Mathematics

5.NF.4a Transcript

This is Common Core State Standards Support Video in Mathematics Standard 5.NF.4a. This standard reads: Apply and extend previous understandings of multiplication to multiply a fraction or whole number by a fraction. Part a states: Interpret the product (a/b) × q as a parts of a partition of q into b equal parts; equivalently, as the result of a sequence of operations a × q ÷ b.

Now this isn't a very long standard compared to some others, but there is 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 are a lot more under the surface to this than you would think at first. Let's take the first part, this whole idea of this product a/b x q being parts of a partition of q into b equal parts. Again it's a little bit confusing, because of the variables. Let's start with our quantity of q, and let's take the easiest possible scenario. Let's have q be 1, and now we want to take our 1 and partition it into equal parts. Make the b a 4. So we cut it up into four equal parts, and we have some quantity a of those equal parts. Let a be 3, so basically what we have here from our a/b x q is 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 x 1. A lot of time we'd think of it as 3/4 of 1 in plain English. Again, this is the simplest scenario, because the whole was 1.

Let's try a different scenario. Let's say the q itself was a proper fraction, in other words, q is a fraction between 0 and 1. Let's take this example where we want 2/5 x 2/3 or in plain English 2/5 of 2/3. we first need to partition the 2/3 into 5 equal parts. Again the 2/3 is the q and 5 is the b. 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. Our task is, well we need two out of those parts. What we need is 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 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. We have a total of 15 parts that we cut into, and we only want 1, 2, 3, 4 of these smaller rectangles—four parts. Our solution is 4/15.

It might be difficult for students to do all those subdivisions especially trying to get them into equal parts. So here's something that you can do to make it a lot simpler. You can make some fraction manipulatives. You know how to find the solutions, so take some clear plastic sheets and you want to make congruent squares. That's important. In other words, all your squares need to be the same size. Then of course you need to have one for 1/2, you need another set for your thirds. You have your 1/3 and 2/3, you have a set for your fourths, 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 tenths. Let's see how this works. Let's say we wanted 2/3 x 3/4. This means that our q is 3/4. We're starting off with 3/4, then we find our other manipulative that's 2/3. So now we need to take the 2/3 and rotate it around to where this one is horizontal compared to the original. The other one being vertical. We need to just slide it over and put one over the other. Notice the result now is this situation, and so our solution is going to be where the shaded areas overlap. Notice again here was our 3/4, but we only want 2/3s of which is indicated this way. So 2/3 of 3/4 would be right here.

Again how do we get our solution? Well again we can do it just by counting. We had let's see 1, 2, 3, 4 and 1, 2, 3, that's 12. So we have 12 total parts, and our solution, the common shaded areas 1, 2, 3, 4, 5, 6. Our solution is 6/12. One thing that's important to notice here, note that in this process when we put one manipulative over the other we now have 1 square manipulative. Notice that now we just have one as our whole and we really have 6/12 of 1. Again that was one of the main reasons, a really critical reason to make sure that your manipulatives, these sheets with the shaded areas, are congruent squares and are the same size squares so that when we rotate one it'll fit exactly over the other. When you do that, when one is over the other, your whole is now 1 because you just have one square.

If the students keep doing these types of problems over and over, they're going to start seeing a pattern. Like this one 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. In this other example, 2 x 3 is 6, and on the bottom is the denominator 3 x 4 =12. 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. There's one scenario that we haven't covered yet. What happens if your q is something bigger than 1? 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 into 12 parts. That's our b.

So we've done that here, but notice that we have to do it by cutting up each individual circle into four parts to get 12 for the total. Now we need some quantity of those equal parts. Let that be five. So here's our situation. We had q as 3. We cut that up into 12 parts, and we wanted five of them, so we have 5/12 x 3, and we've already figured out the standard algorithm. So we convert that 3 to be 3/1 to make it a fit. 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 I can see 1, 2, 3, 4, 5/4 over here. Let's check out what happened. We started off with 5/12 x 3— 5/12 of 3, and the whole was our set of 3. Then we multiplied it and 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 idea that anything can be expressed as something times 1. Now the situation has changed to where when we are dealing with our product, our solution the whole is 1. It's not 3 anymore, so let's review this one more time. We started off with 5/12 x 3, and our whole was a set of 3. When we got our solution 5/4, that's really 5/4 x 1. That changed to where our whole for that is 1. That 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. 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 x 3/4 being 6/12 and 5/12 x 3 being 15/12. But again we ended up with a whole of 1. 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 that's how much you have—2/3 of 3/4 of a gallon. But that doesn't make sense, it's hard to see that. But if we convert it, we would do the multiplication. That makes it to where I can visualize it, with one gallon as the whole. So I would have 6/12 of a gallon, which of course is 1/2 of a gallon. That's what happens when converting this expression to something with respect to one being a whole.

Now let's look at another scenario where our q is something bigger than 1. Let q be a 4, and let's say we wanted 2/3 of that. We have 2/3 x 4. Of course we know it is 8/3. Knowing our standard algorithm is important. 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 the quantity of 4 and we need to partition that 4 into three equal parts. Now we have a little bit of a problem. It's very difficult to look at this and say well there are 4 rectangles here. But where in the world do I draw my lines or whatever to split this up into three equal parts? I don't see it. Well there are three equal parts, and what would happen if we take each of those and cut them up into three equal parts? Maybe now there is something we can deal with here. Now I've got 12 of those smaller parts, and I want to split them up three ways. Of course, 12 ÷ 3 is 4. 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. Here is the first of the three parts, there's the second of the three parts, and here is the third of the three parts. Then of course now with the shaded drawing I can figure out my solution. I have a whole one here, another whole one there, and 2/3 of another here because this is where I had to stop right here. So I've got two 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.

Let's try this. What would happen if we were to take what the standards says? Take the commutative property and reverse it and look at in terms of q times x a/b. We take the example that we were just working with, the 2/3 x 4. Let's reverse it, let's make it 4 x 2/3. 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 about 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, and to get the solution, really all we have to do is a little bit of a 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. Viola I can get my solution. I've got a whole one here, a whole one there, and 2/3 of another. So I’ve got two 2/3 or 8/3. Notice that I don't need this one anymore, so I'll just "x" that out. What about the other example that we did, the 5/12 x 3 with the circles?

What happens when we reverse that and think of it as three 5/12? Well here's what it look like visually. I got one 5/12, another 5/12 here, and another 5/12 over here. Just like we did with the rectangles, it just takes a little bit of imaginative moving. So we take a 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. There we have it. We have a whole one here and 3/12 of another one. We get our one 3/12, which would be 1, 2, 3, and 12 and 3 more that's 15/12.

Again we don't need this one anymore. We can't neglect this last part of the standard, the equivalency as a result of operations a x q ÷ b and a/b x q. What happens here as a result of operations a x q 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 where circles 5/12 x 3 = 5 x 3 over 12. Okay to go back and review this is what happened. We start off with 3 and we have to split it 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 x 3 over 12. Now if we do the standard algorithm, we get 15 over 12. Now that’s 15 over 12, you know 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 is a over b is 1. 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.

Let's go back and look at our standard and redo this side to fit what the standard says as far as interpretation. So now, this says that I've got a whole of one that must be partitioned into 12 parts. So we do that. We've partitioned it, and now we need 15 of those parts. So oh wait a minute. There is only 12 that I can get out of that. So I need another circle, and so there's three more. 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 as we try 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.