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I am trying to prove the following statement from Terence Tao's Analysis 1 book.

Definition of multiplication $ab{+\!+} = ab+b$.

Definition of addition $(a{+\!+})+b=(a+b){+\!+}$.

Let $n$ be a natural number, and let $q$ be a positive number. Then there exists natural numbers $m$, $r$ such that $0 \le r \lt q$ and $n=mq+r$.

Now I attempted to prove the statement by inducting on $n$.

So for the base case $n=0$ $$0=mq+r$$ Since $q$ is positive $m$ and $r$ must be $$m=0$$ $$r=0$$ and we have $$0\le r \lt q$$ as desired.

Now assume true for $P(n)$ i.e. $$n=mq+r$$ and $$0\le r \lt q.$$

Now for $n{+\!+}$ (the successor to n) case.

Need to show $$n{+\!+}=mq+r.$$ Then using the induction hypothesis we have on the left hand side$$n{+\!+}=(mq+r){+\!+}=(mq{+\!+})+r=mq+q+r.$$ Then we use cancellation to arrive at $$mq+q+r=mq+r$$ $$q=0.$$ Which is not possible cause we assumed q is positive. So clearly I have done something wrong because the division algorithm is true.

Meow
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Andrew
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2 Answers2

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We use induction on n. For the base case n = 0 we have 0 = mq + r which implies that m = 0 and r = 0. Thus we have 0 = 0 + 0 and 0 ≤ 0 < q so the base case is done.

Now suppose that for some natural number n we have found natural numbers m, r such that 0 ≤ r < q and n = mq + r. Then n + 1 = mq + r + 1. From the induction hypothesis we know that r < q. Then r + 1 ≤ q. We have two cases: r + 1 < q or r + 1 = q. If r + 1 < q, then we are done since n + 1 = mq + (r + 1) and 0 ≤ r + 1 < q. On the other hand if r + 1 = q, then n + 1 = mq + q so n + 1 = (m+1) q + 0 and 0 ≤ 0 < q. We can close the induction.

J. Perica
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Increment and "$\color{#c00}{\rm carry}$" if need be $\ n\!+\!1 = mq+r\!+\!1\,$ if $\, r < q-1\,$ else $\ (m\!+\!\color{#c00}1)q+0\,$ if $\,r = q-1$

Bill Dubuque
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  • Are you saying that I should be trying to prove $n++=(mq+r)++$? Because I was inducting on $n$. – Andrew Aug 07 '14 at 02:34
  • The above hint is the induction step. You just need to reexpress the algebra using the recursive definitions of addition and multplication. – Bill Dubuque Aug 07 '14 at 02:54