# A high school level proof that $a/b > 0$ [closed]

Is there a high school level proof of the following?

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

• Hi, could you be more explicit about the connection to teaching mathematics and what you know already? Jan 15 at 8:20
• 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. Jan 15 at 14:34
• I'm pretty sure that the answer to this question is "yes", but this question is not about math education. Jan 16 at 15:51

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 $$\frac ab > 0$$.
If $$b>0$$, then $$\frac{1}{b}>0$$. Product of two positives is positive.
• If $b>0$, then $\frac{1}{b}>0$. Why is that? Jan 23 at 10:12