This is Common Core State Standards Support Video for Mathematics; the standard is 5.NF.B.4b. This standard reads: Apply and extend previous understandings of multiplication to multiply a fraction or whole number by a fraction. The particular standard that we’re addressing today is Part B: Find the area of a rectangle with fractional side lengths by tiling it with unit squares of the appropriate unit fraction side lengths, and show that the area is the same as would be found by multiplying the side lengths. Multiply fractional side lengths to find areas of rectangles, and represent fraction products as rectangular areas.
Well first, let’s look at some of the other standards that this is connected to. If you look at standard 5.NF.B.6, it has a connection to this one: Solve real-world problems involving multiplication of fractions and mixed numbers by using visual fraction models or equations to represent the problem. So that’s what we’re doing with 5.NF.B.4b. We are going to be using models to represent what’s happening. Standard 5.NF.B.4b is also related to standard 5.NF.A.1 which states: Add and subtract fractions with unlike denominators including mixed numbers by replacing given fractions with equivalent fractions in such a way as to produce an equivalent sum or difference of fractions with like denominators. So we will be doing that today because we will have to be combining fractions that do have unlike denominators.
Let’s address the first part of the standard: Find the area of a rectangle with fractional side lengths by tiling it with unit squares of the appropriate unit fraction side lengths. So let’s take something fairly simple. Let’s go with 3 1/2 by 2 1/2. Now it’s a good idea to actually have a model that’s larger than 3 1/2 times 2 1/2. Let’s round them up to the next whole numbers. Let’s make it 4 by 3 so that students can see exactly where the fractions come from. So, just by looking at this visually, students can determine what the fractions actually are as well, of course, as the unit squares. So, all students have to do now is just count them up. Okay, so here we have six. So we have that so far, and we have a couple of halves here. So that’s a one. We have a couple more halves here that will make another one. So let’s combine all our whole numbers. So far that’s eight. Now we have the 1/2, so that just makes it 8 1/2. What we have left now is the 1/4, which we combine with the 8 1/2, so we get 8 3/4. So what we really have here is 8 unit squares, eight 1 by 1 squares, and 3/4 of another 1 by 1 square.
Let’s address the next piece of the standard: Show that the area is the same as would be found by multiplying the side lengths. Okay, let’s take 3 1/2 times to 2 1/2, but let’s expand it out. Let’s use our distributive property so that we can see exactly what originates from what based on the partial products. First, we multiply 3 times 2, and if you connect it back to the model, here’s three sets of two, which would give us six. So that takes care of that so far. Next we have to multiply 3 times 1/2, and if we connect that back to our model—oh, there’s 3 halves right there, which would be 1 1/2. Now we have to multiply 1/2 times 2, but it might be better to look at this as 2 halves. That way, when we look at our model—ah, here’s the two 1/2s. So that gives us one. Last but not least, we have to multiply 1/2 times 1/2, and if we connect that to our model, there’s our 1/4. All we have to do now is combine everything. Let’s take our whole numbers like we did a while ago with just the tiling. So that gives us 8. Then we have this 1/2, and all we have left is the 1/4, which we combine with the 8 1/2 to get 8 3/4 like we did earlier. Again, that’s actually 8 unit squares plus 3/4 of another one.
Now let’s use that same model to address the last piece of this standard: Multiply fractional side lengths to find areas of rectangles, and represent fraction products as rectangular areas. Okay, so we’ll start off with 3 1/2 times 2 1/2. If we use our standard algorithm, we convert those to improper fractions. Then we multiply our numerators, multiply our denominators. We get 35/4. But what does this 35/4 really mean? We need to make sense of it and connect it back to our model. Let’s look at 7/2. What’s really happening here is that you have 7 halves. So each one of these lengths like here, that’s a distance of 1/2, and we have seven of those. Likewise, when we look at the other dimension, our 5/2, that’s 5 lengths of 1/2 each. So we multiply 7 times 5. We get 35, but exactly what is that 35? What does it really mean? And the same thing when we multiply 2 times 2 to get 4.
Well let’s start with the 4. Let’s start with the denominator. If we look at our unit square at the bottom, it’s 1 by 1, be it 1 centimeter by 1 centimeter or 1 foot by 1 foot, whatever the dimensions are. Notice that we have 1/4 of it shaded. So we have an area of 1/4. Okay, now it starts to make a little bit more sense. Each one of those really small squares is 1/4 of a 1 by 1 square. So if you count these up, we do have 35 one-fourths. So what the 35 really means—we have 35 area units, and each of those is 1/4 the area of our unit square, which would be of course, 1 by 1. So in this context, multiplication is a process by which the linear units get converted to area units. Let’s take a closer look. Again, let’s say we have 1/2 times 1/2. So here’s 1/2 by 1/2, and when we multiply that together, we get 1/4. That’s 1/4 of our unit square, so should we call it square units instead of just area units?
Let’s take another example, 1/3 times 1/4. So, take our unit square, divide it up into 4 equal lengths one way and then by 3 equal lengths the other dimension. So each one of these is 1/3 by 1/4, which is 1/12 when we multiply it. So what we really have here is 1/12 of a 1 by 1 unit square. Now notice, wait a minute, 1/3 by 1/4—that’s the dimensions of that smaller rectangle, and that’s not a square. So we have to be careful. I think it’s better to use the phrase area unit as opposed to square unit because students might get confused, because again, these units here are not each squares. That’s going to happen. We used the first example; it did turn out to be perfect little squares because they were all 1/2 by 1/2. But that’s not always going to happen. You’re going to have situations like this where you might have something like 1/3 by 1/4.
