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The exercise goes like this:

Prove the division theorem using strong induction. That is, prove that for $a \in \mathbb{N}$, $b \in \mathbb{Z}^+$ there always exists $q, r \in \mathbb{N}$ such that $a = qb + r$ and $r < b$. In particular, give a proof that does not use $P(n−1)$ to prove $P(n)$ when $b > 1$.

I have done a few proofs with strong induction before, but never with a predicate with multiple variables, so I'm unsure how to approach this.

One idea I had, was to use the following as my predicate:

$$P(a,b):= \exists r,q\in\mathbb{N}(a=b\cdot q+r)$$

and then use $\forall b \in \mathbb{Z}^+ .\forall i < a(P(i, b))$ as my first inductive hypothesis, and $\forall a \in \mathbb{N} .\forall i < b(P(a, i))$ as my second, proving them separately. But I'm not sure this is right, as I can't seem to prove it this way.

Am I even on the right track here? Any help would be much appreciated!

Bill Dubuque
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tobloef
  • 155

3 Answers3

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You're overthinking, in my opinion.

Do induction on $a$.

For $a=0$, the statement is true: $0=b0+0$ and $0<b$.

Suppose $a>0$ and that the statement holds for all $c<a$.

If $a<b$, then $a=b0+a$ and $a<b$. If $a\ge b$, then $a-b<a$, so $a-b=bq+r$ with $r<b$; since $a=b(q+1)+r$, we're done.

egreg
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2

You can prove it by strong induction on $a$.

For $a=0$, it is trivial.

Now, consider an arbitrary $a\in\mathbb N$ and assume that each $a'<a$ can be written as $qb+r$, with $r<b$. Now, if $a<b$, you can write $a$ as $0\times b+a$. Otherwise, consider $a-b$. By the induction hypothesis, it can be written as $bq+r$, with $r<b$. But then $a=(q+1)b+r$.

José Carlos Santos
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2

The key idea is that the set $S$ of integers of the form $\, a - n b\,$ is closed under subtraction by $\,b,\,$ hence we can repeatedly subtract $\,b\,$ from $\,a\,$ till we reach an element ("remainder") $\,r<b,\,$ i.e.

Lemma $\ $ If nonempty $\,S\subseteq \Bbb N\,$ is closed under nonnegative subtraction by $\,b,\,$ i.e. $\,s\in S,\,s\ge b\,\Rightarrow\, s-b \in S,\,$ then $S$ contains a natural $< b$.

Proof $\ $ The $\rm\color{#c00}{least}$ $\,\ell\in S\,$ satisfies $\,\ell < b\,$ (else $\,\ell-b\in S\,$ is $\rm\color{#c00}{smaller}$ than $\,\ell)$.

Remark $ $ We used a form of induction known as the $\it\color{#c00}{least}$ number principle, i.e. every nonempty subset $S$ of naturals has a least element (or, equivalently, $\Bbb N$ is well-ordered by magnitude). This can be viewed as a contrapositive form of strong induction (such descent arguments are often phrased in terms of a least counterexample (or minimal criminal), see this Remark).

See here for a proof of the Division Algorithm by normal (not complete or strong) induction.

Bill Dubuque
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