This is Common Core State Standards Support Video for Mathematics,

standard 5.NF.4a.

This standard reads:

Apply and extend previous understandings of multiplication

to multiply a fraction or a whole number by a fraction.

Part a:

Interpret the product

a/b times q

as a parts of a partition of q into b equal parts;

equivalently,

as the result of operations a times q divided by b. Now this isn't a very long

standard compared to some others,

but there's a whole lot to it

First of all it's about fractions.

There are a whole lot of variables in here,

and the reality is there's a lot more under the surface to this

than you would think at first.

So let's take the first part, this whole idea of

this product a/b times q

being

a parts

of a partition of q

into b equal parts.

Again it's a little bit confusing, because of the variables.

Let's start off with our quantity of q, and let's take the easiest possible scenario.

Let's let q be 1.

So now we want to take

our 1

and partition it into b equal parts. Let's let the b be 4,

so we cut it up into four equal parts.

Now we have some quantity a of those equal parts.

Let's let a be 3.

So basically

what we have here from our a/b times q is,

again we started off with 1.

We split it up into 4 parts

and we wanted 3 of them. So we end up with 3/4 times 1,

which a lot of times we think of as

3/4

of 1 in plain English.

Again this is the simplest scenario because the whole was 1.

Now let's try a different scenario.

What if the q itself was a proper fraction?

In other words, q is a fraction between 0 and 1. So let's take this example.

We want 2/5 times 2/3,

or in plain English 2/5 of 2/3.

So we first need to partition the 2/3 into 5 equal parts. Again

the 2/3 is the q

and

5 is the b.

So let's look at this visually so it can make a little bit more sense.

So we're starting off with 2/3,

which is this here.

But we need to split that up into 5 equal parts.

So let's do that.

So there's

1/5,

2/5, 3/5,

4/5, which also takes care of the 5/5.

Now our task is,

well we need two

out of those parts.

What we need to do then, well here's

one of those

parts, here is the second of those parts,

the third, and so forth. But we just need two of those.

So here they are.

This is one

one part,

and this is another part.

Now how do we determine the solution?

Well in a scenario like this, we can actually just get the answer by counting.

Notice that we extended these blue lines all the way over

to make a little bit more sense out of it. We can tell now that we have 1,2

3 this way, and 1, 2, 3, 4, 5 this way.

So we have a total of 15 parts that we cut it into,

and

we only want 1, 2, 3, 4 of these smaller rectangles—four parts.

So our solution is 4/15.

Now it might be difficult for students to do

all those subdivisions especially trying to get them into equal parts and

everything.

So here's something that you can do to make it a lot simpler. You can make some

fraction manipulatives

to find the solutions. So

take

some clear plastic, some clear sheets

and what you want to do is make congruent squares. That's important.

In other words, all you squares need to be the same size.

So then of course you need to have one for 1/2. You need another set for your thirds

so you would have your 1/3 and 2/3.

You have a set for your fourths

so you would have your 1/4, 2/4, and 3/4,

and of course you would continue on

and make the rest for your

fifths, your sixths, your sevenths, and so forth,

probably up to about the tenths.

So let's see how this works.

Let's say we wanted 2/3 times 3/4.

So this means that our q is 3/4, so we're starting off with 3/4.

Then

we find

our other manipulative that is 2/3.

So now what we need to do is take

the 2/3 and rotate it around to where this one is horizontal

compared to the original, the other one being vertical.

Then what we need to do is

just slide it over

and just put one over the other. So notice

the result now is

this situation,

so our solution is going to be

where the shaded areas overlap.

Notice again here was our 3/4, but we only want 2/3 of that, which is

indicated

this way.

So 2/3 of 3/4

would be right here.

Again how do we get our solution? Well we can do it just by

counting.

We had let's see, 1, 2, 3, 4; 1, 2, 3; that's 12.

So we have 12 total parts

and our solution, the common shaded area, is 1, 2, 3, 4, 5, 6.

So our solution is 6/12. Now one thing that's important to notice

here.

Note that in this process, when we put one manipulative over the other

we now just have 1 square manipulative.

So notice that now we just have 1

as our whole. We really have 6/12 of

1.

Again that was one of the main reasons, really the critical reason

to make sure that your manipulatives, these

sheets

with the shaded areas

are congruent squares. They're the same size squares so that when you

rotate one it'll fit exactly over the other.

