Math from the ground up: Division: Divisibility

Division is the opposite of multiplication. When we say that \(4\) divides \(24\), it means that we can subtract \(4\) from \(24\) a certain number of times until we end up with nothing (zero.) This ‘number of times’ we can subtract \(4\) is called the ‘quotient’ of division. The number that is subtracted (\(4\) in this case) is called the ‘divisor,’ and the number being subtracted from is called the ‘dividend.’ In mathematical language, we say that the number \(a\) divides the number \(b\) if \(a\) multiplied by another number \(q\) (the quotient of division) gives us \(b\).

\[ \cssId{equation3.1}{ \begin{equation} \tag{3.1} \color{blue}{ \text{For } a,b,q \in \mathbb{N} \\ a \mid b \equiv b \div a = q \iff b = a \times q } \end{equation} } \]

The notation \(a \mid b\) reads ‘\(\boldsymbol a\) divides \(\boldsymbol b\),’ or ‘\(\boldsymbol b\) is divisible by \(\boldsymbol a\).’ The symbol \(\equiv\) means ‘equivalent to,’ and the symbol \(\div\) is the symbol of division. The left-hand expression in \(\href{#equation3.1}{3.1}\) means that \(a \mid b\) is equivalent to \(b \div a = q\), so if we say \(a \mid b\) we also mean that \(b\) divided by \(a\) equals another number \(q\). Equation \(\href{#euqation3.1}{3.1}\) means that \(a\) divides \(b\) if and only if \(b\) can be expressed as the product of \(a\) and another number \(q\). The symbol \(\iff\) reads ‘if and only if,’ and means (loosely, for now) that we can conclude either one of what's on its sides from the other one. So, if we know that \(a \mid b\), we can conclude that \(b = a \times q\), and if we know that \(b = a \times q\), we can conclude that \(a \mid b\).

Opposites, again!

Division is the opposite of multiplication because, in multiplication, if we add the number \(a\) to itself \(q\) times, we end up with the number \(b=a \times q\). This means we can take away (subtract) the number \(a\) from the number \(b\) repeatedly \(q\) times until we finish \(b\); i.e. until we end up with zero. Although division is essentially repeated subtraction, it is defined in mathematics in terms of multiplication as shown in \(\href{#equation3.1}{3.1}\)

Division is defined in terms of multiplication, although division is essentially repeated subtraction.

\[ \begin{array}{ccccc} ~ & \color{green}{\text{Adding } a \\ q \text{ times}} & b = a \times q & ~ & ~ \\ \color{green}{\begin{array}{c} 1 \times a \\ 2 \times a \\ 3 \times a \\ \vdots \\ (q-2) \times a \\ (q-1) \times a \\ q \times a \end{array}} & \left. \begin{array}{c} \\ \\ \\ \\ \\ \\ \\ \end{array} \right\Downarrow & \overbrace{\begin{array}{ccccccc} 1, & 2, & 3, & \dots , & a-2, & a-1, & a \\ 1, & 2, & 3, & \dots , & a-2, & a-1, & a \\ 1, & 2, & 3, & \dots , & a-2, & a-1, & a \\ ~ & ~ & ~ & \vdots \\ 1, & 2, & 3, & \dots , & a-2, & a-1, & a \\ 1, & 2, & 3, & \dots , & a-2, & a-1, & a \\ 1, & 2, & 3, & \dots , & a-2, & a-1, & a \end{array}} & \left\Uparrow \begin{array}{c} \\ \\ \\ \\ \\ \\ \\ \end{array} \right. & \color{red}{\begin{array}{c} q \times a \\ (q-1) \times a \\ (q-2) \times a \\ \vdots \\ 3 \times a \\ 2 \times a \\ 1 \times a \end{array}} \\ ~ & ~ & ~ & \color{red}{\text{Subtracting } a \\ q \text{ times}} & ~ \end{array} \]

