if $m \lt n$, then $mk < nk$.

Question

How can I prove the following for n,m and k natural numbers (k isn't equal to 0)

if $m \lt n$, then $mk < nk$.

Comments: I'll use induction for k. I have already proven with induction that: if $m \lt n$, then $m+k \lt n+k$

Answer 1

The trick is that for all $c\in\mathbb{R}$, $m<n \Rightarrow m+c<n+c$.

Base case $k=1$:

$m1 = m < n = n1$

Inductive hypothesis: Suppose $m\ell < n\ell$ for some $\ell \in \mathbb{N}$.

Then $m\ell + m < n\ell + m < n\ell + n$

$\Rightarrow m(\ell + 1) < n(\ell + 1)$.

By induction our statement is proven.