How is induction used to prove that $\gcd(an, bn) = |n|\gcd(a, b)$

Question

In the book Elementary Number Theory: Primes, Congruences and Secrets written by William Stein, which can be found at http://wstein.org/ent/, Stein proves the following at page 6/lemma 1.1.17:

Lemma $\ $ For any integers $a, b, n,$ we have $\, \gcd(an, bn) = |n|\gcd(a, b)$

Proof $\ $ We step through Euclid's algorithm for $\gcd(an, bn)$ and note that at every step the equation is the equation from Euclid's algorithm for $\gcd(a, b)$ but multiplied through by $n.$ For simplicity, assume that both $a$ and $b$ are positive. We will prove the lemma by induction on $a + b$. The statement is true in the base case when $a + b \le 2$, since then $a = b = 1$. Now assume $a, b$ are arbitrary with $a \ge b$. Let $q$ and $r$ be such that $a = bq + r$ and $0 \le r < b$. Then by Lemmas 1.1.9-1.1.10, we have $\gcd(a, b) = \gcd(b, r)$. Multiplying $a = bq + r$ by $n$ we see that $an = bnq + rn$, so $\gcd(an, bn) = \gcd(bn, rn)$. Then

$$b + r = b + (a - bq) = a - b(q - 1) \le a < a + b,$$

so by induction $\gcd(bn, rn) = |n|\gcd(b, r)$. Since $\gcd(b, r) = \gcd(a, b)$, this proves the lemma.

My main question concerning this proof is that i have troubles following how $b + r < a + b$ leads to the fact that $\gcd(bn, rn) = |n| \gcd(b, r).$

I think that my troubles comprehending this proof is related to my basic understanding of mathematical induction, I never even knew what it was before stumbling upon this proof. My current understanding of induction is that you prove your statement for a base case, $a + b = 2,$ then you prove that if the statement is true for any one natural number then it is true for the next. However i do not understand how that principle is used in this proof, any explanation is greatly appriciated.

Answer 1

You are correct that mathematical induction consists of three steps:

1. Prove the base case.
2. Assume that the claim is true for some $n\in\Bbb N$ (possibly including $0$ if needed).
3. Use the above assumption to prove the claim for $n+1$.

Let us see this in practice with standard example:

$$1+2+3+\ldots+n = \frac{n(n+1)}{2}\tag{1}$$

Base case: $$1=\frac{1(1+1)}{2}\quad\small\text{(obviously true)}$$

Assume that formula $(1)$ is true for some $n$. Now we need to prove that $$\underbrace{1+2+3+\ldots +n}_{\text{use the assumption on this}}+(n+1) = \frac{(n+1)(n+2)}{2}$$ and we can use the assumption to get $$1+2+3+\ldots +n+(n+1) = \frac{n(n+1)}{2} +(n+1) = \frac{(n+1)(n+2)}{2}$$ so we are done.

Now, in your example, what is used is version of induction called strong induction. The difference is that in step 2, we assume something stronger, but it turns out this is equivalent to "weak" induction (just happens to be more convenient in some cases).

1. Prove the base case.
2. Fix some $n\in\Bbb N$ and assume that the claim is true for all natural numbers $k$ when $k< n$.
3. Use the above assumption to prove the claim for $n$.

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I will now try to clarify the steps in your proof. Let us define $m=a+b\in\Bbb N$. We will do induction on $m$.

Base case is $m = 2$. Since $a+b = 2$ and $a,b>0$, we must have $a=b=1$, so $$\gcd(na,nb) = \gcd(n,n) = |n| = |n|\gcd(a,b).$$

Now let $m\in\Bbb N$ and assume that $$\gcd(na',nb') = |n|\gcd(a',b'),\quad(\forall a',b'\in\Bbb N)\ a'+b' < m.\tag{2}$$ We want to prove that the claim is true for $a,b$ (remember that $a+b=m$).

Using Euclid's division, we have $a = qb + r$, $0\leq |r|<b$, and thus $$\gcd(a,b) = \gcd(b,r).$$ But if we multiply everything by $n$, we get $na = qnb + nr$, $0\leq |nr| < nb$, so we have that $$\gcd(na,nb) = \gcd(nb,nr).$$

Notice that if we can prove that $$\gcd(nb,nr) = |n|\gcd(b,r),\tag{3}$$ we will have $$\gcd(na,nb) = |n|\gcd(a,b)$$ as well. So let us see if we can employ our assumption $(2)$ - for it to work, we must check that $b+r<m$, and you can check with your proof how it is done. Hence, we can apply induction assumption to conclude that $(3)$ is true.

Also, notice that we indeed needed strong induction, since we don't know that $b + r + 1 = m$, which we would want in ordinary induction.