Integers

5.2 Multiplication and Division of Integers

Integer Multiplication

As when we learned to treat multiplication as repeated addition back in whole numbers, the same concepts apply here. Treating negative numbers as opposites cleanly maps onto our different models for mulitplication.

Definition: Integer Multiplication

For integers $a,b$:

  • If $a\ge0$ and $b\ge0$, use whole-number multiplication.
  • If $a<0$ and $b<0$, then $ab=|a||b|$.
  • If exactly one factor is negative, then $ab=-\big(|a||b|\big)$.

Chip Model

The chip model again is the easiest to use for new learners, but also the one that scales the worst. Its great for small multiplications, but becomes too cumbersome to use very quickly if those numbers get too large.

Example: Chip Model Multiplication

Evaluate each of the following using the chip model.

  1. $3(-2)$
  2. $-4(3)$
  3. $-3(-5)$
Show Solution
  1. First up, $3(-2)$, which is read three times negative two. That means we need to add the opposite of $2$ three times. So we'll toss in groups of $-2$ chips, three times.

    $$ \underset{\text{one}}{\underbrace{(\ominus\ominus)}}
    \underset{\text{two}}{\underbrace{(\ominus\ominus)}}
    \underset{\text{three}}{\underbrace{(\ominus\ominus)}} $$

    Which leaves us with $6$ negative chips, or $-6$.

  2. Next, $-4(3)$. Now one way to read this is we're working with groups of $3$ and we need $-4$ of them. Which would mean we need to remove $4$ groups of $3$. But to remove that many, we need to represent our starting point ($0$) with enough $3$'s to remove! So in order to even create the chip representation of the number we have to know what $4$ groups of $3$ is.

    Lets with $0$

    \[ \begin{align*} \oplus \oplus \oplus \oplus \oplus \oplus \\ \oplus \oplus \oplus \oplus \oplus \oplus \\ \ominus \ominus \ominus \ominus \ominus \ominus \\ \ominus \ominus \ominus \ominus \ominus \ominus \end{align*} \]

    Now we remove $4$ groups of $3$.

    \[ \begin{align*} \cancel{\oplus \oplus \oplus}& \cancel{\oplus \oplus \oplus} \\ \cancel{\oplus \oplus \oplus}& \cancel{\oplus \oplus \oplus} \\ \ominus \ominus \ominus& \ominus \ominus \ominus \\ \ominus \ominus \ominus& \ominus \ominus \ominus \end{align*} \]

    And we're left with $12$ negative chips, or $-12$.

  3. Finally, we can answer the age old question - "Why is a negative number times a negative number positive?" Well, $-3(-5)$ means we're working with groups of $-5$, and we need to remove $3$ of them. So we gotta begin at $0$, but with enough groups of $-5$ to remove.

    \[ \begin{align*} \oplus \oplus \oplus \oplus \oplus \\ \oplus \oplus \oplus \oplus \oplus \\ \oplus \oplus \oplus \oplus \oplus \\ \ominus \ominus \ominus \ominus \ominus \\ \ominus \ominus \ominus \ominus \ominus \\ \ominus \ominus \ominus \ominus \ominus \end{align*} \]

    Then we remove $3$ groups of the $-5$ chips.

    \[ \begin{align*} \oplus \oplus \oplus \oplus \oplus \\ \oplus \oplus \oplus \oplus \oplus \\ \oplus \oplus \oplus \oplus \oplus \\ \cancel{\ominus \ominus \ominus \ominus \ominus} \\ \cancel{\ominus \ominus \ominus \ominus \ominus} \\ \cancel{\ominus \ominus \ominus \ominus \ominus} \end{align*} \]

We're left with $15$ positive chips! So $-3(-5)=15$.

MyOpenMath: Chip model for multiplication of integers

Numberline Model

The number line model can be used to show integer multiplication by stepping and turning appropriately. For example, $3(-2) = -6$ can be shown by starting at $0$ walking backwards $3$ times, in increments of $2$.