I have enough ‘nothing’ for everyone

In view of the previous definition, what would be a number that divides zero? In other words, how many times can you take the number \(a\) (which is not zero itself) from zero (i.e. from nothing.) You actually cannot take anything from nothing! This is different from subtraction, where we could take something from the zero and end up with a negative number. Why is it different? Because the end point in division is ending with zero! So, how many times can you take something from nothing and end up with nothing? Exactly! You cannot! In other words, you can do this zero times. Therefore, the quotient of dividing zero by any number (that's not zero itself) is zero; i.e. any number divides zero. This also follows directly from \(\href{http://math-from-the-ground-up.blogspot.ca/2016/09/multiplication.html#equation2.7}{2.7}\)

\[ \cssId{equation3.2}{ \begin{equation} \tag{3.2} \color{blue}{ \text{For } a,q \in \mathbb{N} \\ a \mid 0 \equiv 0 \div a = 0 \iff 0 = a \times 0 } \end{equation} } \]

Division by zero

Now, can zero divide anything? Again, in view of the previous definition, how many times can you take nothing (zero) from any number that is not zero itself, and end up with zero? No matter how many times you take nothing from any positive integer, you will never end up with zero! It actually does not make sense to consider this situation at all. Because zero is the additive identity, it doesn't matter how many times you add it to any number or remove it from any number. As long as you are doing this for a finite number of times (i.e. not endlessly,) you will never ever end up with zero. Therefore, zero does not divide any number at all. Division by zero is undefined in mathematics. Whenever you encounter division, make sure that you are not dividing by a zero value.

Division by zero is undefined.

\[ \text{For } a \in \mathbb{N} \\ 0 \nmid a \]

The previous expression \(0 \nmid a\) means that \(0\) does not divide \(a\). Just like we talked about the \(\in\) symbol, crossing out the symbol that means ‘divides’ gives the opposite meaning.

The funny situation is when you try to divide zero by zero. In essence, this means ‘how many times can you take nothing from nothing and end up with nothing?’ Can you do this one time? Yes, indeed! Can you do it forty times? Of course! Can you do it a million times? Definitely! You can actually do it any number of times, and you will end up with zero, because you are taking zero from zero! The quotient of dividing zero by zero is called in mathematics an ‘indeterminate’ value; i.e. you cannot assign it a single determined value. This is actually beyond the scope of division, but I just wanted to mention it here for completeness and fun.

Closed operations

We know from \(\href{http://math-from-the-ground-up.blogspot.ca/2016/09/multiplication.html#equation2.8}{2.8}\) that any two integers multiplied by each other will result in another integer. An operation in mathematics is said to be ‘closed’ if it results in the same kind of thing that it operated on. For example, in the operation of addition, adding two natural numbers to each others will always result in a natural number. We will never end up with a negative number, for instance, by adding two natural numbers. Therefore, addition is said to be ‘closed’ in natural numbers. For an example from outside mathematics, consider the process of knitting. We start with yarn and do the knitting, ending with some garment. We don't end up with yarn; therefore knitting is not ‘closed’ on yarn. On the other hand, a process like chopping wood is ‘closed’ on wood. We start with wood, and we end with wood, too; albeit more pieces of wood. Multiplication is closed in integers. Any two integers multiplied by each other will result in another integer. Multiplication of two integers will not give us a fraction, for instance. Therefore, if an integer \(b\) can be expressed as the product of two other integers \(a\) and \(q\), then each of those integers divides it. This is essentially what has been stated in \(\href{#equation3.1}{3.1}\) for natural numbers, and it can be extended to include all integers. However, since integers include the zero, we have to refine our definition to exclude the zero as a divisor.