When you do that ,when one is over the other,

again your whole is now 1 because you just have one square.

Now if the students keep doing these types of

problems over and over, they're going to start seeing a pattern.

They're going to start seeing the pattern that,

well like for this one,

it looks like we multiplied the twos to get the 4,

and in the

denominator it looks like we multiplied the

5 and the 3 to get the 15 because the same thing worked here on this other

example, 2 times 3 is 6,

and on the bottom, on the denominator 3 times 4 is 12.

So then they're going to see,

in general that they multiply the numerators and they multiply the

denominators to get their solution,

which of course is your standard algorithm for multiplying two fractions.

So they'll have that down

by using these manipulatives and by figuring it out on their own.

Now there's one scenario that we haven't covered yet.

What happens if your q

is something

bigger

than 1? We've covered one; we've covered

the situation where q was a fraction, but we haven't covered this one

yet.

If q is something bigger than 1, it might be easier to think of q as a

set instead of a whole. It's

easier to think of a set of 3

instead of a whole of 3, but

they are synonymous though.

Your set here is your whole, so 3 is the whole.

We began with this quantity q which was 3,

and let's say we need to cut that

quantity of 3 up into 12 parts. That's our b,

so we've done that here.

But notice that we had to do it by cutting up each individual circle into

four parts to get

12

for the total.

Now we need some quantity a of those equal parts,

and let's let that be 5.

So here's

our situation. We had

q was 3.

We cut that up into 12 parts,

and we wanted five of them. So we have 5/12 times 3,

and

we've already figured out the standard algorithm.

So we convert that 3 to be 3/1 to make it fit,

and so we multiply the numerators, multiply the denominators and we get

15/12. There's a little bit of a problem here.

I don't see

15 anywhere,

but if we simplify the 15/12 to 5/4, okay I can see

that.

I see

1, 2, 3, 4, 5 fourths (5/4) over here.

Now let's check out what happened. We started off with 5/12 times 3,

5/12 of 3,

and the whole was our set of 3.

But then

we multiplied it. We got 15/12,

which was equivalent to 5/4.

Here is the understanding. Here is the heart of the matter,

the heart and soul of this—

this idea of anything

can be expressed as something times 1. Now the situation has changed to where

when we're dealing with our product, our solution,

the whole is 1.

It's not 3 any more. So let's review this one more time. We started off with

5/12 times 3

and our whole was a

set of 3.

Then when we got our solution

5/4,

that's really 5/4 times 1. So that changed to where our whole

for that is 1,

which would look

something like this.

But looking at it this way, our answer of 5/4 makes sense because again here's

4 of them, there's 1 more, so there's 5/4.

So again, what happened was

the whole for the resulting product is 1,

not what you started off with.

So if we look at the examples that we've done,

2/3

times 3/4 being 6/12, and 5/12 times 3 being

15/12.

But again we ended up with

a whole of 1.

So in essence, multiplication that involves a fraction results in changing

the original expression or context

to one where the resulting product will be an expression where the whole is

1.

In other words, let's say that you had 2/3 of 3/4 of a gallon.

A student could say well,

that's how much you have. You have 2/3 of 3/4 of a gallon.

But that doesn't

make sense, at least it's hard

to see that.

But if we convert it, we would do the multiplication. Now it makes it to where

I can visualize it

with 1 gallon as the whole. So I would have 6/12 of a gallon, which of course is

1/2 of a gallon.

So that's what happens. You're converting this expression

to something with respect to 1 being the whole.

Now let's look at another scenario

where our q is something bigger than 1.

Let's let

q be a 4,

and let's say we wanted 2/3 of that. So we have 2/3 times 4,

which of course we know is 8/3 knowing our standard algorithm.

But there's a lot more to this.

So let's look at our standard and follow what it says. So we have to begin with

a quantity of 4,

and we need to partition

that

4 into 3 equal parts.

Now we have a little bit of a problem.

It's very difficult to look at this and say well, there's 4

rectangles here,

but where in the world

do I draw my lines or whatever

to split this up into 3 equal parts?

I don't see it.

Well there's 3 equal parts.

What would happen if we take

each of those and cut them up into 3 equal parts?

Maybe now there is something we can deal with here.

