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$$2\gcd(a, b) \le \gcd(2a,2b)$$ I'm pretty sure it has something to do with Bezout's lemma or it could use prime factorization, but unsure on how to go exactly about it. Please help!

Full Disclosure: This came up on a midterm and something like this may be on the final so I want to make sure I have it down for next time.

Hanul Jeon
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Samyak Jain
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  • it does! to some extent, we were never taught the euclidean algorithm properly so I was hoping for something along the lines of bézouts lemma. – Samyak Jain Nov 10 '20 at 08:43

2 Answers2

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Normally you would prove the equality, as suggested by the link provided in comments. In this case however, you only need to prove one inequality. For this notice $2\gcd(a,b)|2a$ (because $2|2$ and $\gcd(a,b)|a$ by definition) and similarly $2\gcd(a,b)|2b$. So by definition $2\gcd(a,b)|\gcd(2a,2b)$, which implies $2\gcd(a,b)\leq \gcd(2a,2b)$.

Sil
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  • i can't quite wrap my head around why " 2gcd(,)|gcd(2,2) implies 2gcd(,)≤gcd(2,2)" – Samyak Jain Nov 10 '20 at 09:05
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    @SamyakJain If LHS divides the RHS, it can not be greater than RHS,.. – CiaPan Nov 10 '20 at 09:06
  • @SamyakJain Exactly as CiaPan writes, recall the defnition of $m \mid n$, it means $n=km$ (also in your case notice $m,n,k$ are positive) – Sil Nov 10 '20 at 09:11
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It is the special case $\,d = \gcd(a,b)\,$ below.

$$ d\mid a,b\,\Rightarrow\, 2d\mid 2a,2b\,\overset{\!\rm\color{#c00}U}\Rightarrow\, 2d\mid \gcd(2a,2b)\Rightarrow\ 2d\le \gcd(2a,2b)\qquad$$

where we used $\rm\color{#c00}U$ = gcd Universal Property below (provable by Bezout as in the link)

$$ x\mid y,z\!\!\overset{\rm\color{#c00}U\!\!}\iff x\mid \gcd(y,z)\qquad$$

Alternatively employ $\,2\gcd(a,b) = \gcd(2a,2b)\,$ by the gcd Distributive Law.

Bill Dubuque
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