Multiplication

Counting of counting

Patterns are important components of mathematics. Recognizing patterns and using them dates as far back as the concept of multiplication itself. In essence, multiplication is repeated addition with a certain pattern. When we say \(5 \times 20 = 100\), we are actually saying that if we add 5 to itself 20 times, we get 100. So, instead of writing twenty fives with plus signs between them, we simplify this pattern by writing \(5 \times 20\).

\[ \overbrace{5 + 5 + 5 + \dotsb + 5}^{20 \text{ times}}=5 \times 20 \]

The \(\times\) sign is the multiplication sign, and it means ‘add the number before me to itself as many times as the number after me.’ This concept may seem trivial if we are considering patterns of small size like adding 3 to itself 4 times or adding 5 to itself 2 times. With large patterns, however, the importance of multiplication becomes obvious. Consider the product (the result of multiplication) of 176 and 236. Are we going to write 176 repeatedly for 236 times, or are we going to write 236 repeatedly for 176 times? Both cases are painfully redundant, although the second case is less so. But writing the product in the form of multiplication is quite easy.

Multiplication is repeated addition in a certain pattern.

Not only that! Later in this article we are going to examine the properties of multiplication, and we will find that they can indeed simplify more complex calculations. In mathematics, when we build a concept like multiplication on a simpler concept like addition, the new concept does not have necessarily the same properties as the simpler one.

So, for any natural number \(a\), repeatedly adding this number to itself \(b\) times can be expressed as:

\[ \newcommand{\sgn}[1]{\operatorname{sgn}(#1)} \cssId{equation2.1}{ \begin{equation} \tag{2.1} \color{blue}{ \text{For } a,b \in \mathbb{N} \\ \overbrace{a+a+a+ \dotsb +a}^{b \text{ times}} = a \times b } \end{equation} } \]

Commutativity

Let us examine whether multiplication is commutative like addition. In other words, is \(a \times b\) the same as \(b \times a\)? Consider the product \(4 \times 6\) as illustrated below. There are 24 elements in total, arranged in 4 columns and 6 rows. Elements in each column share the same color, and we have 4 colors: blue, red, green and purple. Elements in each row share the same number, from 1 to 6. We can look at those elements as 4 columns, each composed of 6 numbers sharing the same color; or as 6 rows, each composed of 4 of the same number. In either case, we will end up with the same result. So, adding 4 to itself 6 times will result in the same number as adding 6 to itself 4 times. This can be extended to any two natural numbers. Therefore, multiplication is commutative.

\[ \begin{array}{c} 4 \text{ times} & ~ \\ \overbrace{ \begin{array}{} \color{blue}1 & \color{red}1 & \color{green}1 & \color{purple}1 \\ \color{blue}2 & \color{red}2 & \color{green}2 & \color{purple}2 \\ \color{blue}3 & \color{red}3 & \color{green}3 & \color{purple}3 \\ \color{blue}4 & \color{red}4 & \color{green}4 & \color{purple}4 \\ \color{blue}5 & \color{red}5 & \color{green}5 & \color{purple}5 \\ \color{blue}6 & \color{red}6 & \color{green}6 & \color{purple}6 \end{array} } & \left. \phantom{\begin{array}{} \\ \\ \\ \\ \\ \\ \end{array}} \right\} 6 \text{ times} \end{array} \]

Multiplication is commutative.

In a more mathematical way, let us assume that we have two natural numbers \(a\) and \(b\). Each of these numbers is actually the sum of ‘ones’ as many times as the number itself. For instance, the number 7 is the sum of seven ones, the number 18 is the sum of eighteen ones, etc... If we multiply \(a\) by \(b\), this means we are adding \(a\) to itself \(b\) times. If \(a\) is the sum of ones of different colors (like the example above), and we group the ones with the same color together, we will have \(b\) of each color, since \(a\) is repeated \(b\) times, and the ones in \(a\) have different colors. So, we will end up with \(b\) ones repeated \(a\) times, which is \(b\) repeated \(a\) times.

