How can I prove that gcd(a,b) is an integer and gcd(a/b, b/h) = 1?

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SUMMARY

The discussion centers on proving that if \( h = \gcd(a, b) \) for integers \( a \) and \( b \), then \( \gcd\left(\frac{a}{h}, \frac{b}{h}\right) = 1 \). Participants clarify that \( h \) divides both \( a \) and \( b \), confirming that \( \frac{a}{h} \) and \( \frac{b}{h} \) are integers. The assertion that \( \gcd\left(\frac{a}{h}, \frac{b}{h}\right) \) equals 1 is supported by Theorem 1 from "Elementary Number Theory" by Dudley, while the claim that \( \frac{a}{b} \) is an integer lacks justification.

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simo1
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can I get hints on how to prove this question:

let a,b be in Z(integer) with h=gcd(a,b) not equal to zero then [(a divide b), (b divide h)] are in Z. and gcd[ (a divide h, b divide h)]=1
 
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Did you mean "a divide h"? The first part is pretty much the definition of the greatest common divisor ("the greatest integer which divides both numbers", so trivially $h$ divides both $a$ and $b$). What definition of the gcd are you working from? For the second part, suppose that $a/h$ and $b/h$ shared a common factor, so that their gcd was not equal to $1$. Would that not conflict with the definition of $h = \gcd(a, b)$? Can you see the problem?
 
Let me restate the problem to see if I understand it correctly.


Given that a,b are integers, h=gcd(a,b) and h not equal to 0 then
is it true that:
  1. a divided by b and b divided by h are integers
  2. gcd(a divided by h, b divided by h)=1


Statement 2 is true and is given as Theorem 1 in "Elementary Number Theory" by Dudley.

In Statement 1, there is no reason why a divided by b should be an integer.
 
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