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I want to go about doing it instead of like putting numbers in. So the professor did it like this:

$\textbf{Proof:}$

Since $x>0$, let $x$ = $5$

Then $\frac{1}{5}$ > $0$

Proof End.

But I don't know, I don't find it satisfying enough I suppose. Is there actually another way to go about proving it instead of doing it like this or am I just being weird.

Arkilo
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    There are more $x$ such that $x>0$ than just the number $5$. Checking that an example is true is not the same as a proof. To give a proper proof of the proposition, we actually need to know axioms you're working based off. – Thorgott Feb 11 '20 at 11:37
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    Does this question help? – lioness99a Feb 11 '20 at 11:39
  • No that is a bit too rigorous for a non math major – Arkilo Feb 11 '20 at 11:45
  • If you are trying to prove this statement for an abstract algebra course, you will need to prove the fact that $1>0$, and then, by the axioms of a fraction field and the order axioms of integers, you will get your answer. – justadzr Feb 11 '20 at 11:49
  • Does your math teachers have ever stud math in his life ? Such a proof from a "professional" is quite alarming... – Todd Feb 11 '20 at 21:04

2 Answers2

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Suppose $\dfrac{1}{x} \le 0 \implies 1 = x\cdot \dfrac{1}{x} \le x\cdot 0 = 0 \implies 1\le 0$, absurd! .

DeepSea
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This depends on the axioms you have, but in case you'll have to apply the rule of signs (possibly having to prove it first):

If $x$ is positive and $\frac 1 x$ negative, then, by the rule of signs $x\cdot\frac1x=1$ is negative. However, $1$ is positive. Therefore it is impossible.

J. W. Tanner
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Bernard
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