A mathematical operation is said to be ‘closed’ if it results in the same kind of entity on which it operated.

\[ \cssId{equation3.3}{ \begin{equation} \tag{3.3} \color{blue}{ \text{For } a,b,q \in \mathbb{Z}; \quad a \neq 0 \\ a \mid b \iff b = a \times q } \end{equation} } \]

But from \(\href{http://math-from-the-ground-up.blogspot.ca/2016/09/multiplication.html#equation2.8}{2.8}\), we also know that the absolute value of the product of any two integers is not dependent on their sign. If \(a\) is not negative and it divides \(b\), then \(b = aq\) has the same sign of \(q\) (second column from the left below.) If \(a\) is negative and it divides \(b\), then \(b = aq\) has the opposite sign of \(q\), because the negative sign of \(a\) will flip the sign of the product relative to the sign of \(q\) (third column from the left blow.) In either case, \(a\) still divides \(b\) regardless of the sign, although the quotient is different. This means that whether or not \(a\) divides \(b\) is not dependent on the sign at all, and if \(a \mid b\) then \((|a|) \mid (|b|)\). We can, therefore exclude the negative numbers from the definitions to make things easier, knowing that this will not affect the rule.

\[ \begin{array}{c|c|c} ~ & a \ge 0 & a \lt 0 \\ \hline q \ge 0 & b=aq \ge 0 & b=aq \lt 0 \\ q \lt 0 & b=aq \lt 0 & b=aq \ge 0 \end{array} \] \[ \cssId{equation3.4}{ \begin{equation} \tag{3.4} \color{blue}{ \text{For } a,b,q \in \mathbb{Z}^+_0; \quad a \neq 0 \\ a \mid b \iff b = a \times q } \end{equation} } \]

The symbol \(\mathbb{Z}^+_0\) means the positive (hence the small \(+\) sign) integers in addition to the zero (since it is neither positive nor negative.) If we wanted to express the negative integers in addition to the zero, we would have written \(\mathbb{Z}^-_0\). If we wanted to express the positive integers only, we would have written \(\mathbb{Z}^+\). Since zero is not a natural number, but it is still divisible by any other number, equation \(\href{#equation3.4}{3.4}\) is different from \(\href{#equation3.1}{3.1}\) in that it includes the zero as a dividend, but not as a divisor.

The product divisor

If any integer \(c\) is divisible by the (non-zero) product of two other integers \(a\) and \(b\), then each of those two other integers is a divisor of \(c\). This follows directly from definition \(\href{#equation3.1}{3.1}\) and the properties of multiplication.

\[ \text{For } a,b,c,q \in \mathbb{Z}^+_0; \quad a \times b \neq 0 \\ \begin{aligned} (a \times b) \mid c \iff c & = a \times b \times q \\ & = a \times (b \times q) \\ & = b \times (a \times q) \end{aligned} \\ c = a \times (b \times q) \iff a \mid c \quad \text{from } \href{#equation3.1}{3.1} \\ c = b \times (a \times q) \iff b \mid c \quad \text{from } \href{#equation3.1}{3.1} \\ \therefore (a \times b) \mid c \implies a \mid c \quad \text{and} \quad b \mid c \] \[ \cssId{equation3.5}{ \begin{equation} \tag{3.5} \color{blue}{ \text{For } a,b,c \in \mathbb{Z}^+_0; \quad a \times b \neq 0 \\ (a \times b) \mid c \implies a \mid c \quad \text{and} \quad b \mid c } \end{equation} } \]

Notice the direction of the arrow in the last sentence. We cannot imply from \(a \mid c\) and \(b \mid c\) that \((a \times b) \mid c\), because, as we said, \(a \mid c\) does not tell us anything about \(b\), and \(b \mid c\) does not tell us anything about \(a\). A simple example that demonstrates this is \(12 \mid 24\) and \(6 \mid 24\), but \((12 \times 6 = 72) \nmid 24\).

Properties of divisibility

Now let us turn out attention to the properties of divisibility. It is easy to conclude that divisibility is not commutative. If we can give a single example for two numbers that are not divisible by each other, then we can conclude that divisibility is not commutative, because a rule in mathematics has to apply under all circumstances. Take the numbers \(2\) and \(5\), for example. The number \(2\) does not divide the number \(5\) because no matter what you do, you cannot express \(5\) as the product of \(2\) and another whole number. Therefore, divisibility is not commutative.

