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If $c$ is a common divisor of $a$ and $b$ then $c$ divides the greatest common divisor of $a$ and $b$. What can we use to prove this?

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4 Answers 4

First of all, it could be a definition. Otherwise, you may use the Euclidean algorithm and some elementary properties of division, or simply the unique prime factorisation of natural numbers.

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"First, of all it could be a definition." Really? What definition might that be. –  debitanostra Sep 26 '12 at 10:32
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@user42362 E.g. the universal definition of gcd, viz. $\rm\:c\:|\:(a,b)\iff c\:|\:a,b\ \ $ –  Bill Dubuque Sep 26 '12 at 16:26

Hint $\ $ By Bezout's identity, there are $\rm\:j,k\in\Bbb Z\:$ such that $\rm\:gcd(a,b)\, =\, j\:a + k\:b,\:$ which is clearly divisible by every common divisor of $\rm\:a,b.$

Thus a linear common divisor is always greatest, i.e. any common divisor $\rm\:d\:$ of $\rm\:a,b\:$ that is an integral linear combination of them $\rm\:d = j\:a + k\:b\:$ is necessarily the greatest common divisor, since, as above, every common divisor $\rm\:c\:$ divides $\rm\:d,\:$ hence $\rm\:c\le d.$

See here for a proof of Bezout's Identity and discussion of related matters.

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What is your definition for the greatest common divisor? –  quid Dec 18 at 18:35
    
@quid It works for both common gcd definitions, i.e. "greatest" means either wrt (absolute) value, or wrt divisibility, as in the universal definition $\ a\mid b,c\iff a\mid (b,c)\ \ $ –  Bill Dubuque Dec 18 at 18:51
    
Part of your answer is in fact wrong for the former (and for the former with parenthesis and the latter there is not really such a thing as the greatest common divisor, at least not if one want it to be an element). –  quid Dec 18 at 18:54
    
@quid I see nothing wrong. The former definition is intrinsic to Euclidean domains, and the latter to gcd domains. And the abuse of notation (uniqueness of the gcd up to unit factors) is ubiquitous. But this is very far from the question at hand. –  Bill Dubuque Dec 18 at 18:56
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@quid Ah, I see now. That comment was explicitly addressed to the OP, even though it was in the comments of your answer. Today has been so busy that now I don't recall why I placed the comment at that spot (maybe because I thought some readers of your answer might find the links helpful). Apologies for any confusion it may have caused. –  Bill Dubuque Dec 18 at 19:40

Use unique prime factorization: say $m=\displaystyle\prod_{i}{p_i}^{\alpha_i}$ and $n=\displaystyle\prod_i{p_i}^{\beta_i}$, then $\gcd(m,n)$ will contain the exponents $\min(\alpha_i,\beta_i)$.

Meanwhile, if $d$ has exponents $\delta_i$, then $d|m$ means exactly that each $\delta_i \le \alpha_i$. So, this is all a reformulation of the ($\delta_i\le\alpha_i$ and $\delta_i\le\beta_i$ then $\delta_i\le\min(\alpha_i,\beta_i)$) setting.

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This can be shown based on what we are given by showing $a$ and $b$ in terms of a product of $c$. If $c$ is a common divisor of $a$ and $b$, that means a and b are each the product of at least one $c$ and another natural number: $a = c^np$ and $b=c^nq$, with $a,b,c,n,p,q \in \mathbb N$. There are then two possibilities:

  1. Either $c$, or a power of $c$ $c^n$, is the GCD of $a$ and $b$. Trivially, $c$ divides any $c^n$ where $n>0$.
  2. Neither $c$ nor any $c^n$ is the GCD of $a$ and $b$; in that case, $p$ and $q$ have their own GCD $m > 1$, and the GCD of $a$ and $b$ is $c^nm$; again, by inspection, we see that $c$ divides any $c^nm$ to produce $c^{n-1}m$.
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@Downvoter - anything to add or amend? –  KeithS Sep 26 '12 at 17:31

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