Example: Number Line Multiplication

Use a number line to show $5(-3)=-15$.

Show Solution
MyOpenMath: Numberline multiplication of integers

Properties of Integer Multiplication

Fact: Properties of Integer Multiplication

For integers $a,b,c$:

  • Closure: $ab\in\mathbb{Z}$
  • Commutative: $ab=ba$
  • Associative: $(ab)c=a(bc)$
  • Identity: $a\cdot 1=a$
  • Distributive: $a(b+c)=ab+ac$
  • Zero property: $a\cdot 0=0$

As you might have noticed while modeling multiplication of integers here, all of the properties of whole number multiplication hold. The process of adding more or removing more chips will always result in a valid integer. Similarly the process of drawing more arrows in either drection will result in a valid integer. So we're closed under multiplication, $ab\in\mathbb{Z}$.

The order in which we did any of these operations does not change the result, so we're commutative, $ab=ba$.

Grouping our actions in different ways does not change the end result, so we're associative, $(ab)c=a(bc)$.

We have our multiplicative identity of $1$. Drawing a single arrow or adding a single pile of chips leaves the integer unchanged. Similarly not drawing any arrows or adding any chips yields $0$ so we have our zero product property.

The only property that you might have to think a little about is the distributive property, $a(b+c)=ab+ac$. But a quick check might convince you that its still true.

Example: Distribution Property of Integers

Use the chip model to show $4(-2+3)=4(-2)+4(3)$

Show Solution

To have $4(-2+3)$ we'll need $4$ groups of $-2+3$.

$$ \underset{\text{one}}{\underbrace{(\ominus\ominus \oplus \oplus \oplus)}}
\underset{\text{two}}{\underbrace{(\ominus\ominus \oplus \oplus \oplus)}}
\underset{\text{three}}{\underbrace{(\ominus\ominus \oplus \oplus \oplus)}}
\underset{\text{four}}{\underbrace{(\ominus\ominus \oplus \oplus \oplus)}} $$

Now we can rearrange them

$$ \underset{\text{one}}{\underbrace{(\ominus\ominus)}}
\underset{\text{two}}{\underbrace{(\ominus\ominus)}}
\underset{\text{three}}{\underbrace{(\ominus\ominus)}}
\underset{\text{four}}{\underbrace{(\ominus\ominus)}}
\underset{\text{one}}{\underbrace{(\oplus \oplus \oplus)}}
\underset{\text{two}}{\underbrace{(\oplus \oplus \oplus)}}
\underset{\text{three}}{\underbrace{(\oplus \oplus \oplus)}}
\underset{\text{four}}{\underbrace{(\oplus \oplus \oplus)}} $$

And we have $4$ groups of $-2$ together with $4$ groups of $3$, or $4(-2)+4(3)$.

MyOpenMath: Properties of Multiplication

Conviently, we can simplify working with integers greatly by reducing them to whole number multiplication entirely just by following a few rules we can derive from our definition of multiplication.

  1. Positive times positive is just whole number multiplication.
  2. Negative times positive is negative.
  3. Negative times negative is positive.

This way we never have to worry to much about the integer part of multiplication. It all boils down to a few cases of whole number multiplication.

Example: Multiplication sign logic

Evaluate and explain briefly:

  1. $(-8)(7)$
  2. $(-9)(-6)$
  3. $0(-13)$
Show Solution
  1. Negative times positive is negative: $(-8)(7)=-56$.
  2. Negative times negative is positive: $(-9)(-6)=54$.
  3. Any factor 0 gives 0: $0(-13)=0$.

A fast check is to separate sign and size first, then combine.

Example: Distributive simplification

Simplify with no parentheses:

  1. $(-3)(x-2)$
  2. $(a+b)(a-b)$
Show Solution
  1. $(-3)(x-2)=(-3)x+(-3)(-2)=-3x+6$.
  2. $(a+b)(a-b)=a^2-ab+ab-b^2=a^2-b^2$.

