The question: Show that is $\alpha \cdot \alpha = \beta \cdot \beta$, then $\alpha=\beta$.
This is my working, if $\alpha=0$ then the result follows.
Suppose $\alpha \gt 0$, then by the left cancellation law $\alpha = \beta$
But i do no think this is proof of the above question.
Could anyone help me with this? Thank you in advance.
Assume $\alpha>0$. Assume towards a contradiction that we have $\alpha\alpha=\beta\beta$ but $\alpha\neq\beta$, without loss of generality we can assume $\alpha<\beta$. When $\alpha>0$ ordinal multiplication is increasing in the second argument (source: J. Schlöder, "Ordinal Arithmetic"), so $\alpha\alpha<\alpha\beta$. I believe we can prove ordinal multiplication is monotonic in the first argument, i.e. for ordinals $\gamma,\delta,\varepsilon$ we have $\gamma\varepsilon\le\delta\varepsilon$. (I started a proof of this by transfinite induction on the second argument, but unfortunately deleted what I had after thinking it wasn't useful. It was straightforward up to the limit case) If so then $\alpha\alpha<\alpha\beta\le\beta\beta=\alpha\alpha$, a contradiction.