How about associativity?! Divisibility is more of a ‘statement’ rather than an operation. Whe we say \(a \mid b\), we mean to say that \(b\) can be expressed as \(a\) multiplied by another number, but we do not say what number or even try to find it. We do not ‘operate’ on \(a\) and \(b\), but we ‘state’ the relationship between \(a\) and \(b\). Since associativity is about the ‘order of evaluation’ of operations, and divisibility is not an operation, associativity is meaningless in this context. Division, however, is an operation, and we will discuss it later.

From the definition of divisibility \(\href{#equation3.1}{3.1}\), and from \(\href{http://math-from-the-ground-up.blogspot.ca/2016/09/multiplication.html#equation2.6}{2.6}\), we can conclude that every number divides itself, and that \(1\) divides every number.

\[ \cssId{equation3.6}{ \begin{equation} \tag{3.6} \color{blue}{ \text{For } a,b \in \mathbb{Z}^+_0; \quad a \neq 0 \\ a \mid a \\ 1 \mid b } \end{equation} } \]

The chain rule

We will frequently encounter in mathematics the rule that if a certain relation applies to \(a\) and \(b\), and the same relation applies to \(b\) and \(c\), then it applies to \(a\) and \(c\). It is axiomatically easy to conclude, for instance, that if \(a=b\) and \(b=c\), then \(a=c\). A numeric example is that if \(3 \times 4 = 12\) and \(12 = 2 \times 6\), then \(3 \times 4 = 2 \times 6\). This is called the ‘chain rule,’ because each element in a chain of relations links the one before it to the one after it. In the previous example, \(b\) links \(a\) to \(c\). This can extend to more than three elements. Consider the following for a mathematical operation or relation (like divisibility) that has the symbol \(\odot\), which is used here to mean any operation or relation at all, not a specific one.

\[ \text{Given that:} \\ a \odot b; \quad \color{blue}{b \odot c} \implies \color{green}{a \odot c} \\ \text{If } c \odot d, \text{ it is concluded that:} \\ \color{green}{a \odot c}; \quad c \odot d \implies a \odot d \\ \color{blue}{b \odot c}; \quad c \odot d \implies b \odot d \\ \text{Therefore:}\\ a \odot b; \quad b \odot c; \quad c \odot d \implies a \odot b \odot c \odot d \]

With the same line of reasoning, the chain rule can be extended to any number of terms. All what we need to prove is that \(a \odot b; \: b \odot c \implies a \odot c\), and it follows for any number of terms.

\[ \cssId{equation3.7}{ \begin{equation} \tag{3.7} \color{blue}{ \text{If } a \odot b \text{ is defined, and} \\ a \odot b; \: b \odot c \implies a \odot c \\ \text{then, for all } a_n \\ a_1 \odot a_2; \quad a_2 \odot a_3; \quad \dotsc ; \quad a_{n-1} \odot a_n \implies a_1 \odot a_2 \odot \dotsb \odot a_n } \end{equation} } \]

For the definition of divisibility, if \(a \mid b\), then \(b = a \times q\); and if \(b \mid c\), then \(b = c \times q^*\). Therefore, we can substitute for \(b\) in the second equation:

\[ a \mid b \iff b = a \times q \\ b \mid c \iff c = b \times q^* \\ c = (a \times q) \times q^* \\ c = a \times (q \times q^*) \quad \text{from } \href{http://math-from-the-ground-up.blogspot.ca/2016/09/multiplication.html#equation2.10}{2.10} \\ \therefore a \mid c \quad \text{from } \href{#equation3.1}{3.1} \]

Since, we have proved that if \(a \mid b\) and \(b \mid c\) then \(a \mid c\), divisibility follows the chain rule for all values on which it is defined.

\[ \cssId{equation3.8}{ \begin{equation} \tag{3.8} \color{blue}{ \text{For } a,b,c \in \mathbb{Z}^+_0; \quad a,b \neq 0 \\ a \mid b; \quad b \mid c \implies a \mid c } \end{equation} } \]

