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I read the proof from here for convenience, I shall summarise the proof.

Claim: If $f(x)$ and $g(x)$ are two polynomials and $g(x)$ is not a zero polynomial, then there exist polynomials $q(x)$ and $r(x)$ such that $f(x)=q(x)g(x)+r(x)$ and either $r(x)$ is a zero polynomial or $\deg(r) < deg(g)$.

Proof:

If $\deg(f) < \deg(g)$ then this is true quite trivially.

If $\deg(f) \ge \deg(g)$, say $f(x)= \lambda x^n + h(x)$ and $g(x)=\mu x^m+k(x)$ where $\deg(h) < \deg(f) = n$ and $\deg(k) < \deg (g) = m$ and $n \ge m$.

Then $f(x)-(\frac{\lambda}{\mu})x^{m-n}g(x)$ has degree less than $n$, then by induction, this polynomial can be written in the form $g(x)s(x)+r(x)$ for some polynomials $r(x)$ and $s(x)$ with either $\deg(r) < \deg(g)$ or $r(x)$ identically zero and putting things together the claim is proven.

Question

Where is the induction in this proof? Because I always remember proof by induction by having a basis step first and where is the basis step in this proof? Then also, where is the inductive hypothesis (the assumption)?

Thanks in advance!

Bernard
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UnsinkableSam
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    The induction is on the degree of $f$, or on the difference of degrees of $f$ and $g$, if you prefer. – Bernard Jul 19 '19 at 14:38
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    To get the above base case using said degree difference use induction on the following "size" measure $,\delta(f,g):= \max{0,,1!+!\deg f! -! \deg g},,$ with base case $,\delta = 0\iff \deg f < \deg g,\ $ e.g. see here. Often this is done implicitly as in the question (i.e. a suitable size measure is not explicitly stated but is easy to infer). – Bill Dubuque Jul 19 '19 at 15:43

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