3
$\begingroup$

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!

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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 $a=(q+1)b+r$.

$\endgroup$
2
$\begingroup$

The idea is that the set $S$ of integers of the form $\, a - n b\,$ is closed under subtraction by $b$ so we can continually subtract $b$ from $a$ till we reach an element $< b.\,$ Indeed, apply the following.

Lemma $\ $ If nonempty $S\subseteq \Bbb N\,$ is closed under subtraction $\!\ge 0\,$ by $\,b,\,$ i.e. $\,s\ge b\in S\Rightarrow\, s-b \in S\,$ then $S$ contains a natural $< b$

Proof $\ $ The least $\,\ell\in S$ must satisfy $\,\ell < b,\,$ else $\,\ell-b\in S$ would be a smaller element of $S$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.