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I'm reading, but I can't get what induction hypothesis they are talking about on page 10 on at Existence of q and r for nonnegative a and b. What do they mean?

Regards, Kevin

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The induction hypothesis is (indirectly) stated just before the line The theorem holds if $|a|=0$: ‘A proof follows that proceeds by induction on $|a|$’. That’s why the first step is to prove that the theorem is true when $|a|=0$: that’s the smallest possible value for $|a|$. That takes care of the basis step of the induction $-$ of ‘getting the induction off the ground’.

Let $P(a,b)$ be the statement that there are unique integers $q$ and $r$ such that $a=bq+r$, $|r|<|b|$, and $ar\ge 0$. The basis step showed that $P(a,b)$ is true for all $b\in\mathbb{Z}\setminus\{0\}$ when $|a|=0$. Now you assume that $|a|>0$ and that $P(a',b)$ is true for all $b\in\mathbb{Z}\setminus\{0\}$ whenever $|a'|<|a|$; that’s the induction hypothesis that they’re talking about.

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Very clear, I think I get it now :)! – Kevin Dec 1 '11 at 19:40

Here is your statement:

Let $a,b$ be non-negative integers. Then there exists $q,r$ so that $0 \leq r < b$ and $$a=b \cdot q +r \,.$$

This statement is proven by (complete) induction on $a$; this is your $P(a)$.

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I only learned induction on integers. Am I correct? I read this: Suppose it holds for a-b, then it also holds for a. Since b was arbitrary, it holds for all a (in Z). – Kevin Dec 1 '11 at 19:33
Yes, $a,b,q,r$ are integers in this induction statement. And yes, the inductive step is really $P(a-b) \Rightarrow P(a)$. – N. S. Dec 1 '11 at 20:41

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