6.1: Multiplication
- Page ID
- 132893
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Just like how we could think of addition and subtraction from various perspectives (e.g., add on, put together, compare, etc.), there are many ways we can view multiplication. Before continuing, recall that the result of multiplication is called the product of the two values being multiplied together. It is most likely that you first learned multiplication in terms of addition.
Repeated addition
The operation of multiplication on two known values is represented as \(\times\) or \(\cdot\) or with parentheses. If \(n\) is any positive integer and \(m\) is any real number, then \(n \times m\) is equal to the sum of \(n\) addends equal to \(m\). This is the repeated addition perspective of multiplication.
Let's see an example of using the repeated addition perspective to answer a multiplication problem.
Watermelons cost $3 per pound. How much does a watermelon that weighs five pounds cost?
Solution
Based on the context, many children would not even need to view this as multiplication. Since they know there are five pounds, they might draw a diagram with 5 regions, each representing a pound. It might be a watermelon shape divided into 5 regions, or it might be abstract and simply be 5 separate circles or other shapes. Regardless, there are 5 portions, and each portion represents a pound that costs $3. So, when they look for the total cost, a child is likely to simply think of it as \($3+$3+$3+$3+$3=\$ 15\).
Notice that in the example above, we added 5 copies of 3 instead of 3 copies of 5. This is because of the context. Think about how you might write a problem in which a child would naturally solve it by adding \(5+5+5 = 15\) instead.
Are we able to use repeated addition when multiplying decimals such as \(0.3 \times 6=1.8\)?
Solution
Yes, in some cases.
For example, if we were asked to find the total weight of six boxes that weigh 0.3 pounds each, it would make sense to add \(0.3+0.3+0.3+0.3+0.3+0.3=1.8\).
What is 30% of 6 pounds?
\(6 \times 0.3=0.3+0.3+0.3+0.3+0.3+0.3=1.8\) (6 groups of 0.3)
Note that in the example above, the answer was "Yes, in some cases." If we change the context, repeated addition does not make quite as much sense.
Peaches cost $2 per pound. You buy 3.7 pounds of peaches. How much do you have to pay for the peaches?
Solution
We cannot add 3.7 copies of something (see more on this in the next perspective), so we cannot add 3.7 of $2. While it is possible to add 3.7 + 3.7, that does not make sense contextually. So, in order to answer this problem with repeated addition, we have remove the context. This is difficult for young learners because abstraction itself is difficult.
Once the context is removed, we can change the question to one more easily answered.
New question: What is the product of 2 and 3.7?
\(2 \times 3.7 = 3.7 + 3.7 = 7.4\)
Now we can return to the context and answer the original question. 3.7 pounds of peaches cost $7.4.
Another situation in which repeated addition does not work is when we simply have two values that are both not positive integers.
Apples cost $2.99 per pound. How much does 3.2 pounds of apples cost?
Solution
Even with abstraction, there is no way to solve this problem with repeated addition. You cannot add 2.99 copies of 3.2, nor can you add 3.2 copies of 2.99. We must switch to a different perspective or think of the multiplication in a completely abstract way and use an algorithm. For example, most adults would not use any particular perspective and just think of the contextless question, "What is the product of 2.99 and 3.2?" They would answer this by using whatever multiplication algorithm they are most comfortable with. If they use that multiplication algorithm correctly, they would get that \(2.99 \times 3.2 = 9.568\). Returning to the context, we would have that they cost $9.568 or $9.57 when rounded to the nearest cent.
As noted above, there are other perspectives. Let's look at another one.
Portion of a Quantity
If \(n\) and \(m\) are any positive real numbers, then \(n \times m\) is equal to the value that represents an \(n\) portion of \(m\). This is the portion of a quantity perspective of multiplication.
The portion referred to above may be none of, a fractional part of, exactly equal to, or more than the quantity we are considering. Let's look at each situation, starting with none.
Spot's favorite dog treats each have 5 calories. Spot isn't given any treats today. How many calories does Spot get from treats today?
Solution
In this case, we are looking at the product \(0\times 5\) as no portion of 5 calories. Having no portion of the 5 calories at all means Spot gets 0 calories from treats today.
When the portion is greater than 0 but not a positive integer, then we are dealing with a fractional part of a quantity.
Two friends go into business together. Based on the amount of work they each do, Frances gets 30% of all profits and Garfield gets 70% of all profits. The average profit from each sale is $6. How much does Frances get from a profit of $6?
Solution
Here, we are looking for the fractional part (\(30%\) or \(\frac{30}{100}\) or \(0.3\)) of a quantity ($6). We want to find the product \(0.3 \cdot 6\) or, equivalently \(\frac{30}{100} \cdot 6\). One way to do this is to literally use fractions. We think of splitting 6 into 100 portions and take 30 of those portions. Or we can use the reduced fraction \(\frac{3}{10}\) instead to get that our answer is equal to the same value as 3 copies of one-tenth of 6. Let's use that one.
