Ordinal arithmetic proof: $a,b <\omega^2$ $\implies$ $a+b<\omega^2$

Question

i need to prove the following

If $a$, $b$ are ordinals such that $a,b <\omega^2$ $\implies$ $a+b<\omega^2$.

I think it could be done using injective functions but have no idea on how to do it.

Answer 1

Hint: Any ordinal that is strictly smaller than $\omega^2$ can be written as $$ \omega\cdot m + n $$ where $m, n$ are natural numbers.

Answer 2

Developing Arthur's idea, $a,b < \omega^2$ implies $a = \omega \cdot m_1 + n_1$ and $b = \omega \cdot m_2 + n_2$, with $m_1,m_2,n_1,n_2 \in \mathbb{N}$. In consequence, one has : $$ a + b = \omega \cdot m_1 + n_1 + \omega \cdot m_2 + n_2 = \omega \cdot (m_1 + m_2) + n_2 < \omega^2. $$

Alternatively, you may argue straightforwardly $a + b < a + \omega^2 = \omega^2$, where the first inequality is due to $b < \omega^2$ and the last equality to $a < \omega^2$ respectively.