This is Common Core State Standards support video in mathematics. The standard is 5.NBT.B7. This standard states: Add, subtract, multiply, and divide decimals to hundredths using concrete models or drawings and strategies based on place value, properties of operations, and/or the relationship between addition and subtraction; relate the strategy to a written method and explain the reasoning used.
It’s important to see what other standards this connects to, but not so much the connections—we really need to understand what types of numbers. What are our limitations as far as the numbers that we can use here? Standard 5.NBT.B 5 deals with multiplying multi-digit whole numbers using the standard algorithm, so we know that whole numbers are pretty much fair game. Standard 5.NBT.B.6 talks about whole number quotients of whole numbers with up to four-digit dividends and two-digit divisors. So, two-digit divisors would be our maximum.
As far as multiplication, standard 5.NF.B.4 deals with that, but more importantly, it deals with fractions. It looks like we’re limited to multiplying a fraction by a whole number or a fraction by another fraction, and specifically, standard 4a under that, again just reemphasizes one more time that we’re dealing with multiplying a fraction times a whole number or multiplying a fraction times another fraction. Standard 5.NF.B.7 deals with previous understandings of division to divide unit fractions by whole numbers and whole numbers by unit fractions. So here, under that, 7a deals with division of a unit fraction by a non-zero whole number whereas 7b deals with division of whole numbers by a unit fraction.
Now that we have a better idea as to what numbers are allowed, let’s go ahead and get into this standard. Let’s first focus on adding decimals to hundredths. Well, let’s relate this back to whole numbers. Let’s say we have this example. To solve this though, we have to put these numbers vertically, but we can’t just put them any way that we want. This would be incorrect. We teach our students that we have to align them. The critical idea here is that only like items can be combined, so it’s important that we align these vertically to where we make sure that we’re adding ones with ones, tens with tens, and hundreds with hundreds.
Now let’s use those same numerals but let’s throw in decimals. So, let’s say we have 2 plus 34.8 plus .75. When we put these vertically, again, just like with whole numbers, you can’t just put them anywhere that you want. So we were taught that okay, you line up the decimal points, and we also need to fill in our zeros for our different place values. And when we compute this, this is the solution that we get. Now notice again the idea of lining up the decimal points, but there’s something interesting here. Wasn’t it the same case with whole numbers? But we didn’t approach it in terms of decimal points. We just lined up ones with ones and so forth, but it’s the same idea. So it’s not like this is something that’s brand new that applies to decimals.
But the key idea here isn’t the fact that the decimal points are aligned vertically, but really it’s the reason why they’re aligned. Well, the decimal points are aligned vertically to make sure that we’re adding like items, over here, again ones with the ones, tens with tens, hundreds with hundreds. And in the case of decimals, it’s the same exact idea—make sure that we’re adding hundredths with hundredths, tenths with tenths, ones with ones, and tens with tens. But again, it’s why we’re doing it; again, it’s to make sure that we’re adding like items.
Of course, the same idea applies to subtraction. So we have to align the decimal points when subtracting to again, make sure that we’re subtracting like items. So like in this example here, this isn’t going to cut it because, for example, if I’m going to subtract 8 from 3, wait a minute, that’s not right because this is hundredths and the 3 is tenths. So what we have to do is just like with addition, make sure that we align them correctly so that we’re adding, or in this case, subtracting, like items.
Fill in our zero. We can’t take 8 hundredths from no hundredths, so we have to decompose one of our tenths into 10 hundredths. So now we can subtract. We take 8 hundredths from 10 hundredths, so we get two. Then we have to subtract 6 tenths from 2 tenths, oops, can’t do it. So we’re going to have to decompose one of our ones into 10 tenths. So now we have 12 tenths, subtract 6 tenths, and we get six. Then we can subtract 2 ones from 6 ones, and then we can subtract 4 tens from 5 tens; same idea with our decimal points for addition, and it makes sense that our solution has to be 14 and 62 hundredths.
Let’s look at this idea of using concrete models or drawings with place value. Most of you all have these types of manipulatives where you have hundreds and tens and ones. We might be able to adjust a little bit and think of the equivalent idea here of 1 one being 10 tenths, which in turn is 100 hundredths. But physically, there might be a little bit of a problem because when you start getting down to tenths and hundredths, those are very, very small, and most of you may not have those types of class sets of manipulatives. More than likely, what we’ll have to do in some of your classrooms is use drawings or representations rather than physical models.