\[ \begin{aligned} a \times b & = \overbrace{a+a+a+ \dotsb + a}^{b \text{ times}} \\ & = \overbrace{\overbrace{\color{blue}1+\color{red}1+\color{green}1 + \dotsb + \color{purple}1}^{a \text{ times}}+\overbrace{\color{blue}1+\color{red}1+\color{green}1 + \dotsb + \color{purple}1}^{a \text{ times}}+\overbrace{\color{blue}1+\color{red}1+\color{green}1 + \dotsb + \color{purple}1}^{a \text{ times}}+ \dotsb + \overbrace{\color{blue}1+\color{red}1+\color{green}1 + \dotsb + \color{purple}1}^{a \text{ times}}}^{b \text{ times}} \\ & = \overbrace{\overbrace{\color{blue}1+\color{blue}1+\color{blue}1+ \dotsb +\color{blue}1+}^{b \text{ times}} + \overbrace{\color{red}1+\color{red}1+\color{red}1+ \dotsb +\color{red}1+}^{b \text{ times}} + \overbrace{\color{green}1+\color{green}1+\color{green}1+ \dotsb +\color{green}1+}^{b \text{ times}} + \dotsb + \overbrace{\color{purple}1+\color{purple}1+\color{purple}1+ \dotsb +\color{purple}1+}^{b \text{ times}}}^{a \text{ times}} \\ & = \overbrace{b+b+b+ \dotsb +b}^{a \text{ times}} \\ & = b \times a \end{aligned} \] \[ \cssId{equation2.2}{ \begin{equation} \tag{2.2} \color{blue}{ \text{For } a,b \in \mathbb{N} \\ a \times b = b \times a } \end{equation} } \]

Negative numbers

So far, we spoke about multiplication of natural numbers. What about negative numbers?! We can use the properties we have proven previously to investigate this. Let the numbers \(a\) and \(b\) be natural numbers (positive whole numbers); then the number \(-a\) will be a negative whole number. If we multiply \(-a\) by \(b\) we will get:

\[ \begin{aligned} -a \times b & = \overbrace{(-a)+(-a)+(-a)+ \dotsb +(-a)}^{b \text{ times}} \\ & = - \overbrace{\left(a+a+a+ \dotsb +a \right)}^{b \text{ times}} \qquad \text{from }\href{http://math-from-the-ground-up.blogspot.ca/2016/09/addition-and-subtraction.html#equation1.8}{1.8} \\ & = -(a \times b) \qquad \text{from }\href{#equation2.1}{2.1} \\ & = -(b \times a) \qquad \text{from }\href{#equation2.2}{2.2} \end{aligned} \] \[ \cssId{equation2.3}{ \begin{equation} \tag{2.3} \color{blue}{ \text{For } a,b \in \mathbb{N} \\ -a \times b = -(a \times b) = -(b \times a) } \end{equation} } \]

This means that multiplying a negative number by a positive number will result in the product of the absolute values of the two numbers, but with a negative sign. This makes sense, since multiplication is essentially addition, and addition is essentially counting. Multiplying \(a\) by \(b\) essentially means ‘count until \(a\) and repeat this until you've counted \(a\) for \(b\) times.’ It is, therefore, counting of counting, so to speak. So, if we have one negative number multiplied by one positive number, we will count in the negative direction (negative number) as many times as the positive number. That's why we will end up with a negative product. This is easy to understand if we consider the \(-\) sign as an indication to reverse the direction of counting.

If we multiply a positive number by a negative number, however, will the result be the same? Again, we go back to the basics. If \(a\) and \(b\) are positive integers, then \(-b\) is a negative integer, and the expression \(a \times -b\) means ‘count until \(a\) and repeat counting until you've counted \(a\) for \(-b\) times.’ But what is \(-b\) times? If we consider what we've said before that multiplication is ‘counting of counting,’ then \(-b\) will simply mean ‘in the opposite direction.’ Therefore, \(a \times -b\) will give us the same result as \(-a \times b\), because it is counting until \(a\) for \(b\) times in the negative direction.

\[ \cssId{equation2.4}{ \begin{equation} \tag{2.4} \color{blue}{ \text{For } a,b \in \mathbb{N} \\ -a \times b = a \times -b = -(a \times b) } \end{equation} } \]

What if we have two negative numbers multiplied by each other? Again, we will use what we've proved to see what the product will be. If we have two positive integers \(a\) and \(b\), then the numbers \(-a\) and \(-b\) are negative integers.

\[ \begin{aligned} -a \times -b & = -(a \times -b) \qquad \text{from }\href{#equation2.3}{2.3} \\ & = -(-(b \times a)) \qquad \text{from }\href{#equation2.4}{2.4} \\ & = b \times a \qquad \text{from }\href{http://math-from-the-ground-up.blogspot.ca/2016/09/addition-and-subtraction.html#equation1.5}{1.5} \\ & = a \times b \qquad \text{from }\href{#equation2.2}{2.2} \\ \end{aligned} \] \[ \cssId{equation2.5}{ \begin{equation} \tag{2.5} \color{blue}{ \text{For } a,b \in \mathbb{N} \\ -a \times -b = a \times b = b \times a } \end{equation} } \]

So, multiplying two negative numbers will result in a number that is the product of their absolute values, but with a positive sign. This also makes sense if we consider the \(-\) sign as reversal of the direction of counting. Two negative signs indicate reversal of the direction twice, which, as we said before, has no effect since there are only two directions of counting.

