Having trouble with this euclidean algebra proof!
Let $a$ and $b$ be positive integers. Prove that $b$ divides $a$ iff $\gcd(a, b)=a$
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That's false, try $a=2, b=4$ for example. – dxiv Mar 26 '17 at 23:52
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Where did your trouble start? If you show us what you tried, we can help better. Also, please use MathJax: https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – martin.koeberl Mar 26 '17 at 23:52
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I suppose you mean $gcd(a,b)=b$. Let's denote $g\equiv gcd(a,b)$.
In the forward direction, if $b$ divides $a$, then since $b$ divides itself it is a common divisor, therefore $b \leq g$. But if $b < g$ then we get a contradiction since $g$ needs to divide $b$ so cannot be larger than $b$.
In the reverse direction, if $g=b$, then by definition of $gcd$ we have that $b$ divides $a$.
Yoni
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More generally if $,a\mid b,$ then $,d\mid a,b\iff d\mid a,,$ so $,a,b,$ and $,a,$ have the same set $S$ of common divisors $,d,,$ hence the same greatest common divisor (= max $S$), i.e. apply the universal property charatcerizing the gcd. – Bill Dubuque Nov 30 '21 at 00:44