So now I've got 12 of those smaller parts,

and I want to split them up

three ways,

and of course 12 divided by 3 is 4.

So now I can see that

what I can do then is

section this off in chunks of four.

All right, so

that's where I would draw my lines to split it up into three parts.

If we were to do some shading,

this is what it would look like

where here's

the first of the 3 parts,

there's the second of the the 3 parts

and here is the third of the 3 parts.

Then of course now with this,

the shaded drawing, I can figure out my solution. I have a whole

1 here, another whole 1 there and 2/3 of another here, because this is where I had

to stop, right here.

So I've got 2 2/3 or 8/3 like we figured out earlier

with the standard algorithm.

But we still have this problem of

what happens when we have a set and it's very difficult to split it up into

the parts

that they're asking for.

Well let's try this.

What would happen if we take what the standards says but use the

commutative property

and reverse it?

Look at in terms of q times a/b.

So we take the example that we were just working with, the 2/3 times 4.

Let's reverse it. Let's make it 4 times 2/3.

Now keep in mind that for the 2/3 the whole is 1. It's just a regular

2/3 like we are used to thinking of it.

So

this would be one 2/3,

but in this scenario we've got four of them.

So there's our four, so we have four 2/3.

To get the solution,

really all we have to do is just do a little bit of

imaginative moving.

Just slide this over here, take my purple parts and move them over here,

my salmon-colored and move them over here,

and then my last 2/3 and move it over here,

and viola, I can get my solution. I've got a whole 1 here, a whole 1 there, and

2/3 of another. So I've got 2 2/3 or 8/3,

and notice that I don't need this 1 any more. So I'll just "x" that out.

What about the other example that we did, the 5/12 time 3 with the

circles?

What happens when we reverse that

and think of it as three 5/12?

Well here's what it looks like visually.

I've got one 5/12 here, another 5/12 here, and another

5/12 over here.

And just like we did with the rectangles, it just takes a little bit of

imaginative moving.

So we take this 5/12

and move it here,

take the blue-shaded 5/12 and move them over here,

and it's a little tougher with this last 5/12, but it's okay. We take it and

we move two of them here and the other three there.

So there we have it. We have a whole

1 here and 3/12 of another 1,

and we get our 1 3/12, which would be 1, 2, 3, there's

12, and 3 more. That's 15/12,

and again we don't need this one any more.

We can't neglect this last part of the standard,

the equivalently as the result of operations a times q divided by b.

So

a/b times q;

now what happens here as a result of operations

a times q all over b? Notice that we just made this into a fraction bar

instead of the division sign to make it equivalent to what we're dealing

with as far as the representation.

So what happens here, what's the difference?

Well let's take that example we just worked with, the one with a circles,

5/12 times 3

equals

5 times 3 all over 12. To go back and review, this is what

happened.

We start off with 3

and we have to split it up into 12 equal parts. So we've done that,

and then we want five of those parts.

So here's 1, 2, 3, 4, so there's

five of those parts out of that set of three.

And of course this diagram applies to the left side of the equation. So now let's

look at the right-hand side of the equation, the original equation,

5 times 3 all over 12.

Now if we do the standard algorithm

we get 15/12.

Now that 15/12, the numerator indicates how many

of this

bottom number that we have. So it indicates that we have 15 of these.

The problem I'm having is that

where is the 15?

I don't see it, plus

unless otherwise indicated, the whole for any fraction a/b is 1,

and I'm dealing with three circles here not one. So

there's some confusion in the problem here.

So this diagram doesn't

fit here, it doesn't work.

So let's go back and look at our standard

and let's redo this side

to fit what the standard says as far as the interpretation. So now

this says

that I've got a whole of 1

that must be partitioned into 12 parts.

So we do that.

We've partitioned it, and now we need 15 of those parts.

So, wait a minute. There are only 12 that I can get out of that,

so I need another circle and so there's

three more.

So that gives us our total of 15, and connecting back to what we had

originally,

again we don't need this third circle, because we don't have a set of three

anymore. We have a set of one. So even though this was a short standard as far

the length,

there was a lot to it. There was a lot under the surface, but hopefully this

will clarify this.

We tried to dig a little bit deeper and go way beyond just the computation,

because the students need that,

because again they need to really understand what's going on

with the models

for this standard.