Identity element

If we multiply a number by 1, we will get that same number again. This follows from \(\href{#equation2.1}{2.1}\) as follows:

\[ 1 \times a = \overbrace{1+1+1+ \dotsb +1}^{a \text{ times}} = a \\ a \times 1 = \overbrace{a}^{1 \text{ time}} = a \]

The number 1 is the multiplicative identity.

\[ \cssId{equation2.6}{ \begin{equation} \tag{2.6} \color{blue}{ \text{For } a \in \mathbb{Z} \\ a \times 1 = 1 \times a = a } \end{equation} } \]

Therefore, the number 1 is the ‘multiplicative identity,’ since multiplying any number by 1 will result in the same number, and this is the definition of the ‘identity element’ of any mathematical operation as we said before.

Lots of ‘nothing’ is ‘nothing’

What about the zero, then?! Multiplying zero by any number will result in zero, since it is essentially adding zero to itself a finite number of times will only result in zero. If you have lots of nothing, you have nothing still. If we are doing it the other way; i.e. multiplying any number by zero, it means that we are counting this number zero times; i.e. we are not counting at all! So, if we start counting from the zero point, and not count at all, we are still at the zero point.

\[ \cssId{equation2.7}{ \begin{equation} \tag{2.7} \color{blue}{ \text{For } a \in \mathbb{Z} \\ a \times 0 = 0 \times a = 0 } \end{equation} } \]

Multiplication of integers

So, from all of the above, we can conclude that multiplication is defined for all integers, and is commutative for all integers.

\[ \cssId{equation2.8}{ \begin{equation} \tag{2.8} \color{blue}{ \text{For } a,b \in \mathbb{Z} \\ a \times b = b \times a \\ a \times b = \begin{cases} |a| \times |b|; & \quad \sgn{a} = \sgn{b} \\ -(|a| \times |b|); & \quad \sgn{a} \neq \sgn{b} \end{cases} } \end{equation} } \]

We use \(\sgn{a}\) in mathematics to indicate the sign (\(+ \text{ or } -\)) of the number. In mathematical terms, \(\sgn{a}=1\) if the number \(a\) is positive or zero, and \(\sgn{a}=-1\) if the number \(a\) is negative.

\[ \cssId{equation2.9}{ \begin{equation} \tag{2.9} \color{blue}{ \text{For } a \in \mathbb{Z} \\ \sgn{a} = \begin{cases} 1; & \quad a \ge 0 \\ -1 & \quad a \lt 0 \end{cases} } \end{equation} } \]

So, equation \(\href{#equation2.8}{2.8}\) means that the product of any two integers will be positive if their signs are the same, and negative if their signs are different. We know from \(\href{http://math-from-the-ground-up.blogspot.ca/2016/09/addition-and-subtraction.html#equation1.5}{1.5}\) that the absolute value gets rid of the sign; i.e it transforms a negative number into its corresponding positive number (additive inverse), while it has no effect on positive numbers. The absolute value is always positive. So, the product of two positive numbers will be a positive number, and getting the additive inverse (with the \(-\) sign) of that product will be a negative number; hence the rule for defining \(a \times b\) when their signs are different.

Associativity

Now we turn our attention to the property of associativity. Is multiplicatioin associative like addition, or is it left-associative like subtraction? We remember from the discussion about addition that the property of associativity resulted directly as a consequence of the property of commutativity. Since multiplication is commutative, as we've shown above, it is also associative.