Let’s use this idea of concrete models or drawings in connection with place value. Let’s take that same problem that we had a while ago, and here’s our physical representation or drawing. We have 5 tens, 7 ones, and 3 tenths. Before we go on, and I have to be careful about this too, the pronunciation—we have a tendency to say things like 57 point 3 or 42 point 6, 8, and it may not be mathematically incorrect, but it really does take students’ minds away from the idea of place value. So it’s best to just name these for what they are—57 point 3, that’s not 57 point 3. It’s 57 and 3 tenths. Same thing with 42 point 6, 8, no, that’s 42 and 68 hundredths. Again, students can connect a lot better to place value and understand what a number really is; pronounce them like, again, what they are, 57 and 3 tenths in this case.
Now, what we want to do also is connect the physical model to place value. So again, make sure you do something like this, and of course here’s our decimal point so that again, students understand exactly what each of these model. So here’s our situation; we couldn’t take 8 hundredths from no hundredths, so we have to decompose one of our tenths into 10 one-hundredths, and now we can take away eight of those, and here’s what we have. Then we have to subtract 6 tenths from 2 tenths, but we can’t take six from this. So just like we had done previously, we decompose one of our ones here into 10 tenths, and now we can take away six of those. So this is what we’re left with, and now we can take 2 ones from 6 ones, and here’s what we’re left with. Then we have to take 4 tens away from 5 tens, and here’s the result as far as our model. It should be our decimal between the 4 and the 6 to make 14 and 62 hundredths. Also, you can use your physical model or your drawing to connect back and make sure. Again, here’s 1 ten, here’s 4 ones, here’s our 6 tenths, and here’s our 2 one-hundredths.
Now let’s switch over to multiplication and division. Now, it says “to hundredths,” but there’s a little bit of confusion here. By hundredths, are we talking about the products? Are we talking about the factors? Are we talking, does this is apply to the quotient, or also to maybe the divisor or the dividend? An investigation of the standards reveals this statement in the introduction for fifth grade. It states: They compute products and quotients of decimals to hundredths efficiently and accurately. So here’s the answer right here. We can go to hundredths in terms of the result, products for multiplication, and quotients for division. So that adds some clarity, and now we know what our limitations are.
It’s not until sixth grade when you look ahead at those standards where standard 6.NS.B.3 states that students are supposed to fluently add, subtract, multiply, and divide multi-digit decimals using the standard algorithm for each operation. So it’s not until the next grade level that they’re expected to go beyond products or quotients to the hundredths place. Now, at the fifth grade level, it’s really important to focus on ensuring that students understand the placement of the decimal point beyond the memorization of rules. They need to understand the why. Why did this happen with the decimal point? Why are we putting it here instead of there? So that needs to be the focus.
Let’s look at multiplication. Let’s take this example, 37.23 times 5. If we do it this way so we can use our distributive property, we can break it up to its actual value of 30 + 7 + .2 + .03, and all of that times 5; multiply it out and get our partial products. Then we figure out what each of those partial products are, and we combine them all to get our final solution of 186.15. But this was a little bit fast. Students are probably going to be a little bit confused, especially as far as the decimal placement. One of the things that students need to do is to establish the reasonableness of a solution. So in a situation like this, 37.23, well that falls between 37 and 38. So when we multiply these out, just using reasonableness, we know that our answer is going to have to be somewhere between 185 and 190. So our answer of 186.15 makes sense.
Now, students pretty much can handle the whole number part, so let’s concentrate on the decimal section of this, and let’s just look at .23 times 5. Let’s use our standard algorithm and just multiply this out. Okay, so now the question is, where’s the decimal point going to go? Well, if we look at this in terms of money, this would be like 23 cents, which is pretty close to a quarter. If we estimate, well 5 quarters would be $1.25. So, this being 1.15 makes sense, so the decimal placement here would be reasonable. That would be the expectation. We can also look at this and convert this to addition. We could just have .23 five times, and so we combine them all, and we know it’s 1.15. So we know exactly where to put the decimal point.
Let’s take a different approach. Let’s use our distributive property and change this to .2 + .03 times 5, and then we just follow the rules for the distributive property; multiply them out, get our partial products to get our final solution of 1.15. Now let’s tie this back to the idea of fractions. If we take .2 times 5, well, that’s just 5 sets of 2/10. Okay, so we combine all of those, 10 over 10, which is 1. And using that same line of thinking, we have .03 times 5, so we have five 3/100 expressed in fraction form. When we combine all those together, that would be 15/100. So, 15/100 as a decimal would be .15. And now when we look at this all together, take our fractions, we have 1 plus .15, which connects back to our partial product and then our final answer.