Note that the ‘direction of the operation or relation is important for the chain rule. If \(a \mid b\) is not symmetric (i.e. the synonym for commutative, but for relations,) then \(a \mid b\) and \(a \mid c\) does not mean that \(b \mid c\). For the chain rule to apply, the second term in a relation or operation has to be the same first term in the following relation or operation. Because \(a \mid b\) does not imply that \(b \mid a\), the conclusion that \(a \mid c\) in the previous example is wrong. A numeric example makes things more obvious. It is true that \(2 \mid 6\) and that \(2 \mid 8\), but it is not true that \(6 \mid 8\). If the order is observed, though, the chain rule holds. It is true that \(3 \mid 6\) and that \(6 \mid 30\), and it is also true that \(3 \mid 30\).

Reflexive?

We showed that \(a \mid b\) does not imply that \(b \mid a\). But is it always the case that the two statements cannot be said to be true at the same time? In other words, can \(a \mid b\) and \(b \mid a\) be true at the same time? Let us examine what it would mean if divisibility is reflexive for two numbers \(a\) and \(b\).

\[ \text{For } a,b,q,q^* \in \mathbb{Z}^+_0; \quad a,b \neq 0 \\ a \mid b \iff b = a \times q \\ b \mid a \iff a = b \times q^* \\ \text{Substituting:} \\ a = (a \times q) \times q^* \\ a = a \times (q \times q^*) \quad \text{from }\href{http://math-from-the-ground-up.blogspot.ca/2016/09/multiplication.html#equation2.10}{2.10} \\ q \times q^* = 1 \quad \text{from }\href{http://math-from-the-ground-up.blogspot.ca/2016/09/multiplication.html#equation2.6}{2.6} \\ q = q^* = 1 \\ \therefore a = b \]

We concluded in the next-to-last line above that \(q=q^*=1\) because both \(q\) and \(q^*\) are non-negative whole numbers by definition, and there is no value for \(q\) and \(q^*\) that can satisfy \(q \times q^*=1\) other than one itself. This means that if two numbers divide each other, they have to be equal! And because of \(\href{#equation3.6}{3.6}\), if they are equal, then they divide each other; i.e. the relationship goes both ways.

\[ \cssId{equation3.9}{ \begin{equation} \tag{3.9} \color{blue}{ \text{For } a,b \in \mathbb{Z}^+_0; \quad a,b \neq 0 \\ a \mid b; \quad b \mid a \iff a = b } \end{equation} } \]

If I divide you, I divide lots of you

If a positive integer \(a\) divides a non-negative integer \(b\), then it also divides \(b\) multiplied by any integer \(c\). Conceptually, it is easy to imagine how this happens. If you can divide the candy in a certain package into equal groups of, say, eight pieces, then you can also divide the candy in any number of packages into groups of the same size (provided that all packages have the same number of pieces of candy.) Mathematically, the proof follows.

\[ \text{For } a,b,c,q \in \mathbb{Z}^+_0; \quad a \neq 0 \\ a \mid b \iff b = a \times q \\ \begin{aligned} b \times c & = a \times q \times c \\ & = a \times (q \times c) \quad \text{from } \href{http://math-from-the-ground-up.blogspot.ca/2016/09/multiplication.html#equation2.10}{2.10} \\ \end{aligned} \\ \therefore a \mid b \times c \quad \text{from } \href{#equation3.1}{3.1} \] \[ \cssId{equation3.10}{ \begin{equation} \tag{3.10} \color{blue}{ \text{For } a,b,c \in \mathbb{Z}^+_0; \quad a \neq 0 \\ a \mid b \implies a \mid (b \times c) } \end{equation} } \]

If I divide each of you, I divide both of you together

If a positive integer \(a\) divides a non-negative integer \(b\) and another non-negative integer \(c\), then \(a\) divides the sum or difference of \(b\) and \(c\). Again, this is conceptually easy to imagine, just like the previous candy example, especially when we recall that multiplication is essentially repeated addition in a pattern. The mathematical proof follows.

