# Proof that integers (except $1$ and $-1$) don't have an integer multiplicative inverse [closed]

This is a very elementary question, so please forgive me, but precisely because of this reason I haven't found an answer elsewhere:

How can it be proven that, among the integers, only $$1$$ and $$-1$$ have an integer multiplicative inverse? Or is that statement taken as an axiom?

Well, let's assume $$ab=1$$ when $$a,b\in\mathbb{Z}$$. Of course it means that both $$a$$ and $$b$$ are non zero. Because they are integers we can conclude that their absolute values are at least $$1$$. And then $$1=|ab|=|a||b|\geq |a|$$, and in the same way $$1\geq |b|$$. So both $$a$$ and $$b$$ must belong to the set $$\{-1,1\}$$.

It follows from the well-ordering principle for the natural numbers.

Suppose $$ab=1$$ for $$a,b \in \Bbb{N}$$ with $$b>1$$. Then $$c=a(b-1)$$ is a natural number with $$0. So $$\{c,c^2,c^3,\dots\}$$ is a set of natural numbers which has no least element, contradicting the well-ordering principle.

Now, if $$ab=1$$ for $$a,b \in \Bbb{Z}$$, then either $$a,b$$ or $$-a,-b$$ will fit into the proof in the above paragraph...

This is readily proved once you properly define $$\Bbb Z$$ and its operations. We may want to start from $$\Bbb N$$ with its successor map $$S$$ and the Peano axioms, then define addition $$\Bbb N\times \Bbb N\to\Bbb N$$ recursively as $$x+0:=x$$, $$x+Sy:=S(x+y)$$, and then define multiplication recursively as $$x\cdot 0=0$$, $$x\cdot Sy:=x\cdot y+y$$.

Now we see that $$(a,b)\sim (c,d):\iff a+d=b+c$$ is an equivalence relation on $$\Bbb N^2$$, define $$\Bbb Z:=\Bbb N^2/{\sim}$$ as the set of equivalence classes, define $$\overline{(a,b)}+\overline{(c,d)}:=\overline{(a+c,b+d)}$$ and define $$\overline{(a,b)}\cdot \overline{(c,d)}:=\overline{(a\cdot c+b\cdot d, a\cdot d+b\cdot c)}$$ (which involves showing that this is well-defined). The claim now boils down to $$\overline{(a,b)}\cdot\overline{(c,d)}=\overline{(1,0)}\implies a=Sb\lor b=Sa$$. It does take a bit of work, but by going back the way we came can ultimately be reduced to a statement about $$\Bbb N$$ that is provable by induction.

First of all, of course integers, have multiplicative inverses and of course that statement is obviously false as $$a = \frac 12, \frac 13, \frac 14,$$ etc. bear out.

What you mean to say is $$1$$ and $$-1$$ are the only two integers whose multiplicative inverses are also integers.

I.e. If $$a, \frac 1a \in \mathbb Z$$ then $$a = 1$$ or $$a = -1$$.

That is not an axiom.

If $$a \in \mathbb Z$$ then you have five possibilities. $$a > 1$$ or $$a = 1$$ or $$a = 0$$ or $$a = -1$$ or $$a < 1$$.

Case 1: $$a > 1$$. Then if $$\frac 1a < 0$$ we would have $$a*\frac 1a < a*0$$ and $$1 < 0$$ which is a contradiction. We can't have $$\frac 1a = 0$$ because $$0$$ has no multiplicative inverses so $$\frac 1a > 0$$. If $$\frac 1a \ge 1$$ then $$a*\frac 1a \ge a*1 = a$$ or $$1 \ge a$$ which is a contradiction so $$0< \frac 1a < 1$$ and not an integer.

Case 2: $$a = 1$$. then $$\frac 1a = 1$$.

Case 3: $$a = 0$$ then $$\frac 1a$$ is undefined.

Case 4: $$a = -1$$ then $$\frac 1a = -1$$.

Case 5: $$a < -1$$. that is so similar to case 1: I'll leave it to you.

• I don't agree with your first two paragraphs; at least not on a minimally generous reading of the question. Whether a number has a multiplicative inverse depends on which structure you're considering. In the ring of integers, non-units don't have multiplicative inverses. To say that "their inverses aren't integers" doesn't make sense within the structure; it only makes sense within the rationals. Though the question doesn't explicitly say "within the integers", it only talks about integers, and it stands to reason that the intended context is the ring of integers. – joriki Sep 23 '18 at 15:50
• I've made a clarificatory edit to the question's title (regarding the phrase "don't have a multiplicative inverse"). – Anakhand Sep 23 '18 at 16:00
• jariki. You are correct. But it is the burden of the OP to specify what structure is being used. – fleablood Sep 23 '18 at 16:05

It depends upon the way that $$\mathbb Z$$ and $$\times$$ are defined, of course, but I doubt that anyone ever created axioms for the integers such that what you stated is an axiom.

Suppos, for instance, that you define:

• $$\mathbb Z$$ is the set of equivalence classes of $$\mathbb{N}^2$$ with respect to the equivalence relation$$(a,b)\sim(c,d)\iff a+d=b+c;$$
• $$\bigl[(a,b)\bigr]\times\bigl[(c,d)\bigr]=\bigl[(ac+bd,ad+bc)\bigr]$$.

Then your statement becomes a proposition, which can in fact be proved.

As we are interested in 1/a which is just a symbol being adopted to denote the inverse of an element in Z. Further identity element under multiplication in Z is 1. So if any member say ''a' of Z is invertible then there must be a member in Z denoted as 1/a or a^(-1) such that a. a^(-1) =1. Now it's a simple algebra that in Z 1/a is an integer iff a= 1 or -1. Hope it will help you.