There’s another standard that this connects to, standard 5.NF.B.5b. In particular, this statement that deals with explaining why multiplying a given number by a fraction less than 1 results in a product smaller than the given number, and that’s what we’re going to have in a lot of the situations. For example the one that we just did, we have 5 times .23, and so we get our solution of 1.15, which is smaller than 5, and so that makes sense, that if I multiply a number by a fraction less than 1, which is what we have here, you’re going to get a product that was smaller than the original number.
Okay, let’s take another example, .4 times .2, but now let’s try to use some type of concrete representation and focus on place value. One way to look at this is 4 tenths of 2 tenths. Okay, so we’re going to take a unit square and split it up into 10 equal parts, and we’re starting off with 2 tenths. Okay, so we have 2 tenths and we want 4/10 of that. So now we need to subdivide this into 10 equal parts in the other direction, horizontally, and so there’s 4 tenths. But we’re not done by any means because this is .4 of our whole 1, not of the .2. So that’s not what we need. Okay, so we need to do a little bit of manipulation. Let’s get rid of all of that. Now we need to again, convert this to where we are dealing with .4 of just the .2, not of the whole 1. So we need to get rid of that, and now we also need to eliminate this because this is what we want. This is the .4 of the .2, and there we have it. So this is the result. So we have these little units here and we have 1, 2, 3, 4, 5, 6, 7, 8 of them. Okay, but exactly what are these, which ties back to this idea of where’s the decimal point going to go? Well, remember that had split it up this way. It’s 10 by 10, so there were 100 of these. So we have 8 hundredths. So just by using the representation, we are able to figure out where the decimal point goes.
But we can’t always use the physical model to figure that out, so we need to also look at this from other perspectives. It’s always a good idea to connect your decimals back to fractions, so here we convert this and represent it as fractions as 4/10 times 2/10. If we do our multiplication, bingo, there it is. It’s 8/100. So now this verifies that our decimal placement is correct. It is .08, and again this connects back to the fractions and the decimals. It connects them. We have 4/10 here. We have our 2/10 and we have our 8/100.
Let’s use another example, 3.5 times 1.3. For students to get a reasonable estimate of what the answer should be, if we were to round both of those down, we would have 3 times 1, and that would be 3. If we were to round both of them up, we’d have 4 times 2. So, it’s reasonable that our answer has to be between 3 and 8. So let’s go ahead and multiply this out. So we get our first partial product, then we get our second partial product, and we get a 4, a 5, and a 5. We’ve already estimated that our answer has to be between 3 and 8, so we know that the decimal point needs to go here. So it’s going to be 4.55.
Now, students are probably going to start seeing a pattern that, okay, I had one decimal to the right here, one decimal place to the right here, and I ended up with two decimal places to the right in my solution. So they’ll get the idea of the standard algorithm or the standard shortcut, whatever you want to call it, that I had two places to the right in my two factors, so I should have two places to the right of the decimal point in my product. But that still doesn’t really answer the question as to well, why did this happen?
So let’s take another look at this, and let’s use a concrete representation, 3.5 times 1.3. If we look at 3.5 as 3 1/2, what this means is that we’re going to have to have 3 1/2 sets of 1 3/10. Okay, so here’s one set of 1 3/10, and here’s some more. So now we have three sets of 1 3/10 so far, but we need another. But that makes it four sets, and that’s not what we want. We actually want half of this set here okay, and knowing that we are dealing with decimals, let’s split this whole into 10 equal parts. There’s 10 equal parts, so I only want half of them. Okay, so I get rid of five out of the 10, and now I have these 3 tenths over here. Well, I get rid of one but that’s not enough. That just balances out with this one here. So I’ve got to take this one here and get rid of half of that. So I split that up into 10 equal parts, get rid of half of them, and so now, this is what we have. So to figure this out just using our concrete model, what we can now do is this. We can take that one and bring it over here and combine it with these over here. So now we have 10 of those tenths which gives us another one, and I think we have enough now to where we can figure out our solution. We have 4 ones. We have 5 tenths, and we have 5 one-hundredths. So our answer is 4.55.
