The Ring of Integers

Elementary number theory is largely about the ring of integers, denoted by the symbol $\integer$ . The integers are an example of an algebraic structure called an integral domain. This means that $\integer$ satisfies the following axioms:

(a) $\integer$ has operations + (addition) and $\cdot$ (multiplication). It is closed under these operations, in that if $m, n \in \integer$ , then $m + n \in \integer$ and $m\cdot n \in \integer$ .

(b) Addition is associative: If $m, n, p \in \integer$ , then

$$m + (n + p) = (m + n) + p.$$

(c) There is an additive identity $0 \in \integer$ : For all $n \in \integer$ ,

$$n + 0 = n \quad\hbox{and}\quad 0 + n = n.$$

(d) Every element has an additive inverse: If $n \in \integer$ , there is an element $-n \in \integer$ such that

$$n + (-n) = 0 \quad\hbox{and}\quad (-n) + n = 0.$$

(e) Addition is commutative: If $m, n \in \integer$ , then

$$m + n = n + m.$$

(f) Multiplication is associative: If $m, n, p \in \integer$ , then

$$m\cdot (n\cdot p) = (m\cdot n)\cdot p.$$

(g) There is an multiplicative identity $1 \in \integer$ : For all $n \in \integer$ ,

$$n\cdot 1 = n \quad\hbox{and}\quad 1\cdot n = n.$$

(h) Multiplication is commutative: If $m, n \in \integer$ , then

$$m\cdot n = n\cdot m.$$

(i) The Distributive Laws hold: If $m, n, p \in \integer$ , then

$$m\cdot (n + p) = m\cdot n + m\cdot p \quad\hbox{and}\quad (m + n)\cdot p = m\cdot p + n\cdot p.$$

(j) There are no zero divisors: If $m, n \in \integer$ and $m\cdot n = 0$ , then either $m = 0$ or $n = 0$ .

Remarks.

(a) As usual, I'll often abbreviate $m\cdot n$ to $mn$ .

(b) The last axiom is equivalent to the Cancellation Property: If $a, b, c \in \integer$ , $a \ne 0$ , and $ab = ac$ , then $b = c$ .

Example. If $n \in \integer$ , prove that $0\cdot n = 0$ .

$$\matrix{0 \cdot n & = & (0 + 0)\cdot n & \quad\hbox{(Additive identity)} \cr & = & 0\cdot n + 0\cdot n & \quad\hbox{(Distributive Law)} \cr}$$

Adding $-(0\cdot n)$ to both sides, I get

$$-(0\cdot n) + 0\cdot n = -(0\cdot n) + (0\cdot n + 0\cdot n).$$

By associativity for addition,

$$-(0\cdot n) + 0\cdot n = (-(0\cdot n) + 0\cdot n) + 0\cdot n.$$

Then using the fact that $-(0\cdot n)$ and $0 \cdot n$ are additive inverses,

$$0 = 0 + 0\cdot n.$$

Finally, 0 is the additive identity, so

$$0 = 0\cdot n.\quad\halmos$$

Example. If $n \in \integer$ , prove that $-n = (-1)\cdot n$ .

In words, the equation says that the additive inverse of n (namely $-n$ ) is equal to $(-1)\cdot n$ . What is the additive inverse of n? It is the number which gives 0 when added to n.

Therefore, I should add $(-1)\cdot n$ and see if I get 0:

$$\matrix{(-1)\cdot n + n & = & (-1)\cdot n + 1\cdot n & \quad\hbox{(Multiplicative identity)} \cr & = & (-1 + 1)\cdot n & \quad\hbox{(Distributive Law)} \cr & = & 0\cdot n & \quad\hbox{(Additive inverse)} \cr & = & 0 & \quad\hbox{(Preceding result)} \cr}$$

By the discussion above, this proves that $-n = (-1)\cdot n$ .

The integers are ordered --- there is a notion of greater than (or less than). Specifically, for $m, n \in \integer$ , $m > n$ is defined to mean that $m - n$ is a positive integer --- and element of the set $\{1, 2, 3, \ldots\}$ .

Of course, $m < n$ is defined to mean $n > m$ . $m \ge n$ and $m \le n$ have the obvious meanings.

There are two order axioms:

(k) The positive integers are closed under addition and multiplication.

(l) ( Trichotomy) If $n \in \integer$ , either $n > 0$ , $n < 0$ , or $n = 0$ .

Example. Prove that if $m > 0$ and $n < 0$ , then $mn < 0$ .

Since $n < 0$ , $0 - n = -n$ is a positive integer. $m > 0$ means $m = m - 0$ is a positive integer, so by closure $m\cdot (-n)$ is a positive integer.