The second is a classic difference-of-squares structure.

MyOpenMath: Integer Distribution

Integer Division

Going forwards is easy, going backwards is hard. Adding? No problem. Subtraction? Let me get a pencil and some paper. Multiplcation? Easy-peasy. Division? Wheres the calculator?

Chip Model for Division

Division is the reverse of multiplcation. It feels more difficult, but it is the same problem askin in reverse. Recall that one way to interpret divison is by repeated subtraction. For example,

Demonstrate $$2(-5)=-10$$ with a chip model.

tells us that $2$ groups of $-5$ is $-10$. And $$10\div (-5) = -2$$ contains that same information, but in a different orientation.

Example: Chip Model Division

Demonstrate $10\div (-5) = -2$ using a chip model.

Show Solution

$10\div (-5) = -2$ asks us that if you were to start at $10$ and remove groups of $-5$, or start at $0$ and add groups of $-5$, how many times would you need to do it until you got to $0$ or $10$ respectively. Well, adding groups of $-5$ makes us more negative, so we should remove them instead.

Start at $0$

\[ \begin{align*} \oplus \oplus \oplus \oplus \oplus \\ \oplus \oplus \oplus \oplus \oplus \\ \ominus \ominus \ominus \ominus \ominus\\ \ominus \ominus \ominus \ominus \ominus \end{align*} \]

Remove groups of $-5$ until we reach $10$.

\[ \begin{align*} \oplus \oplus \oplus \oplus \oplus \\ \oplus \oplus \oplus \oplus \oplus \\ \cancel{\ominus \ominus \ominus \ominus \ominus}\\ \cancel{\ominus \ominus \ominus \ominus \ominus} \end{align*} \]

We were able to remove two groups of $-5$. So the answer must have been $-2$.

Numberline Model for Division

The numberline model looks exactly the same as its multiplication sibling, the difference is in how you interpret the picture.

Example: Numberline division model

Demonstrate $(-10)\div (-2)=5$ using a numberline.

Show Solution

Starting at $0$ and asking, how many $-2$ arrows can I draw until I hit $-10$, and the answer would be $5$.

Definition and properties of integer division

Just like before, we define division in terms of multiplication. That way we get all the of the nice properties of multiplication without any of the hassel of rederiving them.

Definition: Integer Division

For integers $a,b$ with $b\ne0$, $a\div b=c$ means there is a unique integer $c$ such that

$$a=bc.$$

Warning: Division by Zero

Remember that dividing by zero just isn't possible, for the same reasons it wasn't possible in whole numbers. $-2\div 0$ would ask, how many $0$ could you remove from $-2$ until you reached $0$, and if you tried you're never stop counting.

Because we're borroing all of the properties of mulitplication here, we can just remember our rules and carry on.

  1. Positive times divided by a positive is just whole number multiplication.
  2. Negative times divided by a positive is negative.
  3. Negative times divided by a negative is positive.

Example: Evaluate division using multiplication

Compute each of the following by first rewritting the expression as multiplication.

  1. $12\div 3$
  2. $-12\div 6$
  3. $-27\div 9$
  4. $(ab)\div b$ with $b\ne0$
Show Solution

Find the integer that multiplies by divisor to give dividend.

