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I have read in the book of counter example in Analysis by Bernard R. Gelbaum John M. H. Olmsted ,(page $18$) that $4$ and $2(1+\sqrt{5})$ without Great commone divisor however it's trivial they have no great commone divisor since $2(1+\sqrt{5})$ is not integer, I want some justification for this in the sense why these number were considered to look for them $\gcd$?

amWhy
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zeraoulia rafik
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  • ok thaks, just to show my question – zeraoulia rafik Jan 29 '18 at 20:23
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    Seems that they mean the $\gcd$ of the numbers in the ring $\mathbb Z[\sqrt{5}]$ – Peter Jan 29 '18 at 20:30
  • I want some justification for this in the sense why these number were considered to look for them gcd! And I want askers to take more time to include context (like the fact that we are talking about gcd in $\mathbb Z[\sqrt 5]$ in this question, and to show their attempts to research and work indepedently before dumping here and demanding what they want. – amWhy Jan 29 '18 at 20:35
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    @DietrichBurde: I'm not sure those are duplicates at all -- this question is not clearly about divisibility in a particular number ring. I would interpret "greatest common divisor" in this (lack of) context to mean the generator of the subgroup of $(\mathbb R,{+})$ they generate, if that subgroup happens to be (infinite) cyclic. – hmakholm left over Monica Jan 29 '18 at 20:37
  • $\Phi$ is defined in example $8$ as the ring of numbers of the form $a+b\sqrt{5}$ with $a,b$ integers. The point of example $10$ is to exploit that $\Phi$, unlike the integers, doesn't have the unique factorization property. They use that $4=2\times 2=(1+\sqrt{5})(-1+\sqrt{5})$. Observe that in this ring (like in the integers) $2$ cannot be factored further. There is $2=1\times 2=-1\times -2$, but in this factorization the factors $1$ and $-1$ are units (have inverses). In the integers, if $2\times 2=ab$ then $2$ divides $a$ or $b$, but in $\Phi$, $2$ doesn't divide $\pm1+\sqrt{5}$. – orole Jan 29 '18 at 20:49

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