# Ordinal arithmetic inequality

I have 3 ordinals $\alpha,\beta,\gamma$, $\gamma\neq 0$. Is this implication true?

$$\alpha<\beta\implies\gamma\cdot\alpha<\gamma\cdot\beta$$

I have reason to believie it is not, namely because I can't prove it by induction. (I get stuck in the inductive step as $(\gamma+1)\cdot\alpha\neq\gamma\cdot\alpha+\alpha$). Nevertheless I can't find any counterexample, so perhaps it is true.

Got any hints?

• You actually can prove it by induction. Alternatively note that $\rho \colon \operatorname{Ord} \to \operatorname{Ord}, \alpha \mapsto \gamma \cdot \alpha$ is increasing and $\gamma \cdot \alpha < \gamma \cdot (\alpha + 1)$. – Stefan Mesken Mar 12 '16 at 12:24
• And it's even easier without induction. – Asaf Karagila Mar 12 '16 at 12:26
• (Also the induction shouldn't be on $\gamma$, it should be on $\alpha$ or $\beta$.) – Asaf Karagila Mar 12 '16 at 12:27

Note that $\gamma\cdot\alpha$ is isomorphic to the lexicographic ordering on $\alpha\times\gamma$.
It is very easy to conclude (via inclusion of the products) that $\gamma\cdot\alpha\leq\gamma\cdot\beta$. All you have to show now is that if $\alpha\neq\beta$ then $\gamma\cdot\alpha$ is a proper initial segment of $\gamma\cdot\beta$. That much is easy since $\beta\setminus\alpha\neq\varnothing$.
If you insist on using induction, you should do the induction on $\alpha$ or $\beta$ and not on $\gamma$. Namely, for a fixed non-zero $\gamma$, if $\alpha<\beta$ then $\gamma\cdot\alpha<\gamma\cdot\beta$. It might be the easier solution to do the induction on $\beta$.