Greatest Common Divisor is the least linear combination

Asked Jan 05 '22 at 22:50

Active Jan 06 '22 at 04:29

Viewed 121 times

Let d be the least positive linear combination of a and b. It is not difficult to understand the proof that d is a common divisor of a and b.

Since $d \ge 1$ and is a common divisor of a and b. d must be bigger than other common divisors.

Why d is the greatest? I'm not able to understand this statement. Any help is appreciated.

Follow up question:

Any common divisor is a divisor of d

It makes sense if this is true, d is the gcd. But why is c|d instead of d|c?

edited Jan 06 '22 at 00:07

asked Jan 05 '22 at 22:50

user16971617

3

Any common divisor of $a$ and $b$ divides any (integral) linear combination of them, in particular $d$. – WhatsUp Jan 05 '22 at 22:55
Every common divisor also is a divisor of any $ax+by,$ hence a divisor of $d.$ So there can’t be any common divisor bigger than $d.$ – Thomas Andrews Jan 05 '22 at 22:57
I am not following here. why is any common divisor of a and b divides d? – user16971617 Jan 05 '22 at 23:50
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Any common divisor of $a$, $b$ divides any linear combination of $a$, $b$ – orangeskid Jan 06 '22 at 00:11
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Recall multiples are closed under $\rm\color{#0a0}{linear\ combinations}$ therefore $,c\mid a,b \Rightarrow c\mid \color{#0a0}{ja+kb}! =! \gcd(a,b),,$ so $,c\le \gcd(a,b),,$ i.e. linear (Bezout) common divisors are always greatest. In ideal language it boils down to the universal property of the ideal sum, i.e. $,c\mid a,b!\iff! (c)\supseteq (a),(b) !\iff! (c)\supseteq (a)!+!(b)! =! (a,b)$ – Bill Dubuque Jan 06 '22 at 04:29
Good job with the duplicates @Bill. Underappreciated but necessary work for the site. – Jyrki Lahtonen Jan 06 '22 at 06:39

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