Mathematics

4.NF.3abcd Transcript

This is Common Core State Standards Support Video for Mathematics. The standard is 4.NF.3 and there are actually four components; a, b, c, and d. The first part of the standard states: understand a fraction a/b with a > 1 as a sum of fractions 1/b. Part a: Understand addition and subtraction of fractions as joining and separating parts referring to the same whole. Part b: Decompose a fraction into a sum of fractions with the same denominator in more than one way, recording each decomposition by an equation. Justify decompositions by using a visual fraction model. So for example, 4/7 would be 1/7 + 1/7 + 1/7 + 1/7, and so forth. Part c: Add and subtract mixed numbers with like denominators, by replacing each mixed number with an equivalent fraction, and/or by using properties of operations and the relationship between addition and subtraction. Finally, Part d: Solve word problems involving addition and subtraction of fractions referring to the same whole and having like denominators, by using visual fraction models and equations to represent the problem.

First off, in adding and subtracting fractions, students must understand a huge assumption in the symbolism that is actually stated in Part a, and that’s the idea of the same whole. So for example, if we had 1/6 + 1/6, there’s no way it can be this kind of scenario where we’re talking about two totally different objects and parts. Even when they are the same objects, like here, they’re both circles and their interiors, the slices are not the same size. So I cannot add this sixth with this sixth.

We focus on Part 3a that states: Understand addition and subtraction of fractions as joining and separating parts referring to the same whole. Again, that’s very important. Here we can combine this sixth and this sixth, and likewise over here, I can combine this sixth with that sixth, because they are parts of the same whole. Of course this applies to subtraction also. Let’s say we have 3/6, and we’re going to subtract 1/6. Well we can subtract this because, again, they are equal-sized parts of the same whole, which leaves us in this case with 2/6. The standard does talk about the same whole, but it is important to realize that we can also be talking about congruent wholes. As long as I know that this rectangular figure and this rectangular figure are congruent, then I can combine these equal-sized parts, this 1/6 with this 1/6. Same situation here; as long as I know that these 2 hexagons are congruent to each other and of course, that the parts are divided equally the same way, then I can combine this 1/6 with this 1/6.

Now lets look at Part b: Decompose a fraction into a sum of fractions with the same denominator in more than one way, and it also states to record each decomposition by using an equation and justify the decomposition by using a visual fraction model. Let’s do that here. Let’s start off with 3/4, and then we can break that down to 2/4 and 1/4. We can take the 2/4 and further break that down so that we would have 1/4 + 1/4 + 1/4. This is a good point to interject that initial statement for this standard that states that we need to understand a fraction a/b with a > 1 as a sum of fractions 1/b. So here, "a" would be 3. So basically, this is what it says—that I have this fraction a/b, and it’s just the sum of all these 1/bs. In this case, there are three of them. Just like over here, 3/4 is simply the sum of three separate 1/4s, and this would apply to any fraction regardless of what the numerators and denominators are.

Let’s take the same idea of decomposing a fraction and apply it to mixed numbers. It basically works the same way. Here I have 3, which breaks up to 1 + 1 + 1, and then we have our fraction 2/6. If we take this literally, and we’re going to do this as a sum of fractions, then we can convert all of the ones into, in this case, 6/6 so that we have this diagram and our accompanying equation. But then we can further take the 2/6 and break it up into 1/6 + 1/6. Then by simply counting, the students can tell, okay, we’ve got 6, that is 18, 19, 20. So we have a total of 20/6; 20 one-sixths.

Now with that idea of mixed numbers in mind, we can go on to the next part of this which is standard 3c which involves adding and subtracting mixed numbers with like denominators. Let’s take this example. Now here’s something very, very important. What students must realize and must do is understand that we need to break this mixed number down into its two parts with the whole number part and the fractional part. Again it is very important that students do this. Now we take that problem, and we have our mixed number split up into the whole number parts and the fractional parts, and we have a visual to represent the 1 and the 2/6 and the 2 plus 3/6.

What’s important to do here is to apply one of our properties. Let’s use our commutative property, and take this here and this over here and switch places. Let’s switch these around, and there is an important reason for doing that. By doing that, what we’ve done is, we’ve put all of our wholes together and our fractional parts together. So it’s a lot easier to combine them. Now what students simply need to do is realize that okay, well I have 3. So I have 3 wholes. There are my 3, and then when I combine all my sixths, well I have 1, 2, 3 here and then 2 more sixths here. I have a total of 5/6. There’s my solution 3 5/6.

Now we can take that same problem and approach it from strictly a fractional standpoint. So we just take our ones and split them up into sixths in this case. So we have 6, that’s 18, 20, 23 sixths. Students can just count this up and get the solution in the form of an improper fraction. As an enrichment, what you could do is show why, with the visuals especially, that 3 5/6 and the 23/6 are equivalent. As you can see, the main difference is with the wholes, that we have our three wholes here, and over here they are split up into the fractional parts. Again they can see that three wholes plus another 5/6 are the same thing as 23/6. Of course the 23 refers to the little chunks of 1/6 each, regardless of the shading or the orientation. They are all still the same 1/6, and we have 23 of them.

