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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?

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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
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Yes, that seems correct. –  Jiri Sedlacek Sep 8 '13 at 22:06
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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

1 Answer 1

up vote 0 down vote accepted

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).

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