# Thread: Proof that x^2 = 1 implies x = 1 or x = -1

1. ## Proof that x^2 = 1 implies x = 1 or x = -1

I know this should be really easy, but I can't seem to figure it out. I don't really want an out an out answer if possible, just a starting point to work from.

"Let R be an integral domain, and suppose $x \in R$ satisfies $x^2 = 1$. Prove that we must either have x = 1 or x = -1."

I started off by saying that if we have:

$x \cdot x = 1$

Then x has an inverse, $x^{-1}$ where $x^{-1} = x$.

This means that x is also in the group of units of R, but then I just don't know where to go... part of me wants to ignore the commutative law and say something like:

$x^2 = 1$
$x \cdot x = 1$
$x \cdot 1 = 1 \cdot x^{-1}$

So by right multiplication then x = 1 and $x^{-1} = 1$. But that just seems off.

Thanks for any help.
--
Dave

2. Hmmm...

x^2=1 <==> x^2-1=0 <==> (x-1)(x+1)=0 <==> x=1 or x=-1

?

3. Originally Posted by Also sprach Zarathustra
Hmmm...

x^2=1 <==> x^2-1=0 <==> (x-1)(x+1)=0 <==> x=1 or x=-1

?
Judging by the type of question you'll probably have to prove $x^2 - y^2 = (x-y)(x+y)$ to justify the second step.

4. Originally Posted by Also sprach Zarathustra
Hmmm...

x^2=1 <==> x^2-1=0 <==> (x-1)(x+1)=0 <==> x=1 or x=-1

?
I'm afraid it's not as easy like that: Factorisation is not necessarily unique in a ID. For example Z/4Z is an ID but factorisation $(x-2)^2= x^2$ is not unique.
You probably need to use the fact in this case that roots of $x^2=1$ must be units.

5. Originally Posted by Dinkydoe
I'm afraid it's not as easy like that: Factorisation is not necessarily unique in a ID. For example Z/4Z is an ID but factorisation $(x-2)^2= x^2$ is not unique.
You probably need to use the fact in this case that roots of $x^2=1$ must be units.

Well, it is true that $x^2=1\Longrightarrow (x-1)(x+1)=0$ in any commutative ring, whether this decomposition is unique or not. So we're forced

to deduce that $(x-1)(x+1)=0$ and now we use that the ring is an ID to reach the conclusion.

And $\mathbb{Z}/4\mathbb{Z}$ is not an ID , since in it $2\cdot 2=0\,,\,but\,\,2\neq 0$

Tonio

6. Originally Posted by tonio
Well, it is true that $x^2=1\Longrightarrow (x-1)(x+1)=0$ in any commutative ring, whether this decomposition is unique or not. So we're forced

to deduce that $(x-1)(x+1)=0$ and now we use that the ring is an ID to reach the conclusion.

And $\mathbb{Z}/4\mathbb{Z}$ is not an ID , since in it $2\cdot 2=0\,,\,but\,\,2\neq 0$

Tonio
Ai, yes...I mixed up things. We must have $ab=0\Rightarrow a=0$, or $b=0$.

Thus now (x-1) =0, or (x+1)=0 gives the desired conclusion.

7. It is simple:
x^2=1
x^2-1=0
x^2+x-x-1=0
x(x+1)-(x+1)=0
(x+1)(x-1)=0
x=1 or x=-1

8. Thanks everyone for your help! I made this a lot more difficult for myself than I needed to.

I really appreciate all the responses.

--
Dave