By a property of integers (which you should try proving from the axioms), $m\cdot (-n) = -(mn)$ . Thus, $-(mn)$ is a positive integer. So $0 - mn = -(mn)$ is a positive integer, which means that $0 > mn$ .

The Well-Ordering Property of the integers sounds simple: Every nonempty subset of the positive integers has a smallest element. Your long experience with the integers makes this principle sound obvious. In fact, it is one of the deeper axioms for $\integer$ ; for example, it can be used to proved the principle of mathematical induction, which I'll discuss later.

Example. Prove that $\root 3 \of 2$ is not a rational number.

The proof will use the Well-Ordering Property.

I'll give a proof by contradiction. Suppose that $\root 3 \of 2$ is a rational number. In that case, I can write $\root 3 \of 2 = \dfrac{a}{b}$ , where a and b are positive integers.

Now

$$\root 3 \of 2 = \dfrac{a}{b}, \quad\hbox{so}\quad b\root 3 \of 2 = a, \quad\hbox{and}\quad 2b^3 = a^3.$$

(To complete the proof, I'm going to use some divisibility properties of the integers that I haven't proven yet. They're easy to understand and pretty plausible, so this shouldn't be a problem.)

The last equation shows that 2 divides $a^3$ . This is only possible if 2 divides a, so $a = 2c$ , for some positive integer c. Plugging this into $2b^3 = a^3$ , I get

$$2b^3 = 8c^3, \quad\hbox{or}\quad b^3 = 4c^3.$$

Since 2 divides $4c^3$ , it follows that 2 divides $b^3$ . As before, this is only possible if 2 divides b, so $b = 2d$ for some positive integer d. Plugging this into $b^3 = 4c^3$ , I get

$$8d^3 = 4c^3, \quad\hbox{or}\quad 2d^3 = c^3.$$

This equation has the same form as the equation $2b^3 = a^3$ , so it's clear that I can continue this procedure indefinitely to get e such that $c = 2e$ , f such that $d = 2f$ , and so on.

However, since $a = 2c$ , it follows that $a > c$ ; since $c = 2e$ , I have $c > e$ , so $a > c > e$ . Thus, the numbers a, c, e, ... comprise a set of positive integers with no smallest element, since a given number in the list is always smaller than the one before it. This contradicts Well-Ordering.

Therefore, my assumption that $\root 3 \of 2$ is a rational number is wrong, and hence $\root 3 \of 2$ is not rational.

Finally, I want to mention a function that comes up often in number theory.

Definition. If x is a real number, then $[x]$ denotes the greatest integer function of x. It is the largest integer less than or equal to x.

Lemma. If x is a real number, then

$$[x] + 1 > x \ge [x].$$

Proof. By definition, $x \ge [x]$ . To show that $[x] + 1 > x$ , I'll give a proof by contradiction.

Suppose on the contrary that $[x] + 1 \le x$ . Then $[x] + 1$ is an integer less than or equal to x, but $[x] + 1 > [x]$ --- which contradicts the fact that $[x]$ is the largest integer less than or equal to x. This contradiction implies that $[x] + 1 > x$ .

Lemma. If $x, y \in \real$ and $x \ge y$ , then $[x] \ge [y]$ .

Proof. Suppose $x \ge y$ . I want to show that $[x] \ge [y]$ .

Assume on the contrary that $[y] > [x]$ . Since $[x]$ is the {\it greatest} integer which is less than or equal to x, and since $[y]$ is an integer which is greater than $[x]$ , it follows that $[y]$ can't be less than or equal to x. Thus, $[y] > x$ . But $x \ge y$ , so $[y] > y$ , which is a contradiction.

Therefore, $[x] \ge [y]$ .

Example.

$$[3.2] = 3, \quad [117] = 117, \quad\hbox{and}\quad [-1.2] = -2.$$

(Notice that $[-1.2]$ is not equal to -1.)

Example. Let x be a real number and let n be an integer. Prove that $[x + n] = [x] + n$ .

First, $x \ge [x]$ , so $x + n \ge [x] + n$ . Now $[x] + n$ is an integer less than or equal to $x + n$ , so it must be less than or equal to the greatest integer less than or equal to $x + n$ --- which is $[x + n]$ :

$$[x + n] \ge [x] + n.$$

Next, $x + n \ge [x + n]$ , so $x \ge [x + n] - n$ . $[x + n] - n$ is an integer less than or equal to x. Therefore, it must be less than or equal to the greatest integer less than or equal to x --- which is $[x]$ :

$$[x] \ge [x + n] - n.$$

Adding n to both sides gives

$$[x] + n \ge [x + n].$$

Since $[x + n] \ge [x] + n$ and $[x] + n \ge [x + n]$ , it follows that $[x] + n = [x + n]$ .

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Last updated: June 7, 2008

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