Is the gcd function symmetric?

iamsmooth
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Just a quick theory question. I'd assume it is, but usually the bigger number goes first.

e.g. gcd(10, 5) = 2

but does gcd (5, 10) = 2?

My guess is yes.

Thanks for the help.
 
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Yes. This is apparent from the definition of the words "greatest" and "common".
 
My input is probably trivial but I thought it would be nice to prove it formally.

We need : gcd(x,y) = gcd(y,x)

Suppose gcd(x,y) = d. Then x = nd and y = md for some integers n,m where gcd(n,m) = 1.

So gcd(y,x) = gcd (md,nd) = d because gcd(m,n) =1.

We have shown for all x,y that gcd(x,y) = gcd(y,x).

P.S.
Hopefully no silly errors. -_-'''
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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