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Instructor: Hi class!
Today we're going to take a look at fractions.
And before we get into the operations of fractions,
I want to take a look at the definition of a fraction
and what the different parts of a fraction actually are.
Oftentimes people have a hard time
with adding and subtracting fractions
because they never really understood what a fraction is
and what the different parts represent.
So I want you to take a look at this circle
and pretend that this circle is an apple.
And let's say that you have a friend over
and you decide to share this apple with your friend,
so you cut it down the middle here.
And let's say that your friend eats this part of the apple,
and you save the other part of the apple
for yourself for later.
So you still have this one piece of this apple.
And the apple was divided into one, two different pieces.
So we would say that you have one half [―] of this apple.
Now, we want to examine the different parts of this fraction
right here that we have that's one half [―].
The 1 represents the one piece that you have.
The 2 represents the number of pieces that the apple was cut into.
So you want to think of this [bottom]
as the number of parts the item was divided into,
and this [top] is the number of parts that you actually have.
The other parts were given away, or thrown out or something.
Now, if, when we think about it in terms of that,
if a fraction is actually negative
it doesn't make sense to cut the apple into negative two [-2] parts.
So this 2 would not make sense to be negative [ - ].
You can have [-2], or negative in the bottom of a fraction.
However, in terms of a real life situation,
it doesn't necessarily make sense.
So, a number is negative,
whether the negative is in the denominator of the fraction
[1/-2 ],
out front of the fraction [- 1/2 ]
or up top on the fraction [ -1/2 ].
So, what we want to do is actually put that negative [ - ]
wherever it makes sense.
And it doesn't make sense to cut something into negative parts,
so if a fraction is negative, put the negative [ - ] in the top,
because it does make sense to owe somebody part of an apple.
And that negative [ - ] being up top would say
that you owe somebody one out of the two pieces of the apple.
And that makes sense.
But cutting the apple into 2, into negative 2 pieces [-2 pieces],
would not make sense.
Now the top of the fraction here,
which is the number of pieces that you have
is what we call the 'numerator.'
The bottom of the fraction down here,
which is the number of pieces you cut your item into
is called the 'denominator.'
So, those are the names for these two parts of your fraction.
Now let's take a look at a situation,
knowing that the bottom is how many pieces you cut the fraction into
and the top is how many pieces you have.
And we're going to look at a little situation
where I'm drawing pictures to kind of illustrate
how you would add fractions and why you do the things you do.
You will not have to draw the pictures,
and do what I'm doing on every single problem.
This is just to illustrate why we do
what we do for adding and subtracting.
So we currently have this one apple.
And we have half [1/2] of the apple left.
Let's say you have another apple,
and this time, two of your friends come over.
And you want to share this apple with them.
And so there's a total of three of you if you include yourself.
So you divide the apple into three parts.
And your two friends eat these two parts,
which leaves you with just this one part.
And that one part, since you divided into equal parts,
is one third [ 1/3] of the apple.
So this picture over here is representing one third [1/3].
Now let's say you want to take
the two pieces of apple you have, this one half [ 1/2] from earlier
and this one third [1/3 ], and you want to combine them together
and see how much you have,
to see if maybe you can use it in a recipe.
If you want to do that,
what you're wanting to do is add these two fractions together.
Now to add these two fractions together
you would be putting this piece of the apple here
together with this piece of the apple.
When you place those together,
you actually get kind of a weird shape.
You have your half [1/2] of an apple that looks like that,
and your third [1/3] of an apple that looks like that.
And it's hard to tell from looking at a picture like that
how much of an apple you really have,
and the reason why is because this piece
is not the same size as that piece.
So what we want to do is make it so that the two apples
were divided into the same number of pieces
so that we can tell how much we have.
So you think to yourself,
since this was divided into two pieces and this one three,
what's the smallest number that 2 and 3 go into [ũ]?
And they both go into [ũ] 6.
Since you can't paste pieces back together,
but you could cut them into smaller pieces,
what we're going to do is divide this into 6 pieces,
and this one into 6 pieces.
To make this one into 6 pieces,
we have to take each of these pieces,
and divide them into 3 parts.