Let’s take the examples that we’ve used so far and let’s look at this from the big picture perspective. Let’s go ahead and multiply these out like we were taught using the standard algorithm. We were taught to go ahead and multiply it all out. Yes, we have all these whole numbers and don’t worry about the decimal point yet. Now, if we convert this back to fractions okay, look what happens. On this first example here, 4 times 2 is 8; on the second example 3,723 times 5 gives us the 18,615 there, and then over here, 35 times 13 gives us the 455. So, there’s a direct correlation here that okay, I just multiply here. Don’t worry about the decimal points.
Well, the same thing happens here with the fractions. Okay, multiply the numerators. Now look at the denominators. On this first one, we have 10 times 10, which is 100. So, we have 8 hundredths, and so now we know that our decimal point needs to be over here. We would have .08. Over here, we’re going to have 18,615 divided by 100, so we’d have to move our decimal place. Same thing over here; we’re going to have 100 in our denominator. So there’s a definite pattern here. If we look at this in terms of the fractional representation and connect it back to what happens with our rule as far as what to do with the decimal point, it makes sense. So again, this is what would happen in each one. It just so happens that the denominator for each one of these is 100, but it still connects back to where if I’m dividing by 100, it’s the same thing as taking my decimal point and moving it two places to the left in each case. So again, a very definite connection here and it adds a lot of understanding as to well, why is it that you move the decimal point as many places as you do? So for example, if this over here would have been something to where we would have ended up with 1,000 in the denominator, then we’d be dividing by 1,000, which means we’d have to move our decimal place three places to the left, or if it was 10 in the denominator, we know we’d have to move one decimal place to the left. Again, the connection to the fractions enables us to make sense of what happens as far as why you move the decimal point the number of places that you do.
Now let’s look at division and that’s going to be probably the more difficult of the four operations. Let’s use .2 divided by 4. Connecting this back to the standards at this grade level, remember that dividing by fractions, we were limited to dividing a fraction by a whole number or whole number by a fraction, and that’s exactly what the correlation that we have here with decimals. So we’re going to have a decimal here, .2 divided by a whole number, 4. We’re going to focus on properties of operations also. Now, what happens here is that it’s very important that students are flexible in terms of how they interpret the symbolism.
One of our properties is that division is the same thing as multiplying by the reciprocal. What we have here then is we’re going to take .2 and we just want 1/4 of that, or another way of thinking of it is we have to take 2/10 and split it up into four equal parts, and again, we only want one of those. So we take our unit whole, split it up into 10 equal parts and get rid of eight of them. So, we have 2 tenths. So now we need to split that up into four equal parts. We only want one of those. Now the problem is, what size are these? What do they represent? Okay, I’ve got two of them, but two what? Well, let’s go back. We had tens this way and then if we continue these horizontal lines out, now we can count them out and see how many of these we have, and if we count them out, it’s 10 by 4 so we have 40 of them. So actually what we have now is 2 over 40. We have 2/40. Each one of these little shaded areas is 1/40 of the area of this unit square.
But our problem is we have to express this as a decimal, and we have to deal with subdivisions of tenths, hundredths and so forth. So what we need to do here is subdivide this into 100 parts. But it’s hard to tell here, and the problem is that these red lines represent division into four equal parts, which doesn’t exactly coincide with dividing it into the 100 equal parts. So let’s blow this up a little bit so we can see a little bit better. So here’s what we have. Each one of these is 1/100, so we have four of those. But notice here, these got cut in half, so each one of these is half of 1/100, so you can combine these two though. If I move one of these over here, then I have another 1/100. So now we know that we have a total of 5/100, which as a decimal would be .05, and that is equivalent to our fraction of 2/40.
Now let’s look at this a little bit more algebraically and not be dependent on the physical model. So we ended up with 2/40. Now, if we were to divide that by 1, nothing would change, but now we need to be a little creative and change the 1 to dividing by 2 and dividing by 2. So, 20 divided by 2 is 1, then 40 divided by 2 so it’s 1/20. So we simplified the 2/40, but we got 1/20, which still doesn’t help us as far as what to do in terms of a decimal. Well, now if we multiply by 1, well, let’s see, what do we need to multiply the 20 by get 100? Oh, a 5, so let’s just make this multiplying by 5 over 5, and so we get 5/100 and that corresponds to our previous work. So we know that as a decimal it would be .05.
