A University-Level Understanding of Number with an eye towards elementary education

Divisibility is a fundamental concept in mathematics. It describes one type of fundamental relationship between two integers, as we can think of a factor of a number being a "building block" of the number it is a factor of. Note that in this chapter, we will predominately study divisibility properties of whole numbers, but these can be easily extended to integers. From now on, unless otherwise stated, any number that appears is a whole number; that is, for all numbers \(n\) in this chapter,\(n \in \mathbb{N}_0\text{.}\)

In mathematics, the "divides" symbol is represented as "|" and the "does not divide" symbol is represented as "∤". For example, if \(b\) divides \(a\text{,}\) we write it as \(b | a\text{.}\) If \(b\) does not divide \(a\text{,}\) we write it as \(b ∤ a\text{.}\)

Let’s look at two examples. Consider the numbers 4 and 12. We can say that 4 divides 12, or in symbol form, \(4 | 12\text{.}\) This is because there exists an integer (3 in this case) such that \(4 * 3 = 12\text{.}\) Now consider the numbers 7 and 23. We can say that 7 does not divide 23, or in symbol form, \(7 ∤ 23\text{.}\) This is because there does not exist an integer such that \(7 * c = 23\text{.}\)

We know that \(3 \mid 12\) since \(12 =3 \times 4.\) If there is another number \(n\) for which we know \(12 \mid n\text{,}\) do we need to check whether \(3 \mid n\text{?}\)

INSERT PIC OF THIS DIVISIBILITY

Now that we’ve shown this, we can use it to help up study divisibility of many numbers. We’ll do many rules in the next subsection, but if we look at the example at the beginning of this section, we see that this shows the following:

Theorem 7.1.3.

If a number \(n\) is divisible by \(12\) then it is also divisible by \(3\text{.}\)

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We will talk about relation between two objects, like numbers, or polygons, or sets, or other things. Basically a relation between two objects is any statement involving two objects that is either TRUE or FALSE.

For example, if our objects are sets, then \(\in\) is a relation between them. For example, let \(A:=\{1,2\}, \quad B:=\{1,\{1,2\},3\}, \quad C:= \{1,2,3,4\}\text{.}\) Under the relation \(\in\text{,}\) \(A\) is related to \(B\) since \(A \in B\text{,}\) but \(A\) is not related to \(C\) since \(A \notin C\text{.}\) Divisibility is also a relation. We can say that under divisibility \(3\) is related to \(12\) since \(3 \mid 12\) but \(3\) is not related to \(10\) since \(3 \nmid 10\text{.}\)

This "chain"-like property of relations is incredibly important in mathematics. In fact, it comes up so often that we decided to give it it’s own name. In fact, we’ve talked about this property before; equality has it as well, as we’ve seen in PROPERTIES OF EQUALITY. We’ll talk about it in more generality here.

There are some special cases when considering divisibility. Firstly, any number is divisible by 1. This is because for any number \(a\text{,}\) \(a = 1 * a\text{.}\) Secondly, 0 is divisible by any number (except for 0 itself). This is because for any number \(b ≠ 0\text{,}\) \(0 = b * 0\text{.}\) However, 0 is not divisible by itself, as there is no number \(c\) such that \(0 = 0 * c\text{.}\) Lastly, any number \(a\) is divisible by itself, as \(a = a * 1\text{.}\)

If a number \(a\) is divisible by \(b\text{,}\) then \(a*n\) is divisible by \(b\) for any natural number \(n\text{.}\)

Lets try and prove this now. The idea is that if we know that \(c\) dots can be represented as a rectangle with dimensions \(a\) and \(b\text{,}\) then we can "glue" \(n\) of these together to make a larger rectangle with dimensions \(a\) and \(nb\text{.}\) INSERT PIC OF THIS HERE

Theorem 7.1.8.

If a number \(c\) is divisible by \(a\text{,}\) then \(nc\) is divisible by \(a\) for any natural number \(n\text{.}\)

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Proof.

