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Prove that $(a+b, a-b) \geq (a, b)$

My attempt

Let $(a+b, a-b) = d$ and $(a, b) = c$. Since $c \mid a,b$

$c$ is also a factor of $a+b$ and $a-b$. Thus $c \leq d$.

Is this enough as a proof? It feels kind of skimpy to me, so I was hoping for some feedback and ideas to make it more rigorous from the MSE community. In particular, I feel as though it's not rigorous enough on showing that $c$ is not always equal to $d$ and thus put some actual meaning behind the 'less than' part of $\leq$.

Airdish
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3 Answers3

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Your proof seems to be OK, as every number that divides two numbers, also divides their greatest common divisor. And it's well known that $c \mid d \implies c\le d$, when working in positive integers.

For your second requirement here's how one can generate the cases when the GCDs are not equal. By Euclid's Algorithm we have $(a+b,a-b) = (a+b,2b)$ Now obviously $c \mid (a+b,2b)$ and let $a=ca', b=cb'$ where $(a',b')=1$. Then we have that: $(a+b),2b) = c\cdot(a'+b',2b')$ If a factor divides $b'$ then it can't divide $a'+b'$, as $(a',b') = 1$. So $(a'+b',2b')=1$ or $2$. Hence we have that $d=c$ or $2c$. The second case is obtained when $a=ca', b=cb'$, where $a',b'$ are odd coprime integers. While in every other case $d=c$.

Stefan4024
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Your proof is fine.

Here is a more concise proof:

$(a+b, a-b) = (a+b-(a-b),a-b) = (2b,a-b)= (2b,a+b)$

$(a, b) = (a+b,b)=(b,a+b)$

It is easy to show that $(2x,y) \geq (x,y)$

Therefore $(a+b, a-b)=(2b,a+b)\geq(b,a+b)=(a, b) $

Zau
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  • The OP's proof is both more concise and general: it implies that if $,n_i = j_i a + k_i b,$ for integers $,j,k,$ then $,(a,b)\mid a,b,$ $\Rightarrow,$ $(a,b)\mid n_1,n_2,$ therefore $, (a,b) \le (n_1,n_2),$ because every common divisor is $\le$ greatest common divisor. Further $,(a,b)\mid (n_1,n_2),$ since $,c\mid n_1,n_2\iff c\mid (n_1,n_2),,$ which is the universal property of the gcd (= the definition in general rings). – Bill Dubuque Jul 11 '16 at 14:00
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Your proof is correct, it shows that c divides d, so c may be less than or equal to d. To see the case of equality, take $a=5$ and $b=2$ then $(5,2)= (7,3)=1 $ and we're done.

Burrrrb
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