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We have the following statement:

Prove that if $a=qb+r$ with $0\leq r<b$ then $\gcd(a,b)=\gcd(b,r)$

Notation: $D_{a,b}$ stands for $\{x\in\Bbb{Z}\mid x\mid a,\;x\mid b\}$. So we need to prove $D_{a,b}=D_{b,r}$.

We prove $\subseteq$: for all $x\in D_{a,b}$, $x\mid a$ and $x\mid b$, then $x\mid qb+r$ and $x\mid b$ and $x\mid b$. So $x\mid qb+r$ and $x\mid (-q)b$ and $x\mid b$, then $x\mid(qb+r-qb)$ and $x\mid b$, hence $x\mid r$ and $x\mid b$, thus $x\in D_{b,r}$.

The same for the converse.

But then it says:

We can show not only that the greatest common divisors are equal but that the set of all common divisors of $a$ and $b$ is the same as the set of all common divisors of $b$ and $r$, so the greatest of each set is the same.

It is asking to prove that $\max(D_{a,b})=\max(D_{b,r})$? How can we prove it? Why is it important?

manooooh
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2 Answers2

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No. That sentence says that $D_{a,b}=D_{b,r}$ and that therefore $\max D_{a,b}=\max D_{b,r}$.

José Carlos Santos
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The part in italics basically means that if $\gcd(a,b)=\gcd(b,r)$, then all common factors of a and b are the same as all the common factors of b and r, less than the gcd.

For example, Take numbers 48 and 104. $104=2\cdot 48 +8$. Here, $a=104,b=48,r=8$. So according the the sentence in italics, $\gcd (48,104)=\gcd(48,8)=8 $ and all common factors of 48 and 104:$1,2,4,8(D_{a,b})$ are the same as all common factors of 48 and 8:$1,2,4,8(D_{r,b})$

You can prove both the gcd and the prime factor part by splitting the 2 gcds into their prime factors and equating them. It is in fact quite easy.

Aniket Gupta
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