# Proof of Euclidian Algorithim from Terrence Tao's Analysis 1

In Analysis 1 in chapter two section 3 on multiplication one of the exercises is the proof of the following.

(Euclidean algorithm). Let n be a natural number, and let q be a positive number. Then there exist natural numbers m, r such that 0≤r < q and n = mq +r.

With the hint to induct on n. I am having trouble navigating this proof with only without the idea of subtraction and only cancellation laws for addition and multiplication.

A proof would be helpful for me to reference.

• That's not what's usually called the Euclidean Algorithm, it's usually called the Division Theorem or the Division Algorithm. The Euclidean Algorithm is the one that calculates greatest common divisors (by repeated application of the Division Theorem). If you search this website for "Division Algorithm", you will probably find an answer to your question. – Gerry Myerson Jul 21 at 2:23
• @Gerry Try googling "Euclidean division" and "Euclidean division algorithm", e.g. Hendrik Lenstra wrote "Kummer noted that it would be sufficient to prove uniqueness of factorisation and for this he turned to the method that also Wantzel would apply: the euclidean division algorithm." – Bill Dubuque Jul 21 at 3:19
• We can extend $\Bbb R$ to a larger ordered field $\Bbb R^*$ in which all the basic rules for $+,-,\times, /$ and $<$ still apply, but $\Bbb R^*$ has (among other things) a member $q$ which is larger than every member of $\Bbb Q.$ The Q is not true if "number" means member of $\Bbb R^*$.... If "number" means member of $\Bbb R$ we must use some consequence(s) of the $definition$ of $\Bbb R$ beyond the basics of arithmetic to ensure we are not talking about $\Bbb R^*....$ Topics: Field, Ordered field, Complete ordered field, Archimedean property. – DanielWainfleet Jul 21 at 9:07
• @Daniel, I suspect $q$ is meant to be a positive integer. – Gerry Myerson Jul 21 at 9:13
• @Bill, thanks, but I'm specifically questioning "Euclidean algorithm", and nothing in your comment refers to that phrase. – Gerry Myerson Jul 21 at 9:15

Note that the base case is trivial, that is, for $$n=1$$ and any positive integer $$q$$ we have that $$n=0.q+1$$ if $$q>1$$ ($$n=1.q+0$$ otherwise).

Assume that the claim holds for every natural number $$n\leq N$$. (To be more explicit, it means that given a natural number $$n\le N$$ and a positive integer $$q$$ we have $$n=mq+r$$ such that $$0 \le r < q$$.)

We need to show that the claim also holds for $$n=N+1$$. To this end, recall that $$N=mq+r$$, $$0\le r < q$$ therefore $$N+1=mq+(r+1)$$. Now, analyse the following two cases separately: if $$r then we are already done. Otherwise, $$r=q-1$$ then $$r+1=q$$. Therefore, $$N+1=mq+q=(m+1)q+0$$. This proves that the claim holds for $$N+1$$.

P.S.: I do not have the book you have mentioned with me, so I am not sure how the set of natural numbers is defined. It may so happen that $$0$$ is included in the set or natural numbers. Most part of the proof still goes through, only the base case needs to be modified. In fact, it only makes it simpler.

Hint: the induction step is like adding $$1$$ in radix $$q$$ arithmetic. The analogy is clearer with $$2$$ digits $$r_i$$.

By induction $$\,n\!-\!1\, =\, m\, q^2 + r_1 q + r_0,\$$ all $$\,r_i < q$$

when $$\,r_0 < q\!-\!1\,$$ then adding $$\,1\,$$ yields $$\,r_0 < q\,$$ so we're done.

else $$\ \ \ r_0 = q\!-\!1\,$$ then adding $$\,1\,$$ we get a carry into $$\,r_1,\,$$ handled as above, i.e.

when $$\,r_1 < q\!-\!1\,$$ then adding $$\,1\,$$ yields $$\,r_1 < q\,$$ so we're done,

else $$\ \ \ r_1 = q\!-\!1\,$$ then adding $$\,1\,$$ we get a carry into $$m$$ and we're done (since $$m$$ can be any natural).

So it is essentially a radix $$q$$ increment operation restricted to the least $$2$$ digits (with all higher digits collected into $$m)$$. In the OP we use only the least digit (vs. least two).