So let’s look at another example, 2 3/5 times 1 2/3. Again, let’s start with a larger model. Let’s start and make it 3 by 2. So looking at this dimension, we have one, two, and then our 3/5. Then looking at the other dimension, we have 1 2/3. So vertically, we split this up into thirds. This way, we split it up into fifths, and of course, remember that we’re only dealing with this area in here. That’s going to be our product. We eliminate all of this out here, but we need this diagram so students can again, understand where all these fractions come from.
So let’s walk through our model and see what actually happens. If we take our 2 3/5 times 1 2/3, and let’s use our distributive property and change it to 2 plus 3/5 times 1 plus 2/3. We start off and multiply 2 times 1, and so here are 2 ones, so we have 2 unit squares. So we have that so far. Then we have to multiply 2 times 2/3. Now these dimensions here are one-thirds, and if we look at the model and try to connect it back to what’s happening with the multiplication, here’s 2/3 here and here’s 2/3 there. So there are our 2 two-thirds. Let’s go ahead and shade them in a different color so students can tell what’s what. Now we continue on. We have to multiply 3/5 times 1, and we connect that to our model. Here’s 3 one-fifths right here. So we have 3/5 of our 1 by 1 unit square. Shade those in, again in a different color.
Then the last partial product, 3/5 times 2/3—we look at the bottom right. Here’s our 1 by 1 unit square, and it’s split up in five equal pieces one way and three equal pieces the other for a total of 15. So each one of these smaller rectangles is 1/15 of our unit square. There’s going to be 15 of those. So we have 6/15 for our product here for the last multiplication. So now we go ahead and multiply it out, although we’ve already kind of informally done that with our physical model. Two times 1, okay, there’s our 2, and then 2 times 2/3 in simplified form as a mixed number would be 1 1/3—3/5 times 1 and then 3/5 times 2/3 is 6/15. We combine those whole numbers to be a 3. So we’ve taken care of that, and it connected back to our model.
Now let’s go ahead and work it using this method. We get a common denominator so that we can combine these so that we have fifteenths, 5/15 plus 9/15. Okay, now our 6/15, that’s our last partial product. That’s represented physically down here, and the 1/3 here we had converted to 5/15. So we convert our model to match what we have with fifteenths. Then this is our 3/5 times 1, which again was still 3/5. We converted that to 9/15, so there they are. So when we combine all the fifteenths, we have 20 of them. Convert our 20/15 to a mixed number, so it’s 1 5/15. Now let’s connect this to the physical representation. This 1 here, what we can do is take for example, these 3/15 and move them here and these 3/15 and move them here. And that is a whole 1, so we can combine the 3 and the 1, we have 4. Now all that’s left is our 5/15 there, which of course, simplifies to the original 1/3. So our final solution would be 4 1/3 in simplest form.
Let’s retrace our steps and maybe take a slightly different approach this time. Let’s take our 6/15, and this time, let’s simplify that to 2/5. Now we can combine 3/5 and 2/5. That’s a nice easy addition because that’s just going to be a whole of 1. A way to connect this slightly differently to our physical model—we can take this 1/15 and move it over here, and take this 1/15 and move that down there, and look what we have. So this 3/15 is 1/5. This 3/15 is 1/5, and when we slide them up in here, that gives us a whole of 1, which corresponds to this 1 here. We can now connect all of our whole numbers. We have one, two here, so that takes care of that. We have these 3 one-thirds here, which takes care of another whole number, another 1. Then we already saw that these fifteenths and these fifths can be combined to be another whole. So there’s a whole of 4. So everything is accounted for except for the 1/3, which of course is right there. So now we made a connection from our abstract, from our multiplication over here to the actual physical model here. So our final solution is 4 1/3, and again, we’ve made connections so that students can see exactly what is represented where and where all these different pieces and all these different fractions come from.
Let’s take a slightly different approach, and again focusing on this last part of the standard that deals with multiplying fractional side lengths to find areas of rectangles and represent those fraction products as fractional areas. This time, we’re going to use our standard algorithm. Change the 2 3/5 and the 1 2/3 to improper fractions, do our multiplication, and we get 65/15. We’ve already had a little bit of experience with this, so we have a better notion of what represents what. So the 65 is actually 65 of these 1/15. We have 65 fifteenths. Each one of those 1/15 is actually an area amount, and again there are 65 of those in the total diagram. Again, keep in mind that the dimension of each one of these is 1/3 by 1/5. Again, the temptation is going to be to say 65 square units, and each of those is 1/15 of a unit square. But again, they’re not really little squares. They’re actually just little rectangles. So again, we have to exercise caution when we use that phrase square units. It might be a little bit more accurate to say 65 area units, and again, each one of those is 1/15 of a unit square.
When we simplify 65/15, we get 4 1/3, which is the exact same product that we got earlier when we used the distributive property. But let’s connect this to this diagram to see where the different pieces come from. If we were to take this section here, these 5/15, and move them over here, we’ll make a unit square. And if we take these fifteenths here and move them here and here, there are six of them that we could fit in these six open spaces. So we’ll have a unit square here. So now here’s what we’ve created. We have our 1, 2, 3, 4 unit squares and then we have 1/3 of another unit square. So we have a total of 4 1/3 square units. And to be absolutely sure again that students understand what this really says: we have 4 unit squares and 1/3 of another unit square, and again use caution with the phrase square units.
Connecting this to the standards for mathematical practice, our first four are listed here. Looking back at the activities that we did, we can safely say that we have applied numbers 2, 3, and 4. We’ve reasoned abstractly and quantitatively. We constructed viable arguments and critiqued the reasoning of others, especially if you have students do this in groups, group assignments. We did model the mathematics. If you look at the other four standards for mathematical practice, 5, 6, 7, and 8, in our activities we addressed number 6 and number 7. We did attend to precision and we looked for and made use of structure.