Which now makes the apple into 1, 2, 3, 4, 5, 6 parts.
The equivalent of doing that down here on our fraction
is multiplying the top and bottom by 3 [1x3/2x3 ].
That makes this fraction into 3 over 6 [3/6].
And if we look at our picture,
notice now we have 1, 2, 3 pieces out of the 6 total.
Now let's look at this other apple over here.
To get 6 pieces over here,
we have to divide [ũ] each of these 3 pieces into 2 pieces.
The equivalent in our fraction down here
is multiplying the top and the bottom by 2 [1x2/3x2],
and that makes this fraction 2 over 6 [2/6].
And notice if we look at our picture,
we have 1, 2 of the 6 pieces,
which is the fraction 2 over 6 [2/6 ]
Now these two apples here have the same size shapes
for all of the pieces in them.
So if we take a look at the diagram up here
and make it match these ones,
we would take this piece here and divide into 3 [ũ 3],
and this one and divide it into 2 [ũ 2].
Now each of the pieces are one sixth [1/6] of the apple,
and if you count them we have
1
2
3
4
5 of them.
The equivalent down here with our fractions
is to add the 3 and the 2 [3+2/ ]
in the numerators of our fractions,
which gives us a 5 [= 5/].
Now the denominator down here
is a 6 on both of them [ 3/6+2/6 ].
It's going to stay a 6 for our answer,
because that 6 is representing the size of each of the pieces
in relation to the original object.
And each of the pieces stayed one sixth [1/6] of the apple
when we shoved them together into this one picture over here.
They didn't change into 12 pieces, and they stayed 6.
A lot of people want to add these and make it a 12.
But the shape of the pieces
when we added them together did not change.
So they're still one sixth of the apple [1/6 apple].
So what we end up with is 5 pieces
that are all one sixth [ 1/6 ] of the apple,
and if you look at your picture,
you've got 1, 2, 3, 4, 5 pieces that are all one sixth.
So this here represents how much you have at the end [5/6].
Now it's a lot of work
in order to draw these pictures every time
to see what's happening.
This example here was just to illustrate what's going on
when you're adding and subtracting fractions,
so that you know why you're doing the steps that you want to do.
So let's go ahead and take a look at
what are the steps for adding and subtracting fractions.
First of all, when you're subtracting
we're going to be applying, a definition of subtraction
where we change all our subtraction
into adding the opposite
and then thinking about positives and negative numbers.
And the video on positive and negative thinking
will explain that to you.
So in most cases we're going to be thinking about subtracting
as adding the opposite of the number.
A negative number's going to be money you owe to people,
and a positive number's going to be money that you have.
So subtracting is really going to end up
being the same as adding.
You'll see that happen when I do my first subtraction problem.
But our definition,
or, our steps are going to be the same for both of them.
The first thing you're going to do is get common denominators.
That was what we did
when we multiplied the top and bottom of each fraction
by values that made it so the denominators matched.
The second thing you're going to do is add,
I'm going to put in parentheses (or subtract)
because it's really the same thing;
we're going to be doing the same thing... the numerators.
The steps for adding and subtracting are really the same thing.
It's just whether you are going to be
using the operation for addition [+]
or using the operation for subtraction [ - ].
And it's really just these two steps
for adding and subtracting them.
Now, let's take a look at these two steps
with a different problem.
Let's say that we have four fifths [4/5 ]
and we're adding to it, two thirds [4/5+2/3 ].
The first step up here is to get common denominators.
That means we want these bottom numbers here to match.
Which was what we were doing
when we were changing them into 6's
so that the apples had the same number of pieces
in the previous example.
So we need to think to ourself,
okay, what's the smallest thing 5 and 3 go into?
And they both go into 15. [15 ũ 5, 15 ũ 3]
So we need to make these into equivalent fractions
that both have 15 in the denominator.
So think to yourself for this first fraction:
what could I multiply 5 by to make it a 15 [5 x _ =15]?
And I multiply 5 by 3 [5 x 3 = 15] to make that happen.
So I'm going to multiply the 5 by 3 [5 x 3].