\(\frac{1}{10}\) of \(6\) = \(\frac{6}{10}\)
3 copies of \(\frac{6}{10} = \frac{6}{10}+\frac{6}{10}+\frac{6}{10}\)
\(=\frac{18}{10} = \frac{9}{5} = 1 \frac{4}{5}\)
Since \(\frac{4}{5} = \frac{80}{100}\) that portion of a dollar is 80¢. So, Frances' share of the profit is $1.80.
You might have noticed that the repeated addition perspective popped up in the problem solution above. That does sometimes happen! It is okay to mix perspectives.
When the portion is exactly 1, then we have a very simple situation.
Toni has 16 ounces of water in his water bottle. He drinks all of it. How much water did he drink?
Solution
This question is essentially asking for the product of 1 and 16. This is because 1 represents the entire portion of the quantity at hand. Just like how \(1 \times 16 = 16\), the answer to the question is that Toni drank 16 ounces of water.
Finally, we can have a portion greater than one.
Last year, Lupe had to pay $24 for a parking permit to park in the "A" parking lots on her school's campus. This year's parking permit prices are 150% of last year's prices. How much will the same parking permit to park in the "A" parking lots cost this year?
Solution
In this case, we are looking for 150% of the price $24. There are a few ways this can be done, but one is to imagine 24 one-dollar bills. 150% of those would be all 24 plus an additional amount equal to 50% of 24. Thus, the new price is $24+$12 = $36.
Note that in none of these cases did we use values that are negative. We will deal more with negative values in the future. But keep in mind that the portion of a quantity perspective of multiplication does not apply very well to products involving negative numbers! Also note that although these kinds of problems can get quite difficult, it is possible to create problems for young learners that encourage viewing multiplication this way. For example, children learn "half" much earlier than they learn the general concept of fractions. So, surprisingly young children can understand what the following problem is asking.
Gabrielle took 20 minutes to do her hair yesterday morning. It only took her half as long to do it this morning. How long did it take her to do her hair this morning?
Solution
One way a child might solve this is to use a clock model. They might shade in 20 minutes of a clock (such as one made from an image such as this blank clock face with numbers provided by the Florida Center for Instructional Technology) and use that to find what half of 20 minutes is. Then they can count out the 10 minutes that they get in order to find the answer: Gabrielle took 10 minutes to do her hair this morning.
Arrays and Area
While the previous two perspectives can be adapted to fit some situations involving non-integers, we still haven't discussed a perspective of multiplication that can be apply to multiplying any two positive real numbers. We also have not discussed a perspective that is inherently visual or physical. This is where the array and area perspective of multiplication comes in.
If \(n\) and \(m\) are any positive integers, then \(n \times m\) is equal to the number of objects that form an array of \(n\) columns and \(m\) rows. This is the array perspective of multiplication.
If \(n\) and \(m\) are any positive real numbers, then \(n \times m\) square units is equal to the value that represents the area of a rectangle of length \(n\) units and width \(m\) units. This is the area perspective of multiplication.
The array perspective pairs well with the repeated addition perspective because we can think of the number of objects in an array as a collection of equal-sized groups.
A crafter puts his beads into small piles. There are 3 rows of piles, and each row has 2 piles in it. How many piles did he make?
Solution
If a child draws or directly models this situation, they are may solve this problem the same way they solve a repeated addition problem. In other words, they might look at a drawing like the one below and simply add 2+2+2 = 6.
The child could count the 6 circles individually instead of thinking of it as addition, however. In any case, the answer the child will find is that there are 6 piles of beads.
As noted in the example above, arrays do not have to be thought of as models of repeated addition. But they are more likely to be used that way than area.
Model and find the areas of rectangles that are...
- 3 units by 2 units.
- 2 units by 3 units.
Solution
A rectangle that is 3 units by 2 units is sketched below.
We can see that since there are 6 small squares that form this rectangle, the area of the rectangle is 6 square units.
A rectangle that is 3 units by 2 units is sketched below.
We can see that since there are 6 small squares that form this rectangle, the area of the rectangle is 6 square units.
The previous example illustrates an additional advantage of the array or area perspective: simply changing the orientation of a picture models the commutativity of multiplication. That is, rotating an \(n \times m\) rectangle or array to form an \(m \times n\) rectangle or array shows that \(n \times m = m \times n\).
Let's finish this perspective up with an example illustrating the usefulness of the area perspective with non-integers.
A rectangular room is 5.7 meters long and 4.8 meters wide. The carpet you like costs $20 per square meter. How much will it cost to buy the carpet for this room?
Solution
We can model this by drawing a rectangle in which we can see 5 whole units plus 0.7 of a unit along one dimension and 4 whole units plus 0.8 of a unit along the other dimension. The rectangle below shows the whole units along with the partial units.
A benefit to using such models is that even learners who have not learned how to multiply with decimals can use a model like this to estimate the answer. They can see the 20 complete small squares and note that putting together the partial squares can form a few more complete squares. It is even possible to use small grid lines for them to not just estimate the answer but find it exactly!
For now, we will simply multiply since the purpose of this example is to emphasize the fact that the product \(5.7 \times 4.8\) can be modeled with a rectangle, unlike an array or repeated addition.
\(5.7 \times 4.8 = 27.36\)
So, the cost of the carpet is $20(27.36) = $547.20.