Let’s take another approach. Let’s take our .2 divided by 4 and change it to multiplying by the reciprocal, which would be 1/4. We’re using our properties of operations here. But let’s go ahead now and just deal with fractions. So we’d have 2/10 times 1/4, which is 2/40, which puts us back where we were earlier, and we would finish it off using the same process. Okay, another look—we have 2/40, and if we look at this in terms of our division algorithm, we would be taking 2 and we’re dividing it by 40, and we already know the rules as far as our standard algorithm for division. Forty will not go into 2. Don’t forget your decimal point comes straight up. Forty still doesn’t go into 20 and then we go another decimal place, but 40 does go into 200 five times, and we get the solution that we got previously, 5/100 or .05.
What we can also do is take our original representation as a fraction (we have .2 over 4), and if we do this in terms of the standard algorithm, bring our decimal point up. Four does not go into 2. Four does go into 20 however, 5 times, and so again, we end up with the 5/100 yet again. Let’s switch things around this time. Now we’re going to have a whole number divided by a decimal. Okay, 4 divided by .2, how should we interpret this? Let’s use a concrete model. Well, we have 4 wholes and we’re dealing with tenths, so basically what this is asking is how many sets of .2 are there in 4? So we split each one of these up into 10 equal parts, but we don’t want just .1, we want .2. So, if we split this up into sets of .2, here’s what we have, but we make them different colors to differentiate them. Then we do the exact same thing with the rest of our wholes. So we count them up. We have a total of 20 sets of .2 each.
Now if we tried to do this using our standard algorithm for division—okay, well, let me see. Two into 4 goes twice; okay, a little confused. We know the answer is 20, but if we didn’t know that already, where’s the decimal going to go? It doesn’t come straight up. It’s over here, so it’s confusing. So this is a little bit of a problem. It’s really hard to determine what to do with the decimal place when you’re dividing by something that is a decimal. If your divisor contains a decimal, then this totally changes the picture. So now what? What would be a good way to alleviate this? Well, as a fraction, this would be 4 over .2, but we already saw that that was a problem. Well, we know we can multiply by 1, and what’s messing this up is having .2 in the denominator. Well, we can multiply by 10 and that would fix that problem. So this then converts to 40 over 2, and this sure makes it a lot easier in terms of the division by the standard algorithm.
So if we add a 0, move our decimal place one, which was indicative of multiplying by 10. We have to move the decimal place here too, which was what happened over here when we multiplied .2 times 10. So now there is no confusion. I knew exactly what to do with the decimal point. My decimal point is here. I did the division. The decimal point just comes straight up. But as far as a model, 40/2 is a little bit different than what we had earlier because now we have 40. So we’ll represent those with 40 of these little rods. Okay, so we’ve got 40 of those, but we want to know how many sets of 2 there are. So there they are, count them up. We get 20, but it’s a little bit different because now again, the question was how many twos are there in 40?
Take another example: We have 13.4 divided by .08. What makes this difficult is all of these decimals, so let’s do something about it! What we can do is multiply the numerator and the denominator by 100. That’s multiplying by 1, so it doesn’t change the value. So if we do that we do ourselves a favor. We got rid of all of our decimals. So if we do that, an equivalent problem to 13.4 divided by .08 would be 1,340 divided by 8, and this is a lot simpler in terms of figuring out our solution, where the decimal point goes, and so forth. So what we’ve done then is, we multiplied both of these by 100, and that adds understanding to well, why is it that I moved my decimal place two places, okay? Well now again, it’s a little bit more obvious because again, if you connect it to what happened over here we multiply both by 100, which again connects it back to why we’re moving the decimal place two places here and two places here because we converted this to an equivalent problem, 1,340 divided by 8. So now, again the solution is a lot simpler. We know exactly what to do with our decimal place. We do our division, and voila, there’s our solution. It’s 167.5. The last part of the standard—we’ve been pretty much doing this all along. We’ve been relating the strategy to a written method and explaining the reasoning used.
As far as the standards for mathematical practice, if we look at the first four and then connect that back to the activities that we have done here, we have done 2, 3, and 4. We’ve reasoned abstractly and quantitatively. We’ve constructed viable arguments and critiqued the reasoning of others. We’ve modeled with the mathematics, and then if we look at the last four standards for mathematical practice, based on the activities that we did, we used appropriate tools strategically. We did attend to precision, and we did look for and express regularity in repeated reasoning. That really helped us, number 8, to make sense of why we do what we do with the decimal points with the four operations.