  1. $12\div3=4$ because $3\cdot4=12$.
  2. $-12\div6=-2$ because $6\cdot(-2)=-12$.
  3. $-27\div9=-3$ because $9\cdot(-3)=-27$.
  4. $(ab)\div b=a$ because $b\cdot a=ab$ and $b\ne0$.
MyOpenMath: Rewriting Division

Scaffolding Algorithm with Integer Signs

When our physical models start to become too tedious to use, we rely on algorithms to evaluate division. Together with our rules on how to manage signs for integer division, we shouldn't have any problems.

$$-819\div 9$$

isn't an issue because we can simply note that the result will be negative and evaluate the whole number division and change the sign at the end. First, set the sign aside and divide $819\div 9$:

\[ \require{enclose} \begin{array}{rll} 91 && \\[-3pt] 9 \enclose{longdiv}{819}\kern-.2ex \\[-3pt] \underline{81\phantom{0}} && \hbox{($9 \times 9 = 81$)} \\[-3pt] \phantom{0}09 && \hbox{(bring down 9)} \\[-3pt] \underline{\phantom{0}09} && \hbox{($9 \times 1 = 9$)} \\[-3pt] \phantom{00}0 && \end{array} \]

So $819\div9=91$. Since the original problem has one negative factor, the result is negative:

$$-819\div 9=-(819\div 9)=-91.$$

Example: Scaffold-style integer division

Evaluate:

  1. $-2310\div 3$
  2. $1365\div (-7)$
  3. $-31395\div 5$
Show Solution

Set aside the sign, divide the magnitudes using the scaffold, then apply the sign at the end.

1. $-2310\div 3$: one negative factor, so result is negative. Divide $2310\div3$:

\[ \require{enclose} \begin{array}{rll} 770 && \\[-3pt] 3 \enclose{longdiv}{2310}\kern-.2ex \\[-3pt] \underline{21\phantom{00}} && \hbox{($3 \times 7 = 21$)} \\[-3pt] \phantom{0}21\phantom{0} && \hbox{(bring down 1)} \\[-3pt] \underline{\phantom{0}21\phantom{0}} && \hbox{($3 \times 7 = 21$)} \\[-3pt] \phantom{00}00 && \hbox{(bring down 0)} \\[-3pt] \underline{\phantom{00}00} && \hbox{($3 \times 0 = 0$)} \\[-3pt] \phantom{000}0 && \end{array} \]

$$-2310\div 3=-(2310\div 3)=-770.$$

2. $1365\div(-7)$: one negative factor, so result is negative. Divide $1365\div7$:

\[ \require{enclose} \begin{array}{rll} 195 && \\[-3pt] 7 \enclose{longdiv}{1365}\kern-.2ex \\[-3pt] \underline{\phantom{0}7\phantom{00}} && \hbox{($7 \times 1 = 7$)} \\[-3pt] \phantom{0}66\phantom{0} && \hbox{(bring down 6)} \\[-3pt] \underline{\phantom{0}63\phantom{0}} && \hbox{($7 \times 9 = 63$)} \\[-3pt] \phantom{00}35 && \hbox{(bring down 5)} \\[-3pt] \underline{\phantom{00}35} && \hbox{($7 \times 5 = 35$)} \\[-3pt] \phantom{000}0 && \end{array} \]

$$1365\div(-7)=-(1365\div 7)=-195.$$

3. $-31395\div 5$: one negative factor, so result is negative. Divide $31395\div5$:

\[ \require{enclose} \begin{array}{rll} 6279 && \\[-3pt] 5 \enclose{longdiv}{31395}\kern-.2ex \\[-3pt] \underline{30\phantom{000}} && \hbox{($5 \times 6 = 30$)} \\[-3pt] \phantom{0}13\phantom{00} && \hbox{(bring down 3)} \\[-3pt] \underline{\phantom{0}10\phantom{00}} && \hbox{($5 \times 2 = 10$)} \\[-3pt] \phantom{00}39\phantom{0} && \hbox{(bring down 9)} \\[-3pt] \underline{\phantom{00}35\phantom{0}} && \hbox{($5 \times 7 = 35$)} \\[-3pt] \phantom{000}45 && \hbox{(bring down 5)} \\[-3pt] \underline{\phantom{000}45} && \hbox{($5 \times 9 = 45$)} \\[-3pt] \phantom{0000}0 && \end{array} \]

$$-31395\div 5=-(31395\div 5)=-6279.$$

A close look at the distributive property

One final note on the difficulty of going backwards: undoing the distributive property is divison. Students generally grasp the forward direction

$$2(x+y)=2x+2y$$

but struggle when using the same property if its in the other direction.

$$2x+2y=2(x+y).$$

Algebra students will know this as factoring because it is an application of the fundamental theorem of arithmetic. If we have two numbers that are both being multiplied by $2$, then we can divide them by $2$ to decompose the number, or factor the expression.

