# Intro to proof in real analysis 1

This is what I have to prove: For elements $x, y$ in an ordered field, if $0 < x < y$ then $y^{-1} < x^{-1}.$

My proof: $0 < x < y$
multiply $x^{-1}$ on the left of both sides to get $x^{-1}0 < x^{-1}x < x^{-1}y$
$0 < 1 < x^{-1}y$
then multiply $y^{-1}$ to the right of both sides to get $0y^{-1} < 1y^{-1} < x^{-1}yy^{-1}$
And then we get $0 < y^{-1} < x^{-1}$.

Are there any problems with this proof?

-
By the looks of it, your proof is fine. The only improvement I could suggest is to use the slightly more technical terms of "pre" and "post" multiplying, rather than saying "multiplying on the left/right side". – Andrew D Sep 8 '13 at 21:57
Yes, that seems correct. – Jiri Sedlacek Sep 8 '13 at 22:06
It is correct, but be aware this only wors because of $x,y > 0 \Leftrightarrow x^{-1}, y^{-1} > 0$. – AlexR Sep 8 '13 at 22:13
Do I have to assume that x^-1 > 0 and y^-1 > 0 for the proof to work? – user87274 Sep 8 '13 at 22:17

As stated in the comment by AlexR, the crucial part of the argument is to notice that because $x,y>0$ it follows that $x^{-1},y^{-1}>0$. Indeed, since $1>0$, and since $x\cdot x^{-1}=1$, it follows that $x$ and $x^{-1}$ are either both positive or both negative, and the claim follows.
Now the argument is simply multiplying the inequality $0<x<y$ by the positive element $x^{-1}y^{-1}$ (on the left or on the right, it doesn't matter since any field is commutative).