Numeration Systems and Whole Number Operations
3.4 Multiplication of Whole Numbers
Defining Multiplication
Multiplication is really just a shortcut for repeated addition. If we get into a situation where we need to add something like
$$4+4+4+4+4$$
we can handle this, but its more convienient to say, I have $4$ being added to itself $5$ times. This in turn gets shortened to $5$ times $4$, and symbolically becomes $5\times 4$.
Definition: Multiplication of Whole Numbers
For whole number $a$ and natural number $n$,
$$ n\times a=\underbrace{a+a+\cdots+a}_{n\text{ times}} $$
A note here, the symbol $\times$ is useful to remind young students that the operations is like $+$, but notation becomes confusing later on. Once students start covering algebra topics, $4\times x$ just isn't distinct enough to use. Some opt to use a $\cdot$ for multiplication, $4\cdot x$. And often, multiplication will become an implied operation. $4x$ is implying that there are $4$ copies of $x$ being added.
The reason why we don't start with this notation is because the implied multiplication is only useful once students understand variables. One cannot write $4\times 5$ as $45$. It is also true that $4\cdot 5$ isn't ideal because it can look too similar to $4.5$ when students are learning the notation.
Just know that if you see a $4\times x$, a $4\cdot x$, or an implied multiplciation $4x$, that they all mean the same thing: the number $x$, $4$ times.
Repeated Addition Model
Example: Repeated-addition model
Show $3\times4$ as repeated addition.
Show Solution
$$ 3\times4=4+4+4=12 $$
So three groups of four make 12.
Counting with physical objects is another great way to tackle multiplication. If we have several groups of rows or columns of items, we can use multiplication to count them. This lines up nicely with out repeated addition.
MyOpenMath: Physical model for multiplication
Number Line Model
Our number line model works nicely with multiplication. To show $3\times 4$ we only need to add $3$ arrows of length $4$ and see where it lands.
Area Model
The area model is nice because it gets students thinking about more complicated ideas to come. If we take our traditional number line and then cross it with another number line we can make a lattice. Each square made by the points in this lattice have an area of 1 and we can use them to count and multiply.
MyOpenMath: Area model for multiplication
Multiplication Properties
With a new operation comes several new properties. Lets have a look.
Theorem: Whole Number Multiplication Properties
For all $a,b,c\in\mathbb{W}$:
- Closure: $a\times b\in\mathbb{W}$
- Commutative: $a\times b=b\times a$
- Associative: $(a\times b)\times c=a\times (b\times c)$
- Identity: $a\times 1=a$
- Zero property: $a\times 0=0$
We have some returning favorites here. Whole numbers are closed under multiplication. This means no matter which two whole numbers you choose, when you multiply them together you are guarenteed another whole number.
Multiplication of whole numbers is both commutative and associative. This means that it doesn't matter how we group the numbers or how we arrange them. As long as only multiplication is happening between them, the result will be the same.
We have a multiplicative identity as opposed to our additive one from before. Any whole number multiplied by $1$ will remain the same.
There is also the strangest of the bunch, multiplication by zero will always result in zero. $0\times 4$ means we have $0$ copies of $4$ that need to be added... meaning we have nothing.
MyOpenMath: Identify multiplication property
When we combine multiplication and addition together we get a new and interesting property.
Theorem: Distributive Property
For all $a,b,c\in\mathbb{W}$,
$$ a(b+c)=ab+ac $$
and in the whole-number setting where defined,
$$ a(b-c)=ab-ac $$
To verify that the distributive property holds we can do a quick check. Lets evaluate $2(3+4)$.
$$2\times (3+4)$$ means we have $2$ copies of $(3+4)$.
$$(3+4)+(3+4)$$
Our parenthesis are not doing anything here, we can drop them and rearrange the terms (using which property?).
$$3+3+4+4$$
We've got $2$ copies of $3$ and $2$ copies of $4$ so we can group those back up with multiplication.
$$2\times 3 + 2\times 4$$
And this is exactly what the distributive property says.
$$2\times(3+4)=2\times 3 + 2\times 4$$
MyOpenMath: Verify the distributive property
The distributive property isn't just a means to simplify some expression, its a great way to help with mental arithmetic.
Example: Distributive computation
Compute $20\times13$ using the distributive property.
Show Solution
At first this seems daunting, but we can choose to break up either of those numbers. In particular $13$ might be better off as $10+3$
$$ 20\times (10+3) $$
Now we can use our distributive property.
$$20\times 10 + 20\times 3$$
There, thats much better.
$$200 + 60=260$$
So the next time you're doing some multiplication with numbers close to a multiple of 10, try breaking them down to be that multiple of 10 plus a little bit!