Example: Factor an expression

Suppose $a,b\in \mathbb{Z}$ and we have the expression $$3ab^2+9a^2b.$$ Factor the expression by writing as a product.

Show Solution

The first thing we notice, is that both terms are divisible by $3$. That means we can rewrite it as $3$ times some expression where each of the terms have been divided by $3$.

$$3ab^2 \div 3 = ab^2$$ Why? Because $3(ab^2)=3ab^2$

$$9a^2b\div 3 = 3a^2b$$ Why? because $3(3a^2b)=9a^2b$.

So we can rewrite the expression as

\[ \begin{align*} 3ab^2+9a^2b&=3(ab^2+3a^2b)\\ \end{align*} \]

But thats not all of the factors those numbers have in common. They both have an $a$ and a $b$ that could be divided out front.

\[ \begin{align*} 3ab^2+9a^2b&=3(ab^2+3a^2b)\\ &=3ab(b+3a) \end{align*} \]

What we've done is division, but disguised as the distributive property. We've shown

$$(3ab^2+9a^2b)\div (3ab)=(b+3a)$$

Why? Because $$(3ab)(b+3a)=3ab^2+9a^2b$$

The reverse direction is just a bit tougher to see for the unpracticed.

MyOpenMath: Factoring expressions

Special Cases

I'll also note here that thare are a few cases of the distributive property that appear so frequently that we like to make a big deal about them.

Fact: Difference of Squares

Given $a,b\in \mathbb{Z}$,

$$a^2-b^2 = (a+b)(a-b)$$

The difference of squares is really just a single case of the distributive property, but it ends up being incredibly useful as a tool for solving problems, or even as a tool for mental arithmetic.

Example: Using the difference of squares

Evaluate $(42)(38)$ by rewriting it as a difference of squares.

Show Solution

At first glance, this seems like we could use some multiplication algorithm to knock this out, no problem. And we could, but the point here is that both of those numbers are the same distance away from a pretty nice number.

$$(42)(38)=(40+2)(40-2)$$

Its a difference of squares. We know after going through the long process of distributing that out, we'll end up at

$$40^2-2^2$$

Which is far easier to compute.

$$1600 - 4 = 1596$$

MyOpenMath: Difference of squares

So being able to identify when you're dealing with a difference of squares is a very strong skill to have. You can use either side of the relationship to your advantage depending on the problem at hand.

MyOpenMath: Difference of squares

Ordering The Integers

As with each of the types of numbers we've discussed so far, integers have a property in that they are ordered. This means that given any two integers, we can decide one of three things about them.

Definition: Integer Inequality

For $a,b\in \mathbb{Z}$, exactly one of the following are true:

  1. $a<b$ means there exists an integer $k>0$ such that $a+k=b$.
  2. $a>b$ means there exists an integer $k>0$ such that $b+k=a$.
  3. $a=b$

This can sometimes be confusing to new learners when comparing negative numbers. The number line helps us visualize the relationship, as numbers that are smaller are to the left of the reference number. Numbers that are larger are to the right of the reference number.

Example: Comparing integers

Decide which is larger using a number line and write the correct inequality.

  1. $-7$ and $-3$
  2. $5$ and $-2$
  3. $0$ and $-1$
Show Solution
  1. $-3>-7$ because it lies to the right.
  2. $5>-2$ because any positive is greater than any negative.
  3. $0>-1$ because 0 is to the right of -1.