Multiplication is associative.

\[ \text{For } a,b,c \in \mathbb{Z} \\ \begin{aligned} a \times b \times c & = (a \times b) \times c \\ & = b \overset{\curvearrowleft}{\times} a \times c \qquad \text{from }\href{equation2.7}{2.7} \\ & = b \times c \overset{\curvearrowleft}{\times} a \qquad \text{from }\href{equation2.7}{2.7} \\ & = (b \times c) \times a \\ & = a \overset{\curvearrowleft}{\times} (b \times c) \text{from }\href{equation2.7}{2.7} \end{aligned} \] \[ \cssId{equation2.10}{ \begin{equation} \tag{2.10} \color{blue}{ a \times b \times c = (a \times b) \times c = a \times (b \times c) } \end{equation} } \]

Distributivity: a brand new property

Previously, we talked about addition and subtraction, and we showed that subtraction can indeed be treated as addition of the additive inverse, so the two operations of addition and subtraction can be reduced to only the operation of addition; i.e. we could be dealing with only one operation on number, which is addition. We now have two different operations at least, which are addition and multiplication. If we have three integers \(a,b,c\) in the form of \(a \times (b + c)\), can we express this in any other way? How does multiplication interact with the (simpler) operation of addition. Let us examine this using what we've proven so far.

\[ \begin{aligned} \color{red}a \times (\color{blue}b+\color{green}c) & = \overbrace{(\color{blue}b+\color{green}c) + (\color{blue}b+\color{green}c) + \dotsb + (\color{blue}b+\color{green}c)}^{a \text{ times}} \quad \text{from } \href{#equation2.1}{2.1} \\ & = \overbrace{\color{blue}b+\color{blue}b+ \dotsb +\color{blue}b}^{\color{red}a \text{ times}} + \overbrace{\color{green}c+\color{green}c+ \dotsb +\color{green}c}^{\color{red}a \text{ times}} \quad \text{from }\href{http://math-from-the-ground-up.blogspot.ca/2016/09/addition-and-subtraction.html#equation1.1}{1.1}, \href{http://math-from-the-ground-up.blogspot.ca/2016/09/addition-and-subtraction.html#equation1.2}{1.2} \\ & = \color{blue}b \times \color{red}a + \color{green}c \times \color{red}a \quad \text{from }\href{#equation2.1}{2.1} \\ & = \color{red}a \times \color{blue}b + \color{red}a \times \color{green}c \quad \text{from }\href{#equation2.2}{2.2} \end{aligned} \tag{*} \]

So, we had two operations, one of them is an outer operation (which is multiplication in this case) and another is the inner operation (which is addition in this case.) We could ‘distribute the outer operation on the terms of the inner operation. In this case, we distributed the multiplication by \(a\) on the operations of addition, which are \(b\) and \(c\). This can be proven for subtraction, too.

\[ \begin{aligned} a \times (b - c) & = a \times (b + (-c)) \quad \text{from }\href{http://math-from-the-ground-up.blogspot.ca/2016/09/addition-and-subtraction.html#equation1.6}{1.6} \\ & = a \times b + a \times (-c) \quad \text{from (*)} \\ & = a \times b - a \times c \quad \text{from }\href{#equation2.4}{2.4} \end{aligned} \] \[ \cssId{equation2.11}{ \begin{equation} \tag{2.11} \color{blue}{ a \times (b \pm c) = a \times b \pm a \times c } \end{equation} } \]

An outer mathematical operation is said to be ‘distributive’ over an inner, different mathematical operation if and only if the outer operation can be distributed over every term of the inner operation without changing the value of the expression.

The sign \(\pm\) is pronounced ‘plus or minus,’ and it means that the operation applies for both addition and subtraction, or both the plus and the minus sign (in case the sign is used to indicate the direction of counting rather than addition or subtraction.) The reverse of this sign is \(\mp\) and is pronounced ‘minus or plus.’ It is used in instances where the \(\pm\) sign is used and signs are to be reversed, so the plus becomes a minus and vice versa. The reader is to verify the following:

\[ -a \times (b \pm c) = a \times b \mp a \times c \]

Because addition is both associative and commutative, the previous property is extended to any number of terms added together.

\[ \begin{aligned} a \times (b+c-d) & = a \times [b+(c-d)] \\ & = a \times b + a \times (c-d) \\ & = a \times b + a \times c - a \times d \end{aligned} \] \[ \cssId{equation2.12}{ \begin{equation} \tag{2.12} \color{blue}{ \text{For } a,x_n \in \mathbb{Z}; \quad n \in \mathbb{N} \\ a \times (x_1 \pm x_2 \pm x_3 \pm \dotsb \pm x_n) = a \times x_1 \pm a \times x_2 \pm a \times x_3 \pm \dotsb \pm a \times x_n } \end{equation} } \]

The previous equation reads that for any integer \(a\) and group of integers \(x_n\) expressed in subscript notation, with the subscript being a natural number, the product of \(a\) and the sum of the numbers \(x_n\) or their difference is the sum or the difference, respectively (i.e. in the same order), of the products of \(a\) and each of the numbers \(x_n\). In layman's speech, if \(a\) is multiplied by a group of numbers within parentheses, and the only operation between the parentheses is addition or subtraction, we can distribute the multiplication of \(a\) on each of the numbers within the parentheses.

