# If $\gamma >0,$ then $\alpha < \beta$ implies that $\gamma \ . \alpha<\gamma \ . \beta$ and that $\alpha \ . \ \gamma \leq \beta \ . \ \gamma.$

Show that if $\gamma >0,$ then $\alpha < \beta$ implies that $\gamma \ . \alpha<\gamma \ . \beta$ and that $\alpha \ . \ \gamma \leq \beta \ . \ \gamma.$ (All operations are cardinal operations). Could anyone advise on the correct approach to this problem? And why $\leq$ cannot be replaced with $<?$

Thank you.

This can't be right for cardinal operations, because cardinal arithmetic is commutative (both addition and multiplication), and moreover $\gamma\cdot\alpha=\max\{\gamma,\alpha\}$.

If we take $\alpha,\beta,\gamma$ to be initial ordinals, which are often taken to be cardinals, then considering the above inequalities using ordinal arithmetics is true. This can be proved by induction on $\beta$.