Let’s use the diagram above as a guide. Assuming the if-part is true, we assume \(a \mid c\text{.}\) This means that \(c=ab\) for some natural number \(b\text{.}\) We need to show that the then-part is true; that is, \(nc=a( )\text{,}\) where the brackets contain some integer. Noting that the LHS of the second equation is the LHS of the first equation multiplied ny \(n\text{,}\) let’s multiply the first equation by \(n\text{:}\)

\begin{equation*} n \times (c) = n \times (ab). \end{equation*}

Using commutativity and associativity:

\begin{equation*} nc = a(nb). \end{equation*}

Since both \(n\) and \(b\) are natural numbers (we can write this as \(n,b \in \mathbb{N}\)), their product \(nb \in \mathbb{N}\text{.}\) Thus we’ve shown that the then-part must be true, and the theorem is proved.

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If a number \(a\) is divisible by \(c\) and a number \(b\) is divisible by \(c\text{,}\) then \(a+b\) and \(a-b\) are also divisible by \(c\text{.}\)

Let’s try and prove this. It’s similar to Theorem 1, and in fact, Theorem 1 can be looked at as a repeated application of Theorem 2. Let’s say we know that \(a\) is a factor of both \(b\) and \(c\text{.}\) Then we can express both \(b\) and \(c\) dots as a rectangle, with one of the dimensions being \(a\text{,}\) and the other we’ll call \(x\) and \(y\text{,}\) respectively. Then in the addition case, we can "glue" the two rectangles together to make a rectangle with dimensions \(a\) and \(b+c\text{.}\) INSERT PIC OF THIS

Theorem 7.1.10.

If a number \(b\) is divisible by \(a\) and a number \(c\) is divisible by \(a\text{,}\) and say that \(b \geq c\text{,}\) then \(b+c\) and \(b-c\) are also divisible by \(a\text{.}\)

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Proof.

We’ll only prove the case for addition, and leave the case for subtraction as an exercise later. Assuming the if-part is true, assume that \(b=ax\) and \(c=ay\) for some natural numbers \(x,y\text{.}\) We need to show that \(b+c=a( )\) where the bracket is some natural number. Adding both equations together,

\begin{equation*} b+c=ax+ay. \end{equation*}

Factoring out the a from the RHS,

\begin{equation*} b+c=a(x+y). \end{equation*}

Since \(x,y \in \mathbb{N}\) we know their sum \(x+y \in \mathbb{N}\) too. Thus we have shown the then-part must be true and the theorem is proved.

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If a number \(a\) is divisible by \(c\) but a number \(b\) is not divisible by \(c\text{,}\) then \(a+b\) and \(a-b\) are not divisible by \(c\text{.}\)

Like the other two theorems, let’s try and get an idea for a proof via a diagram first. This is a good opportunity for you to try to come up with a diagram that shows this situation before looking below. Try and draw this result, at least for the additive case, and then check if your idea is close to the picture hidden below:

INSERT HIDDEN PIC HERE, MAYBE BY AN EXERCISE OR SOMETHING.

It looks like when you "glue" the two rectangle-like objects together, the remainder that doesn’t make a full row in the non-divisible number is still there when the objects are added. This is essentially the entire idea.

Theorem 7.1.12.

If a number \(b\) is divisible by \(a\) but a number \(c\) is not divisible by \(a\text{,}\) then \(b+c\) and \(b-c\) are not divisible by \(c\text{.}\)

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Proof.

Again, we do the additive case, and leave the subtraction case to an exercise later. Let \(a \mid b\) and \(a \nmid c\text{.}\) This means that \(b=ax\) for some natural number \(x\text{,}\) and \(c=ay+r\text{,}\) where \(y\) is a natural number, and thinking about it as a remainder, \(r\) is a natural number such that \(r \leq a-1\text{.}\) We need to show that \(a \nmid b+c\text{,}\) so we need to show that \(b+c\) has a remainder. Adding both equations together, we get

\begin{equation*} b+c = ax+ay+r. \end{equation*}

Factoring out an \(a\) from the first two terms on the RHS,

\begin{equation*} b+c = a(x+y)+r. \end{equation*}

Since \(x+y \in \mathbb{N}\) and \(r \in \mathbb{N}\) with \(r \leq n-1\text{,}\) we have written \(b+c\) in quotient-remainder form, and we know there is exactly one way to do this. And since \(r \neq 0\text{,}\) then \(a \nmid b+c\text{.}\)

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If you look at the previous two theorems, you’ll see that they cover when \(a\) divides both numbers \(b\) and \(c\text{,}\) and when \(a\) divides exactly one of these and not the other. A natural question we can ask is "what happens if \(a\) divides neither \(b\) nor \(c\text{?}\)"