MyOpenMath: Using the distributive property
Multiplication Algorithms
Standard Algorithm
The standard algorithm is a tried and true replication of a physical model for addition. If we represent a number as units, longs, and flats, then multiplying a number is making that many copies of every physical object. The process is in the cleanup by snapping together all those blocks into their appropriate chunks.
Example: Partial-product view
Compute $12\times4$.
Show Solution
First we line the place values up
| 10's | 1's | |
|---|---|---|
| 1 | 2 | |
| x | 4 | |
We start by multiplying the ones together. $4\times 2=8$
| 10's | 1's | |
|---|---|---|
| 1 | 2 | |
| x | 4 | |
| 8 |
Then we multiply the ones in the bottom times the tens in the top. $4\times 1 = 4$
| 10's | 1's | |
|---|---|---|
| 1 | 2 | |
| x | 4 | |
| 4 | 8 |
What we're mimicing here is really the distributive property. We first duplicated those 2 ones by 4, and then the one ten by 4. $$ 12\times4=(10+2)4=4(10)+4(2)=48 $$
This is exactly what the standard algorithm is doing under the hood.
As usual, things get harder when we overshoot the base we're working in. As soon as one of these product end up as more than 10 we have to carry.
Example: A harder multiplication
Evaluate $456\times 25$
Show Solution
Line 'em up.
| 100's | 10's | 1's | |
|---|---|---|---|
| 4 | 5 | 6 | |
| x | 2 | 5 | |
We start by focusing on that 5, and we'll multiply it by every place in the upper number. This handels all of our 1's. First up, $5\times 6=30$ leaves us with $0$ ones and $3$ tens to be carried.
| 100's | 10's | 1's | |
|---|---|---|---|
| 4 | 5 (+3) | 6 | |
| x | 2 | 5 | |
| 0 |
No we have to multiply the $5\times 5$ and remember to add those $3$ extra tens we had from the previous step. $5\times 5 + 3 = 28$. We're left with $8$ tens that will stay, but there are $2$ hundreds that need to be carried over.
| 100's | 10's | 1's | |
|---|---|---|---|
| 4(+2) | 5 | 6 | |
| x | 2 | 5 | |
| 8 | 0 |
To finish off the one's we have $5\times 4 + 2 = 22$. We can carry over the 2 extra thousands, but there is no real need because there is nothing in that column to add. We can just write it down.
| 100's | 10's | 1's | |
|---|---|---|---|
| 4 | 5 | 6 | |
| x | 2 | 5 | |
| 2 | 2 | 8 | 0 |
Next up were' gonna multiply the $2$ in the $25$ by everything, this will take care of our tens. A note here, because we are multiplying by tens here, our sum that we will record will begin in the 10's place. We end up shifting over to the left each time we add a digit and that is because we're multiplying by a larger power of 10 each time.
| 100's | 10's | 1's | |
|---|---|---|---|
| 4 | 5 | 6 | |
| x | 2 | 5 | |
| 2 | 2 | 8 | 0 |
| 9 | 1 | 2 |
Verify each of those multiplications for yourself. Now that we're done duplucating its time to add straight down to finish the multiplication. This is the part where we're gathering up all of the partial sums.
| 100's | 10's | 1's | ||
|---|---|---|---|---|
| 4 | 5 | 6 | ||
| x | 2 | 5 | ||
| 2 | 2 | 8 | 0 | |
| 9 | 1 | 2 | ||
| 1 | 1 | 4 | 0 | 0 |
Hence, $456\times 25=11400$
What the algorithm embodies is this:
$$456\times 25$$ $$(400+50+6)\times (20+5)$$ $$400\times 20 + 400\times 5 + 50\times 20 + 50\times 5 + 6\times 20 + 6\times 5$$
Its just a nice and neat table to keep all of these multiplications together.
Example: Multiplication in a different base
Evaluate $312_{\text{five}}\times 43_{\text{five}}$ using the standard algorithm.
