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I'm not sure how to prove that if $xy > 0$ then $x > 0$ and $y > 0$ or $x < 0$ and $y < 0$ just by using the ordered field axioms. Can someone perhaps help me see this through?

lithium123
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2 Answers2

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Since $xy > 0$, $x \neq 0$, and $y \neq 0$. So if $x > 0$, then if $y < 0$, then $xy < 0(y) = 0$, contradiction. Thus $y > 0$. You can prove the other case similarly.

DeepSea
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In an ordered field, one of the axioms is that for every $x$, either $x=0$, $x>0$ or $x<0$. So given $x$ and $y$ we have a few cases:

  1. $x=0$ or $y=0$;
  2. $x>0$ and $y<0$;
  3. $x<0$ and $y>0$;
  4. $x>0$ and $y>0$;
  5. $x<0$ and $y<0$.

In case $1$ we get $xy=0$, in cases 2. and 3. we get $xy<0$ and in cases 4. and 5 we get $xy>0$.

Therefore, the only possibility for $xy>0$ is if we are in either case 4. or 5., which is the implication you want.

Luiz Cordeiro
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