How to prove that $x=y=1$ or $x=y=−1$ if $xy=1$?

Question

I'm struggling with my current home assignment. If $x,y\in \mathbb{Z}$ and $xy=1$ how to prove that $x=y=1$ or $x=y=-1$? To achieve this, only the following properties can be used.

$\forall x,y,z\in \mathbb{Z}:$

1. $(x+y)+z=x+(y+z)$;

2. $x+y=y+x$;

3. $x+0=x$;

4. $x+(-x)=0$;

5. $(xy)z=x(yz)$;

6. $xy=yx$;

7. $x\cdot 1=x$;

8. $x(y+z)=xy+xz$;

9. $xy=0 \implies x=0 \lor y=0$;

10. $x<y \veebar x=y \veebar x>y$;

11. $x<y \land y<z \implies x<z$;

12. $x<y \implies x+z<y+z$;

13. $x<y \land z>0 \implies xz<yz$;

14. $0 \neq 1$;

15. $x+z=y+z \implies x=y$;

16. $-(-x)=x$;

17. $-(x+y)=(-x)+(-y)$;

18. $x\cdot 0=0$;

19. $z\neq 0 \land xz=yz \implies x=y$;

20. $(-x)y=-xy=x(-y)$.

A hint would be greatly appreciated.

As far as I understand, this is an exercise from this book: The Real Numbers and Real Analysis, Bloch, Ethan D.

Answer 1

If $xy = 1$ with elements in a commutative ring, then $x,y$ are invertible in $R$.

In the ring of integers, the invertible elements are $\pm 1$.

Answer 2

You might try this: Let's prove it first, for $x>0,y>0$, later we'll prove the general case. Fix $x>1,x\in\mathbb{N}$ then we'll prove by induction $\nexists y\in\mathbb{N}, xy=1$, that is consider $A=\{y\in\mathbb{N}:xy>1 \}$

1)$1\in A$ Clearly, $x1=x>1$

2) Suppose $n\in A$, then $(n+1)x=nx+1>1+1>1$

Therefore, by the principle of induction $A=\mathbb{N}$

You have to prove now that if $x=1$ then the only solution is $y=1$, I believe the reasoning is very similar as above.

Now, clearly $x$ and $y$ have the same sign, otherwise if $x>0,y<0$ by the rules you said we can use $xy=yx<0<1$. So if $x<0\implies y<0$, choose $z=-x,z'=-y$ and you are in the above proved case.

Hopefully I made myself clear, otherwise feel free to ask and edit whatever you want. Have a nice day! :)

Answer 3

First note that $x, y$ must be non-zero, and have the same sign.

Then note that if $x, y$ is a solution, so is $-x, -y$, as $x y = (-x) \cdot (-y)$. Thus we may assume $x, y > 0$.

Now note that if $x > 1$, then $1 = x y > y > 0$, but there are no integers between $0$ and $1$.