But if I do that, and only that,
I'm changing the value of this fraction.
So what I do is I multiply the top by 3 also.
Because three over three is one. [ 3/3=1].
And this is the equivalent of multiplying by 1;
we're just multiplying by 1
so that it looks in a little different form
that gets us where we want.
And when you multiply by 1
it doesn't change the value of the number.
So 4 times 3 gives us 12 [4 x 3 = 12].
And 5 times 3 gives us 15 [5 x 3 = 15].
So four fifths is the same thing as twelve fifteenths [ 4/5 = 12/15].
Because we multiplied the 4 by 3 [4 x 3]
and we also multiplied the 5 by 3 [5 x 3],
which was the equivalent of multiplying by 1 [ 3/3 = 1].
Now let's take a look at our second fraction.
We need to make this 3 a 15.
To make the 3 a fifteen we multiply by a 5 [3 x 5 = 15].
So, I'm going to multiply the 3 by a 5 [3 x 5],
but I don't want to change the fraction,
so I also have to multiply the 2 by a 5 [2 x 5].
That's the equivalent of multiplying by 1
because 5 over 5 is 1 [5/5 = 1].
2 times 5 gives us 10 [2 x 5 = 10].
And 3 times 5 gives us 15 [3 x 5 = 15].
Now we have common denominators
because the bottom numbers here match.
So we're going to add our numerators,
which is the 12 and the 10, and that gives us 22 over 15.
[ 12/15 + 10/15 = 22/15]
Now we do want to see if we can reduce,
but since there's nothing that will divide evenly
into the 22 and the 15,
we can't reduce this fraction and we're done.
Now many of you are probably wanting
to switch it to a mixed number.
However, there's no need to do that when you're in Algebra,
because if you switch to a mixed number
and then you have to plug that number in
to an algebraic expression,
you'll have to switch back to an improper fraction anyways.
It is perfectly fine to leave your answers like this
in an improper fraction.
I would actually prefer this answer
rather than switching it to a mixed number.
Only switch to a mixed number if you're asked to do so.
If you're asked to reduce, that does not mean a mixed number.
That just means see if something divides evenly
into both the top and the bottom of the fraction.
Now let's look at a subtraction problem.
Let's say that you have five sixths [5/6],
and you're subtracting from it three fourths [5/6 - 3/4].
The first thing you want to do is get your common denominators.
So we're going to go ahead and take a look
at the denominators of 6 and 4. [/6 and /4]
The smallest value both of those go into is 12
[12 ũ 6 and 12 ũ4]
So we're going to turn them both into 12 [ /12 and /12].
Now you might have been thinking 24 [/24 ].
If you were thinking 24 that would be fine,
it's just not the smallest,
which means you would have to reduce your answer at the end.
We're going to use 12,
because 6 and 4 go into 12 [12 ũ 6 and 12 ũ4],
and 12 is smaller than 24.
24 would have been fine;
you just would have had to reduce at the end.
So now what we're going to do is try and turn the 6 into 12
[6 x ?=12]
To turn it into 12 we have to multiply by 2 [6 x 2 = 12].
So I'm going to multiply the top and bottom here by 2.
That gives me 10 over 12 [5 x 2/ 6 x 2 = 10/12].
And then on the second fraction here to turn the 4 into 12,
I have to multiply by 3.
So I'm going to multiply the top and the bottom here by 3.
That gives me 9 over 12 [ 3 x 3 / 4 x 3 = 9/12].
And now we're subtracting these.
What you want to think of is this subtraction here
is only going with the 9 up here.
So this is like 10 minus 9 [10-9],
which is actually the same thing as 10 plus a negative 9
[10 +(-9)].
You want to think of it as you have 10 dollars
and you are going to spend 9 dollars.
So you're going to have 1 left over.
Which gives you a 1 in the numerator
and the denominators stay the same;
we don't do anything with those.
So the answer here is one twelfth. [ 10/12 - 9/12 = 1/12]
And that is what you do when you add or subtract fractions.
Now multiplying fractions is slightly different.
We're going to do something different in multiplying fractions.
