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Is there a high school level proof of the following?

If $a,b > 0$ then $a/b > 0.$

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    $\begingroup$ Hi, could you be more explicit about the connection to teaching mathematics and what you know already? $\endgroup$
    – Tommi
    Jan 15, 2023 at 8:20
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    $\begingroup$ Technically, if you start with the axioms of ordered field, you can immediately get it from the identity $(a/b)b=a$ arguing that two other possible cases $a/b=0$ and $a/b<0$ lead to a contradiction. You do not need to say the scary fancy words "field" and "axiom" here, but you still need to postulate something to be able to derive the statement unless you want to get mired in the intricacies of the formal construction of the real field from positive integers, which I certainly would not recommend at the high school level. $\endgroup$
    – fedja
    Jan 15, 2023 at 14:34
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    $\begingroup$ I'm pretty sure that the answer to this question is "yes", but this question is not about math education. $\endgroup$
    – Jasper
    Jan 16, 2023 at 15:51

2 Answers 2

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Suppose to the contrary that $\frac ab$ were negative. Then $b \cdot \frac ab$ would be the product of a positive and a negative number, which would be negative. But we know that $b \cdot \frac ab = a$, which is positive, so this can't be right. Similarly, if $\frac ab$ were zero, then $b \cdot \frac ab = a$ would have to be zero as well, which isn't true. So the only possibility left is that $\frac ab > 0$.

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  • $\begingroup$ This looks perfect to me. $\endgroup$
    – Simd
    Jan 23, 2023 at 13:48
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If $b>0$, then $\frac{1}{b}>0$. Product of two positives is positive.

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  • $\begingroup$ If $b>0$, then $\frac{1}{b}>0$. Why is that? $\endgroup$
    – Dominique
    Jan 23, 2023 at 10:12

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