Multiplication is distributive over addition and subtraction.

The property of distributivity can be handy sometimes. Consider evaluating the expression \(23 \times 15 + 23 \times 45 + 23 \times 40\). Of course, we can multiply 23 by 15, and 23 by 45, and 23 by 40, then add the products together. But using distributivity, we can also say that:

\[ \begin{aligned} \color{red}{23} \times 15 + \color{red}{23} \times 45 + \color{red}{23} \times 40 & = \color{red}{23} \times (15 + 45 + 40) \\ & = 23 \times 100 \\ & = 2300 \end{aligned} \]

It saves a lot of time and effort to do the calculation the previous way! Instead of doing three operations of multiplication and two operations of addition, we did two additions and one multiplication. It also turned out that this particular multiplication is a piece of cake! Also, consider something like \(73 \times 99\). We can express 99 as the difference between 100 and 1. Therefore:

\[ \begin{aligned} 73 \times 99 & = 73 \times (100 - 1) \\ & = 73 \times 100 - 73 - 1 \\ & = 7300-73 \\ & = 7227 \end{aligned} \]

More can be done with the property of distributivity. Consider evaluating the product \(24 \times 49\). Each of these numbers can be expressed in terms of two other numbers with addition or subtraction: \(24\) as \(25-1\) and \(49\) as \(50-1\). Therefore, the product \(24 \times 49\) can be expressed as \((25-1) \times (50-1)\). Now each of the parenthesized (between parentheses) expressions can be distributed over the terms of the other. Let us distribute the first over the second. The result can be distributed further, simplifying the calculations.

\[ \begin{aligned} 24 \times 49 & = (25-1) \times (50-1) \\ & = 50 \times (25-1) - 1 \times (25-1) \\ & = 50 \times 25 - 50 \times 1 - 1 \times 25 + 1 \times 1 \\ & = 1250 - 50 - 25 + 1 = 1176 \end{aligned} \]

Notice how we used the property of commutativity when distributing \(25-1\) over both \(50\) and \(-1\) to make the expression look nicer. Since \((25-1) \times 50\) is the same as \(50 \times (25-1)\), we chose to write the individual number \(50\) first instead of the parenthesized \((25-1)\). Notice also how the negative sign stuck to the \(-1\) when we multiplied it by the contents of the parentheses. We that the negative sign should stick to the number following it in the previous article. The same principle applies here. That's why we had \(+1\) at the end instead of \(-1\), because we multiplied \(-1 \times -1\), which gives us \(+1\) as stated above in \(\href{#equation2.8}{2.8}\).

Let's agree on something

Of course, the significance of the property of distributivity does not stop here. We will see later how this property helps a lot in mathematics. Finally, when doing mathematics with symbols (i.e. \(a, b,x,z\) rather than \(2,76,-31,29872\),) it is often too ‘crowded’ to write the sign of multiplication over and over again, especially that multiplication is a common mathematical operation. It is the convention, therefore, that whenever the sign of the mathematical operation is omitted, it is assumed to be multiplication. This can be used if and only if there is no chance of confusion. We are not allowed to write \(425\) and assume there is multiplication, because in this case we have no idea if this is just the plain number \(425\) or the number \(4\) multiplied by \(25\), or the number \(42\) multiplied by \(5\). As said before, mathematical expression can have only one meaning. However, if we have two variables (symbols) \(a\) and \(x\) that we have defined already, we are allowed to write \(ax\) to mean \(a \times x\). We are even allowed to write \(5x\) to mean \(5 \times x\). But expressions like \(4a6x\) are highly discouraged, since there's potential for confusion with \(4a_6x\), with \(a_6\) expressed in subscript notation (and this is only one way of confusing things in expressions like this.) The general rule is: ‘whenever there is a chance of confusion, write it down.’ So, the previous example can be written as:

\[ \begin{aligned} 24 \times 49 & = (25-1)(50-1) \\ & = 50(25-1) - 1(25-1) \\ & = 50 \times 25 - 50 \times 1 - 1 \times 25 + 1 \times 1 \\ & = 1250 - 50 - 25 + 1 = 1176 \end{aligned} \]

Where no confusion is possible, the sign of multiplication can be omitted and is assumed implicitly.

