Let $\gamma,\alpha\neq 0$ be ordinals. Then there exists a unique ordinal $\beta$ and a unique $\rho<\alpha$ such that $\gamma=\alpha\cdot\beta+\rho$.
Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!
My attempt:
Let $\beta=\max\{\delta \in {\rm Ord} \mid\alpha\cdot\delta\le\gamma\}$. Since $\alpha\cdot\beta\le\gamma$, there is a unique $\rho$ such that $\gamma=\alpha\cdot\beta+\rho$. We have $\rho<\alpha$. If not, $\alpha\le\rho\implies\alpha\cdot(\beta+1)=\alpha\cdot\beta+\alpha\le\alpha\cdot\beta+\rho=\gamma\implies$ $\alpha\cdot(\beta+1)\le\gamma$. This contradicts the maximality of $\beta$.
To prove uniqueness, let $\gamma=\alpha\cdot\beta_1+\rho_1 =\alpha\cdot\beta_2+\rho_2$ with $\rho_1,\rho_2<\alpha$. Assume that $\beta_1<\beta_2\implies\beta_1+1\le\beta_2\implies\alpha\cdot(\beta_1+1)\le\alpha\cdot\beta_2$. Then $\alpha\cdot\beta_1+\rho_1<\alpha\cdot\beta_1+\alpha=$ $\alpha\cdot(\beta_1+1)\le\alpha\cdot\beta_2\le\alpha\cdot\beta_2+\rho_2\implies\alpha\cdot\beta_1+\rho_1<\alpha\cdot\beta_2\le\alpha\cdot\beta_2\implies\gamma<\gamma$. This is a contradiction and thus $\beta_1=\beta_2$. It follows immediately that $\rho_1=\rho_2$.