equal to 1 plus the number of sig figs
of n.
Alright, so the last thing ready to go over
is logarithms and exponents that aren't like
whole numbers. If we have log 100 for example,
then we just write this (the answer is two
obviously cuz a hundred is 10 to the second
power) but we have to write 2.0 if we want
to include sig figs because a hundred has
one sig fig, but we have to write one decimal
place for the answer. So let's say so like
a general formula, I guess. So if it says
log n equals m the number of sig figs in n
n is equal to the number of decimal places
in m. So that would be log base 100, like
for example, let's say log 238. Try doing
that on your own and see what you get. So
you should get something like 2.3765. Now
since this has three sig figs, our answer
should have three decimal places so we leave
it as 2.377. Hopefully that's what you got.
So this isn't just for base-10 so like when
I say log its base 10, right? It could also
work for like literally any logarithm; so
let's say like ln38 so the answer for that
is also like the number sig figs in 38 should
be equal to the number of decimal places in
our answer so if we get ln38,the answer is
equal to 3.637. And since this has two sig
figs, our answer should have two decimal places.
So we write 3.64 because it's actually 3.637,
but we round up because we have to have the
two decimal places. Okay so the final thing
we're going to go over is exponents. So in
exponents, we already went over things like
2.75 to the 6th power where 6 is an exact
whole number with infinite amount of sig figs.
The difference here is like we're going to
do numbers such as 10 to the negative 7.85th
power or something. So if we see this exponent
here, it has two numbers after its decimal
point. Therefore our answer is going to have
two sig figs, so it should be 1.4 x 10 to
the negative 8th, so 2 sig figs. Basically
to sum it up, the number of decimal places
in the exponent is equal to the number of
sig figs in answer. So hopefully that makes
sense!
