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  • a,b ∈ Z
  • b>0
  • q,r ∈ R
  • a=qb+r with 0 <= r < b

Prove that GCD(a,b)=GCD(a,r)

My answer:

  • Let GCD(a,b)=x and GCD(a,r)=y
  • x,y ∈ Z+
  • a/x,b/x ∈ Z
  • a/x=q(b/x)+r/x ; (Dividing both sides by x) ; holds a=qb+r
  • b/y ∈ Z
  • a/y=q(b/y)+r/y ; (Dividing both sides by y)
  • To hold this a/y ∈ Z
  • So y must be also an divisor of a

Is my way to reach the answer correct or does this have another method?

Bill Dubuque
  • 272,048
Ushan Samarakoon
  • 1
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    The standard way is to prove, for any integer $d$, that $d$ is a common divisor of $a$ and $b$ if and only if $d$ is a common divisor of $a$ and $r$. (You've come close to doing that, although you'd need to prove equivalence and not just a one-way implication; also my preference is to keep all expressions integers and use facts about divisibility of linear combinations of integers.) From that proved equivalence, it follows directly from the definition of greatest common-divisor that $\gcd(a,b)=\gcd(a,r)$. – Greg Martin Mar 19 '20 at 17:42
  • I've corrected your title, but please use MathJax. – J.G. Mar 19 '20 at 17:46
  • Okay thanks @GregMartin – Ushan Samarakoon Mar 20 '20 at 14:19
  • I didn't know about MathJax. I'll use it. Thanks:) @J.G. – Ushan Samarakoon Mar 20 '20 at 14:19

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