This is Common Core State Standards Support Video for Mathematics,
This standard reads:
Apply and extend previous understandings of multiplication
to multiply a fraction or a whole number by a fraction.
Interpret the product
a/b times q
as a parts of a partition of q into b equal parts;
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
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
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.
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
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
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.
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
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,
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
So here's something that you can do to make it a lot simpler. You can make some
to find the solutions. So
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.
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
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
So 2/3 of 3/4
would be right here.
Again how do we get our solution? Well we can do it just by
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
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
Again that was one of the main reasons, really the critical reason
to make sure that your manipulatives, these
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
than 1? We've covered one; we've covered
the situation where q was a fraction, but we haven't covered this one
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
for the total.
Now we need some quantity a of those equal parts,
and let's let that be 5.
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,
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
but if we simplify the 15/12 to 5/4, okay I can 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.
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
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,
times 3/4 being 6/12, and 5/12 times 3 being
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
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.
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
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
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
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
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
that they're asking for.
Well let's try this.
What would happen if we take what the standards says but use the
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.
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
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
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
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.
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
5 times 3 all over 12. To go back and review, this is what
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
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
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
So that gives us our total of 15, and connecting back to what we had
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
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.