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I need help proving this simple fact.

If $a,b,c \in \mathbb{Z}$, all $a,b,c>0$, and $a>b$, then $ca>b$.

How do I prove this?

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up vote 1 down vote accepted

Observe $cb \geq b$ iff $cb - b \geq 0$ iff $b(c-1) \geq 0$.

This last inequality holds, since $b$ and $c-1$ are both nonnegative.

Similarly, we have $ca > cb$ iff $ca - cb > 0$ iff $c(a - b) > 0$.

This last inequality holds, since $c$ and $a-b$ are both positive (the latter from $a>b$).

Putting it all together: $ca > cb \geq b$, hence $ca > b$. QED

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HINT: If $a, c > 0$ and $a, c \in \mathbb{Z}$, then $ca \geq a$.

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