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I've searched a lot but I am not able to find a simple explanation. If you can, please explain it in an intuitive manner. Thanks

Sarthak
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  • because $gcd$ means greatest common divisor – reuns Jul 20 '16 at 21:04
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    It depends on the definition we use for gcd. In the most elementary work, gcd is defined as greatest common divisor. Then we need the theorem that $d$ is a gcd of $a$ and $b$ if and only if $d$ divides $a$ and $b$, and any common divisor of $a$ and $b$ divides $d$. – André Nicolas Jul 20 '16 at 21:08
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    More precisely, there are two concepts of $d=\gcd(n,m)$, which (and this takes some work to show) give exactly the same result: The first is that $d$ is the largest integer that divides $n$ and $m$; the second is that $d$ is the (unique positive) integer such that if $e$ is an integer that divides $n$ and $m$, then $e$ divides $d$. – Dominic Wynter Jul 20 '16 at 21:26
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    The way to prove this is by using division with remainder (the Euclidean algorithm, if you want to be precise). – Dominic Wynter Jul 20 '16 at 21:28
  • As a sidenote, the notation that we use to say "$d$ divides $n$", is $d|n$. – Dominic Wynter Jul 20 '16 at 21:29
  • @MonstrousMoonshine 'some work to show' ? no this is obvious. I'd say the only non-obvious fact is that the Euclid algorithm gives the $gcd$. – reuns Jul 20 '16 at 21:30
  • I have no clue how much background op has. – Dominic Wynter Jul 20 '16 at 21:41

2 Answers2

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Hints:

  • Consider a divisor of the greatest common divisor. Does it divide both of the original numbers?

  • Consider a non-divisor of the greatest common divisor. Does it divide both of the original numbers?

Henry
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Hint $\,\ a,b\,$ and their gcd $\,(a,b)\,$ have the same set of divisors by the following

Lemma $\ \ d\mid a,b\iff d\mid (a,b)\ \ \ $ [Universal gcd Property]

${\bf Proof}\quad\ d\mid a,b\,\Rightarrow\, d\mid (a,b) = ia\!+\!jb\,$ for some $\, i,j\in\Bbb Z,\,$ by Bezout.

$\qquad\qquad\, d\mid (a,b)\mid a,b\,\Rightarrow\, d\mid a,b\ $ by transitivity of $ $ "divides".

Bill Dubuque
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