Let $d = \gcd(a,b)$. If $a = da'$ and $b = db'$, show that $\gcd(a',b')=1$.
So far I concluded that $d$ divides both $a$ and $b$, and their remainders are zero. I don't know what to do next, someone please help me. Thank you.
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Bing
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2You mean $a=da'$, I think, instead of $a=d'a$. If gcd$(a',b') =c >1$, then $a'=ca'', b'=cb''$. So...? – Krish Sep 07 '17 at 08:03
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Have you heard of Bezout's identity? – Arthur Sep 07 '17 at 08:04
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Thank you for the correction, and yes I have heard of Bezout's identity – Bing Sep 07 '17 at 08:13
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The easiest way to continue is to do what is called a proof by contradiction which should look something like this:
- Assume that $\gcd(a',b')\neq 1$
- Do some math
- Arrive at a nonsensical conclusion
- From 3., you can conclude that $\gcd(a', b')=1$.
In your case, assume that $\gcd(a', b')=g\neq 1$. Then, define $a'' = \frac{a'}{g}$ and $b'' = \frac{b'}{g}$.
Now:
- Can you prove that $a''$ and $b''$ are integers?
- Can you show that $a'=ga''$ and $b'=gb''$?
- Can you now find a number that divides both $a$ and $b$ and is larger than $d$ (Hint: think about what $a$ is equal to if $a=da'$ and $a'=ga''$... and do the same with $b$).
- Can you now see a contradiction?
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If $gdc(a', b')$ was $d'>1$ then $dd'$ would divide both a and b. So d would not still be the Greatest Common divisor.
fhn
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