\[ \text{For } a,b,c,q,q^* \in \mathbb{Z}^+_0; \quad a \neq 0 \\ a \mid b \iff b = a \times q \\ a \mid c \iff c = a \times q^* \\ \begin{aligned} b \pm c & = a \times q \pm a \times q^* \\ &= a \times (q \pm q^*) \quad \text{from } \href{http://math-from-the-ground-up.blogspot.ca/2016/09/multiplication.html2.11}{2.11} \end{aligned} \\ \therefore a \mid (b \pm c) \quad \text{from } \href{#equation3.1}{3.1} \] \[ \cssId{equation3.11}{ \begin{equation} \tag{3.11} \color{blue}{ \text{For } a,b,c \in \mathbb{Z}^+_0; \quad a \neq 0 \\ a \mid b; \quad a \mid c \implies a \mid (b \pm c) } \end{equation} } \]

It is easy to show that \(\href{#equation3.11}{3.11}\) extends to involve as many dividends as you want.

\[ \text{For } n \in \mathbb{N}; \quad a, b_n \in \mathbb{Z}^+_0; \quad a \neq 0 \\ a \mid b_1; \quad a \mid b_2 \implies a \mid (b_1 \pm b_2) \\ a \mid (b_1 \pm b_2); \quad a \mid b_3 \implies a \mid \left( (b_1 \pm b_2) \pm b_3 \right) \\ a \mid \left( (b_1 \pm b_2) \pm b_3 \right); \quad a \mid b_4 \implies a \mid \left( \left( (b_1 \pm b_2) \pm b_3 \right) \pm b_4 \right) \\ \vdots \\ \]

Since both addition and subtraction are left-associative, as we have shown before, parentheses here have no added value, and they clutter the expression.

\[ \cssId{equation3.12}{ \begin{equation} \tag{3.12} \color{blue}{ \text{For } n \in \mathbb{N}; \quad a, b_n \in \mathbb{Z}^+_0; \quad a \neq 0 \\ a \mid b_1; \quad a \mid b_2; \quad \dotsc ; \quad a \mid b_n \implies a \mid (b_1 \pm b_2 \pm \dotsb \pm b_n) } \end{equation} } \]

Hell, yeah! I can divide lots of you put together

If a positive integer \(a\) divides a non-negative integer \(b\) and another non-negative integer \(c\), then it also divides the sum or the difference of \(b\) multiplied by an integer and \(c\) multiplied by an integer. This is a direct consequence of what we proved so far.

\[ \text{For } a,b,c,s,t \in \mathbb{Z}^+_0; \quad a \neq 0 \\ a \mid b \implies a \mid (s \times b) \quad \text{from } \href{#equation3.10}{3.10} \\ a \mid c \implies a \mid (t \times c) \quad \text{from } \href{#equation3.10}{3.10} \\ a \mid (s \times b); \quad a \mid (t \times c) \implies a \mid (s \times b \pm t \times c) \quad \text{from } \href{#equation3.11}{3.11} \\ \] \[ \cssId{equation3.13}{ \begin{equation} \tag{3.13} \color{blue}{ \text{For } a,b,c,s,t \in \mathbb{Z}^+_0; \quad a \neq 0 \\ a \mid b; \quad a \mid c \implies a \mid (s \times b \pm t \times c) } \end{equation} } \]

Extending \(\href{#equation3.13}{3.13}\) to involve as many dividends as you want is not difficult. It actually directly follows from \(\href{#equation3.13}{3.13}\) and \(\href{#equation3.12}{3.12}\).

\[ \cssId{equation3.14}{ \begin{equation} \tag{3.14} \color{blue}{ \text{For } n \in \mathbb{N}; \quad a,b_n,k_n \in \mathbb{Z}^+_0; \quad a \neq 0 \\ a \mid b_1; \quad a \mid b_2; \quad \dotsc ; \quad a \mid b_n \implies a \mid (b_1k_1 \pm b_2k_2 \pm \dotsb \pm b_nk_n) } \end{equation} } \]

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