Multiplying fractions is going to require us to do something
that is similar to reducing fractions.
So let's take a look at reducing a fraction
and then we'll take a look at how to multiply
because there's going to be a very similar process
right at the beginning of our multiplication.
So let's say you have something like 25 over 15 [25/15].
If you were asked to reduce this,
you would be wanting to take a look
at the numerator and the denominator
to see if there's a common number that goes into both numbers.
And I notice for 25 and 15 that 5 divides into both of those.
[25 ũ 5 and 15 ũ 5]
so, what you would do is you would divide
the top and the bottom of this fraction by that same number.
So since 5 went into both of them
I would divide the top and the bottom by 5,
and that would leave me with 5 over 3 [25 ũ 5/15 ũ 5=5/3]
Now, another way to think of this is to take the 25 over 15 [25/15]
and split the 25 into its factors of 5 times 5 [25=5x5]
and 15 into its factors of 3 times 5 [15=3x5 ].
When you have a number divided by itself, you get 1.
So you can think of this 5 over this 5 here as just 1. [ 5/5=1]
These can cancel each other out.
And what's left here is 5 over 3 [25/15 = 5/3].
Notice we got the same thing here as we did here
And we did it a little bit different way each,
in both situations.
Both of them are fine,
whoever works for you is what you'll want to do.
So this process of reducing here
by dividing the top and bottom number by the same thing
is going to be kind of similar
to what we're going to be doing when we're multiplying.
We're going to be doing what's called "cross canceling."
And that is different than cross-multiplying
so don't get the two confused.
When you are multiplying fractions
the first thing that you want to do is cross cancel if possible.
You won't always be able to do cross canceling.
Cross canceling is like reducing the problem up front.
And it makes it much, much easier
to deal with your problem if you do this.
So it's like finding those common factors that you can cancel
like we showed in the previous problem up front,
so that when you get to the end
you don't have to look for them with much larger numbers.
Now let's say we have a problem that's 36 over 27
times 18 over 24 [36/27 x 18/24 ].
What you want to do to cross cancel
is take a look at all the values in the tops of the fractions,
or in the numerators,
and all the values in the bottoms of the fractions,
or the denominators,
and you want to strategically pick
something in the top of either fraction
and match it up with something in the bottom of either fraction.
But they need to be two values that have a common factor,
which means something divides into both of them.
So, for instance, if I look at the 18 and the 27,
I notice 9 divides evenly into both of those.
So what I'm going to do, I'm going to kind of
reduce up front by dividing 9 into both of those,
and getting smaller numbers to deal with.
Now I'm going to color code these
so you can see which ones had common factors that we divided.
Because what we're going to do is we're going
to divide the 9 into the 27, cross it out,
and write the answer of 3 down right next to it.
And then I'm going to divide the 9
into the 18 that I paired up with 27
and I'm going to write the answer down [36/ 3 x 18/24 ]
after doing 18 divided by 9, which is 2, right next to it.
The 3 and the 2 are what's left in my problem [36/ 3 x 2/24 ].
I've reduced the 18 and the 27 down to those values.
Now let's look at what's left.
If we look at what's left, [36/ 3 x 2/24 ].
I notice that 12 goes into 36 and also 24.
So I'm going to divide 12 into 36, which gives me 3 [36ũ12=3],
and write the answer down. [3/ 3 x 2/24 ]
I'm also going to divide 12 into 24, which is 2 [24ũ12=2],
and write the answer down [3/ 3 x 2/2 ].
Now I'm going to look at what's left [3/ 3 x 2/2 ].
Now, keep in mind we cross canceled diagonally here,
and diagonally here.
When you cross cancel,
you can either cross cancel diagonally, or up and down,
as long as you match
something in the denominator with something in the numerator.
So I notice right here that 3 goes into 3 [3ũ3],
and it also goes into this 3 up here,
so I can actually cross cancel those two
by dividing them both by 3.
And 3 goes into 3 once, and this other 3 once also [ 3/3=1].
Now that's much smaller numbers than what we started out with,
with the 36 and the 27.
Let's take a look over here and,
and notice that the 2's can cancel.