Notice how we omitted the multiplication sign between the parenthesized expressions, and between the individual numbers \(50\) and \(-1\) and the following parenthesized expression by which they are multiplied. We could not, however, omit the multiplication sign between the individual numbers.

Operator precedence

We discussed the ‘order of evaluation’ of mathematical operations in the previous article, and we concluded that addition is associative while subtraction is left-associative. This means that you can evaluate a chain of terms added to each other in any order you want, while subtraction is evaluated—by consensus—from left to right only. This posed no big problem because we can always evaluate addition from left to right as well, so if we have many terms in a chain of addition and subtraction, we are safe if we evaluate them from left to right. Now that we have another operation, multiplication, what would the order of evaluation be? For instance, if we have some expression like \(2 \times 4 + 5 \times 3\), how should we evaluate it? Should we just proceed from left to right evaluating as we go? If we do that, the previous expression will be evaluated as \(2 \times 4 + 5 \times 3 = 8 + 5 \times 3 = 13 \times 3 = 39\). However, this is the wrong way to do it! Again, the consensus in mathematics, in order to avoid confusion, is to give multiplication priority over addition and subtraction, and give the latter two the same priority. This means that multiplication is evaluated before we do addition and subtraction.

Multiplication has priority in evaluation over addition and subtraction.

Therefore, the previous expression is correctly evaluated—by consensus—as follows:

\[ 2 \times 4 + 5 \times 3 = 8 + 15 = 23 \]

Notice that we did all operations of multiplication first, then we started evaluating addition and subtraction. The order in which mathematical operations are to be performed is called ‘operator precedence,’ and it will be discussed in more detail later. For now, it suffices to say that the highest priority is for whatever between parentheses or brackets, and that multiplication take priority over addition and subtraction. Giving parenthesized expressions the highest priority gives us a means to override the agreed-upon operator precedence. Say that in the previous example we actually wanted to add \(4\) to \(5\) first before multiplying the result by \(2\), then finally we wanted to add \(3\). We can ‘force’ that by putting the addition between parentheses, and since parentheses have the highest priority, we have to evaluate what's inside them first.

\[ 2 \times (4+5) \times 3 = 2 \times 9 \times 3 = 54 \]

When we have more than one set of parentheses, we must evaluate them from inside out. Consider the following example:

\[ \begin{aligned} 3 \times \color{blue}(2+\color{red}(5-2\color{red}) \times \color{red}(4-3\color{red})\color{blue}) & = 3 \times \color{blue}(2+3 \times 1\color{blue}) \\ & = 3 \times \color{blue}(2+3\color{blue}) \\ & = 3 \times 5 \\ & = 15 \end{aligned} \]

Notice in the previous example that we evaluated the red (innermost) parenthesized expressions first, then we had to work out what was inside the blue (outermost) parentheses, and while doing that we maintained the priority of multiplication over addition and subtraction.

For all of you, and each of you, too!

We discussed in \(\href{http://math-from-the-ground-up.blogspot.com/2016/09/addition-and-subtraction.html#equation1.10}{1.10}\) why the absolute value of the sum of integers is not equal to the sum of their absolute values. What about multiplication? The absolute value of the product of integers is equal to the product of their absolute values. From \(\href{#equation2.8}{2.8}\) above, we know that only the sign of the value of the product of two integers is affected by their signs. The product will be either \(|a| \times |b|\) or \(-(|a| \times |b|)\). If we take the absolute value of that product, we will get rid of the sign, and it will be \(|a| \times |b|\) regardless of the signes of either \(a\) or \(b\). This can be extended easily to more than two integers, because multiplication is both associative and commutative.

\[ \cssId{equation2.13}{ \begin{equation} \tag{2.13} \color{blue}{ \text{For } n \in \mathbb{N}; \quad a_n \in \mathbb{Z} \\ |a_1 \times a_2 \times a_3 \times \dotsb \times a_n| = |a_1| \times |a_2| \times |a_3| \times \dotsb \times |a_n| } \end{equation} } \]

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‘Math from the ground up’ by Rafeek Mikhael licensed under Creative Commons Attribution-NonCommercial-ShareAlike International 4.0 license.

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