Let’s do a subtraction example. Let’s say we have 3 4/5, and we’re going to subtract 1 1/5. We draw our visuals. We have 3 4/5, and we’re going to take away 1 1/5. So we do that. We take away 1 whole, and then we take away 1/5. This is what we have remaining, and of course, students can just count and tell that there’s 1, 2 (wholes), and then 1, 2, 3 fifths. We can also do this strictly from a fractional standpoint. Again that just involves taking our wholes and splitting them up into 5 equal parts. But we have a little bit of a problem because we’re going to subtract 1 1/5. So basically break it down to 2 subtractions where you’re going to subtract 5/5, and then another 1/5. So then we do that with our visuals. There’s our result. Then students can simply count that we have 5, 10, 13 fifths total. Just like the previous example, we can also compare and see why it is that 2 3/5 would be equivalent to 13/5 using the visuals.Let’s look at Part d which involves using some real-life contexts, solving some problems involving addition and subtraction of fractions. Let’s look at this first problem: A recipe calls for 1 3/4 quarts of chocolate milk. However you’re expecting a lot of guests and need to double the recipe to make twice the amount. How much chocolate milk will you need? A good practice would be, don’t just stop with just asking how much chocolate milk you’re going to need. Put the additional requirement on there that they need to model the context with visual and numeric representation, and then after doing that, find the solution, because they need to show their thinking. One thing they need to realize is that when I double the amount I’m just adding something to itself. Here I’m going to add 1 3/4 with another 1 3/4. We do that important piece of breaking the mixed number up into its component parts of the whole and the fraction.

Now we can do some visuals to help us out. Here we have our chocolate where again, we have one whole quart here and 3/4 of a quart here. So now we can solve the problem. Just like before, we used our commutative property and changed the order of these two around so that we would have our wholes together and then are fractional parts over here. An interesting piece to this is that if I add 3/4 and 3/4, I can already tell that’s more than 1. But here’s something that the students can do to solve this with the visual. All they simply need to do is just take 1/4 from here and put it over here so that the result will look like this. Basically what we did here was to use that idea of compose and decompose, because all we did was take the 3 here, take 1 away from it and make it a 2. Then take that 1 that we took away, add it over here so that it became 4/4. So we did just a little bit of composing and decomposing here with the numbers to get them what we need them to be. Now of course, students can simply count. We’ve got 1, 2, 3 whole quarts and an additional 2/4, so our solution is 3 2/4 quarts. You could do this by converting this all to fractions and everything and get 14/4, but in a real-life context like this, you’d be better off just sticking with the mixed numbers, because having to get 3 2/4 quarts of chocolate makes a lot more sense in real life than getting 14/4 quarts of chocolate.

Let’s look at another problem. A family is driving on a vacation trip. At the start of the trip, the gas gauge is indicating a full tank. After 2 hours of driving, the gas gauge now looks like this. So how much gas did they use? And again, don’t just stop with how much gas did the car use. Model the context with visual and numeric representations, and then find the solution. With that in mind—this was our context. If we put those two ideas together, this is what happened. We started off with a full tank, and this is where it’s marking now. Now what the students need to do is relate the idea of the empty tank being 0 and a full tank being 1. Then figuring out that well, each of the markings, there’s 8. So these are eighths.

So label the values accordingly. Then we realize, okay this is how much gas was used, and the green is how much gas is left. Now we put that into a plain English statement that reflects the relationship. The remaining gas plus the gas that was used would be our full tank of gas. Then numerically we were at 3/8, and we know that the full tank would be one full tank or 8/8. Here’s our scenario. We’re looking for how much gas we used. Then of course, students can simply compute this and figure out that the difference is 5/8. We don’t know how many gallons this is. All we know is that we used 5/8s of one tank of gas. That is our solution.

Let’s try another problem. A rain gauge indicates that last night’s rainfall was 4 1/3 inches. The sun shone brightly for several days causing evaporation of the water in the rain gauge. The gauge now reads 1 2/3 inches. How many inches did the rain gauge drop? Again, just like before, you want students to be held more accountable and do something like this where they have to model the context with some type of visual representations, and of course the equations, the expressions that show what’s happening symbolically. We start off. This is what we started off with, and this is what we have now, as far as the gauge. So, our problem is we have to find out how much it dropped, which would be this lavender shading here. That is again, our task.

If we start doing some labeling, we started off with 4 1/3, and we dropped to 1 2/3. In order to find the difference, we need to subtract 4 1/3 minus 1 2/3. Now, something like this with a good visual, the students can pretty much solve it by counting. So some students might just count like this. They’ll start here. Okay, we’ve got 1, 2, 3, 4, 5, 6, 7, 8, and this is in chunks of 1/3. So our solution is 8/3. Now you have some students that might be able to see this a little bit differently. What they might see is that from here to here, that’s a distance of 1 whole, and then from here to here, that’s another distance of 1. So that’s 2, and then, I still have this right here and this right here. Both of those are thirds, and so that’s 2 2/3. That’s another way of being able to count and get the solution.

Let’s look at this from a computation standpoint, the 4 1/3 minus 1 2/3. Well, the whole number part of it is fine. We just take away 1 from the 4, but the problem now is I’m supposed to take away 2/3 from 1/3. So I don’t have enough thirds. I’ve only got 1 of them, and I have to take away 2. So what we can then is the simple idea of converting one of these wholes into a fractional component, which of course, in this case I have split it up into three parts. I have 3/3. Now I do have plenty of thirds available to subtract two of them. We do that, and this is the result. The students can see that we have 1 and another 1. That’s 2 2/3. So we were able to do this with a visual. And from a computation standpoint, in the old days, we would call it borrowing and so forth. So students were able to do this simply by using this idea of composing and decomposing, which in this case just involved taking 1 whole and decomposing it into 3 pieces, 3/3, and then we were able to do the problem.

There was a lot to these standards, and hopefully this helped clarify the meaning for you.