So I can go ahead and divide 2 into 2 here and get 1 [2ũ2=1],
and 2 into 2 here and get 1 [2ũ2=1].
Now it's going to be much easier to multiply this problem.
I'm going to have 1 times 1, which is 1 [1x1=1],
and 1 times 1 which 1 [1x1=1] in the denominator.
Notice what I just did is step 2,
which is to multiply straight across.
If you cross canceled everything that was possible
up here in step number 1
the answer when you multiply straight across will be reduced.
However we don't normally write things as 1 over 1.
We actually just write them as 1,
when you get that for your answer [ 1/1=1].
So you would want to write that as just =1.
Let's take a look at one more multiplying
and then we'll go on to divide.
Let's say that our problem was 25 over 24 times 32 over 35
[25/24 x 32/35 ].
Now, we're going to pair up values that have common factors.
I notice right away that 25 and 35 are both divisible by 5.
[ 25ũ5, and 35ũ5] So let's go ahead and divide those.
I'm going to divide 5 into 25 to get 5, [25ũ5=5]
and 5 into 35 to get 7. [35ũ5=7].
So now I'm left with the 5 and the 7 [5/24 x 32/7].
Now I notice that the 24 and the 32
are both divisible by 8 [24ũ8 and 32ũ8].
So let's divide 8 into 24 and get 3, [24ũ8=3]
and 8 into 32 and we get 4 [32ũ8=4].
Now I look at the values that are left [5/3 x 4/7].
5 and 4 are in the numerator, and 3 and 7 are in the denominator.
Nothing divides into any of those four numbers evenly.
So nothing goes into a 5 or a 4 and also goes into a 3 or a 7.
So now we can do step 2 and multiply straight across.
5 times 4 is 20 [5x4=20].
And 3 times 7 is 21 [3x7=21].
That was much easier to do than to multiply 25 times 32 [25x32]
and reduce down to a 20,
and 24 times 35 [24x35] and reduce down to the 21.
So the cross canceling makes it much,
much easier to do your problems
by giving you smaller numbers to deal with when you multiply.
Now let's take a look at dividing.
If you can multiply, you can divide
because the first step that you want to do when you divide
is to take the reciprocal of the divisor
and switch to multiply.
And then, you just multiply.
Which we just learned how to do.
So if you know how to multiply you basically can divide.
Now let's look at what
taking the 'reciprocal of the divisor' means.
Reciprocal means that you flip the fraction over.
So, basically, the divisor is the second fraction,
the fraction you're dividing by.
So you take the fraction you're diving by,
which is always the second fraction.
You flip it over and you switch to multiply.
So let's take a look at a problem like that.
Let's say we have 49 over 25 [49/25]
and we're dividing by 14 over 35 [49/25 ũ 14/35].
The fraction we're dividing by is the second fraction here [14/35].
So we're going to take this second one
and flip it over [35/14] and switch this operation to multiply.
The first fraction stays the same [49/25].
The second one flips over [35/14].
And instead of divide we now have multiply [49/25 x 35/14].
And since we already know how to multiply
we just look for cross canceling.
And notice that I can pair up the 25 with the 35
by dividing them both by 5 [25ũ5 and 35ũ5].
5 goes into 25 fives times, [25ũ5=5]
and 35 seven [35ũ5=7].
Now I notice that I can pair the 49 here up with the 14.
7 goes into both of those [49ũ7 and 14ũ7].
So I'm going to divide 7 into 49 to get 7 [49ũ7=7].
And 7 into 14 to get 2 [14ũ7=2 ]. [7/5 x 7/2]
Now many of you may want to pair up these two 7's here,
but both of them are in the numerators.
You have to have one of them
in the bottom of a fraction, or the denominator,
and one of them in the top, or the numerator.
They can't both be on the same level.
One has to be up here,
and one has to be down here to match them up.
So we have canceled everything that we can cancel.
So now we're going to multiply straight across,
and 7 times 7 is 49 [7 x 7 =49].
And 5 times 2 is 10 [5 x 2 =10].
And remember, improper fractions are perfectly fine
so our answer is 49 over 10 [ = 49/10 ].