Show Solution
Line the places up.
| 25's | 5's | 1's | |
|---|---|---|---|
| 3 | 1 | 2 | |
| x | 4 | 3 | |
Start by multiplying that 3 in the bottom by every place in the top number. We're multiplying 1's so we'll stay in the 1's box and carry to the 5's box if we need. $3_{\text{five}}\times 2_{\text{five}} = 11_{\text{five}}$. So we write down 1 and carry the 1.
| 25's | 5's | 1's | |
|---|---|---|---|
| 3 | 1(+1) | 2 | |
| x | 4 | 3 | |
| 1 |
Next up is $3_{\text{five}}\times 1_{\text{five}}=3_{\text{five}}$. Plus that 1 we carried and we can write down $4_{\text{five}}$
| 25's | 5's | 1's | |
|---|---|---|---|
| 3 | 1 | 2 | |
| x | 4 | 3 | |
| 4 | 1 |
The last of the 1's is $3_{\text{five}}\times 3_{\text{five}} = 14_{\text{five}}$.
| 25's | 5's | 1's | |
|---|---|---|---|
| 3 | 1 | 2 | |
| x | 4 | 3 | |
| 1 | 4 | 4 | 1 |
Now we move to the $4$ in the bottom number and we'll start our totals in the 5's column because we're multiplying by 5's. $4_{\text{five}}\times 2_{\text{five}}=13_{\text{five}}$
| 25's | 5's | 1's | |
|---|---|---|---|
| 3 | 1(+1) | 2 | |
| x | 4 | 3 | |
| 1 | 4 | 4 | 1 |
| 3 |
Next up, $4_{\text{five}}\times 1_{\text{five}}=4_{\text{five}}$ plus that 1 we carried leaves us with $10_{\text{five}}$
| 25's | 5's | 1's | |
|---|---|---|---|
| 3(+1) | 1 | 2 | |
| x | 4 | 3 | |
| 1 | 4 | 4 | 1 |
| 0 | 3 |
Finally $4_{\text{five}}\times 3_{\text{five}}=22_{\text{five}}$ plus that 1 we carried leaves us with $23_{\text{five}}$.
| 25's | 5's | 1's | ||
|---|---|---|---|---|
| 3 | 1 | 2 | ||
| x | 4 | 3 | ||
| 1 | 4 | 4 | 1 | |
| 2 | 3 | 0 | 3 |
Now we total our partial products.
| 25's | 5's | 1's | ||
|---|---|---|---|---|
| 3 | 1 | 2 | ||
| x | 4 | 3 | ||
| 1 | 4 | 4 | 1 | |
| 2 | 3 | 0 | 3 | |
| 3 | 0 | 0 | 2 | 1 |
And our final answer is
$$312_{\text{five}}\times 43_{\text{five}}=30021_{\text{five}}$$
Boy! These alogrithms sure seem tough when we're working in a numbering system we're not familiar with.
MyOpenMath: Multiplication in a different base
Lattice Multiplication
The lattice algorithm for multiplication is similar to is addition counterpart, except that we'll be writing the numbers being multiplied on the outside of the box.
Example: Lattice Multiplication base 10
Find $234\times89$ using a lattice.
Show Solution
Set up the lattice with $2,3,4$ across the top and $8,9$ down the side. In each box, write the two-digit product with tens above the diagonal and ones below.
| 2 | 3 | 4 | |
|---|---|---|---|
| 16 | 24 | 32 | 8 |
| 18 | 27 | 36 | 9 |
Now add along diagonals and we get:
$$1,9,17,12,6$$
But since we're in base 10 we gotta carry all those 10's.
$$1,9,17+1,2,6$$ $$1,9,18,2,6$$ Carry again. $$1,9+1,8,2,6$$ $$1,10,8,2,6$$ Once more. $$1+1,0,8,2,6$$ $$2,0,8,2,6$$ And there we have it!
$$234\times 89 = 20826$$
Example: Base 5 lattice multiplication
Compute $4312_{\text{five}}\times114_{\text{five}}$ using a lattice.
Show Solution
Set up the lattice with $4,3,1,2$ across the top and $1,1,4$ down the side.
| 4 | 3 | 1 | 2 | |
|---|---|---|---|---|
| 04 | 03 | 01 | 02 | 1 |
| 04 | 03 | 01 | 02 | 1 |
| 31 | 22 | 04 | 13 | 4 |
Now sum the diagonals. The digits come out to
$$0,4,20,12,10,12,3$$
Now we carry any extra groups of 5 over. Remember this is all in base 5.
$$0,4,20,12,10+1,2,3$$ $$0,4,20,12,11,2,3$$ $$0,4,20,12+1,1,2,3$$ $$0,4,20,13,1,2,3$$ $$0,4,20+1,3,1,2,3$$ $$0,4,21,3,1,2,3$$ $$0,4+2,1,3,1,2,3$$ $$0,11,1,3,1,2,3$$ $$1,1,1,3,1,2,3$$
$$ 4312_{\text{five}}\times114_{\text{five}}=1113123_{\text{five}} $$