Solving $\omega +\gamma =\omega \cdot 2$ and dividing ${\omega }^2+\omega +3$ by $\omega$

Two computations with the two "inverse" operations of this page.

Left subtraction. The equation $\omega +\gamma =\omega \cdot 2$ has exactly one solution, and it is $\gamma =\omega$. Existence and uniqueness are For $\alpha \le \beta$ there is exactly one ordinal $\gamma$ with $\alpha +\gamma =\beta$, applicable because $\omega \le \omega \cdot 2$; finding the solution is the computation $\omega \cdot 2=\omega +\omega$.

Division with remainder. Dividing ${\omega }^2+\omega +3$ by $\omega$ gives

${\omega }^2+\omega +3=\omega \cdot (\omega +1)+3,3<\omega ,$

so the quotient is $\omega +1$ and the remainder is $3$, and by the uniqueness in For $\alpha >0$ every ordinal $\beta$ is $\alpha \cdot \xi +\rho$ with $\rho <\alpha$, in exactly one way there is no other answer. The step that does the work is ${\omega }^2+\omega =\omega \cdot (\omega +1)$, which is left distributivity.

Uniqueness is what makes "the answer" meaningful. Both computations exhibit a solution and then quote a uniqueness theorem. Without For $\alpha \le \beta$ there is exactly one ordinal $\gamma$ with $\alpha +\gamma =\beta$ the equation $\omega +\gamma =\omega \cdot 2$ would only be known to have a solution; without For $\alpha >0$ every ordinal $\beta$ is $\alpha \cdot \xi +\rho$ with $\rho <\alpha$, in exactly one way the pair $(\omega +1,3)$ would be an answer among possibly many. Both theorems are proved from left cancellation, which is the one cancellation law ordinal addition has.

The equation on the other side has no solution at all. There is no $\gamma$ with $\gamma +\omega =\omega +1$, because $\gamma +\omega$ is a limit ordinal for every $\gamma$ (claim (g) of Monotonicity of ordinal $+$ and $\cdot$: strictly increasing and continuous in the right argument, weakly increasing in the left, with left cancellation, and the identities $0+\beta =\beta$ and $1\cdot \beta =\beta$) while $\omega +1={\omega }^{+}$ is a successor. So ordinal subtraction genuinely exists only on the left, and the same asymmetry is what forces the quotient in For $\alpha >0$ every ordinal $\beta$ is $\alpha \cdot \xi +\rho$ with $\rho <\alpha$, in exactly one way to be written on the right of $\alpha$.

Reading the division off the Cantor normal form. ${\omega }^2+\omega +3$ has normal form with exponents $2>1>0$ and coefficients $1,1,3$ (Cantor normal form: every nonzero ordinal is ${\omega }^{{\beta }_0}\cdot c_0+\cdots +{\omega }^{{\beta }_{k-1}}\cdot c_{k-1}$ with ${\beta }_0>\cdots >{\beta }_{k-1}$ and each $c_i$ a nonzero natural number, in exactly one way). In this instance, dividing by $\omega ={\omega }^1$ has put the term of exponent $0$ into the remainder and lowered each of the two remaining exponents by one: ${\omega }^2$ became $\omega$ and $\omega$ became $1$, which is exactly the quotient $\omega +1$, while $3$ is the constant term. That pattern is what the general division algorithm is doing, but no general statement of it is claimed here.