Little confusion about the greatest common divisor

Question

[Note: question was updated and clarified so comments & answer are a bit out of sync - Bill D.]

I have little confusion about the claim (ii) below. Although the claim employs the hypothesis $\gcd(b,c)=1$, I think I have proved it without this, but I know that without $(b,c)=1$ this is not true. My question is: why was I able to prove it without this hypothesis?

For $a, b, c \in \mathbb Z$, prove that

ii. if $\gcd(b, c) = 1$, then $\gcd(a, bc) = \gcd(a, b)\gcd(a, c)$

This is the way i tried to prove this:

$(a,b)(a,c) = (a\color{#c00}{(a,b,c)},bc)\ [= (a,bc),$ if $,\color{#c00}{(a,b,c) = 1}].\ $ Is that what you mean by "proved without $,(b.c) = 1?$ — Bill Dubuque, Jan 22 '20 at 06:07
Please show your proof of ii that does not depend on $(b,c)=1$. Consider $a=4,b=4,c=4$ where $(a,bc)=(4,16)=4, (a,b)(a,c)=(4,4)(4,4)=16$ — Ross Millikan, Jan 22 '20 at 06:16
For i., if $\gcd(a,b)=1$ and $b|a$ then $|b|$ is a common divisor of $a$ and $b,$ so $|b|\le \gcd(a,b)=1.$ — DanielWainfleet, Jan 22 '20 at 06:44

Bill Dubuque · Accepted Answer · 2020-01-22T19:31:47.500

3

Theorem $\ \ (a,b)(a,c)\, =\, \overbrace{(a,bc)}^{\large e}\iff \color{#c00}{(a,b,c,bc/e) = 1}\ \,$ [e.g. if $\,(a,b)= 1$]

$\begin{align}{\bf Proof}\quad\ \ \ (a,b)(a,c)\, &=\, ((a,b,c)a,\,bc)\\[.2em] &=\, e\,((a,b,c)\color{#0a0}{a/e,\,bc/e})\\[.2em] &=\, e\,\color{#c00}{((a,b,c),\,bc/e)}\,\ \ {\rm by}\, \ \ (\color{#0a0}{a/e,bc/e}) = (a,bc)/e = 1 \end{align}$

The first equality is by gcd "polynomial" arithmetic, i.e. applying the gcd associative, commutative and distributive laws. The second equality factors out $\,e\,$ using the gcd distributive law, and the final equality uses a version of Euclid's Lemma.

Remark $ $ Replying to a query in a comment: an error in your argument is the following inference

$$(a,b)(a,c) = n_0 a + y_0 bc\ \Rightarrow\ n_0 a + y_0 bc = (a,bc) $$

But those coef's $\,n_0, y_0\,$ need not be the same as the Bezout coef's for $(a,bc).\,$ So your argument with no hypotheses on $\,a,b,c\,$ fails (as we know it must by the above theorem).

edited Jan 22 '20 at 19:31

answered Jan 22 '20 at 06:45

Bill Dubuque

272,048

Why is my way wrong? – jvj Jan 22 '20 at 06:48
1

@Akalanka Why do you think your way is wrong? Your question does not ask about your way. Instead it asks how to prove it without using $,(b,c)= 1,,$ which is what I did above, showing that it holds more generally if $,(a,\color{#c00}{b,c},bc/e) = 1\ \ $ Yoru case $,(\color{#c00}{b,c})=1,$ is a special case. – Bill Dubuque Jan 22 '20 at 06:53
Of course but i proved this without that property – jvj Jan 22 '20 at 06:55
1

@Akalanka See the Remark for one error. – Bill Dubuque Jan 22 '20 at 07:07
Thanks i got it – jvj Jan 22 '20 at 07:13

Little confusion about the greatest common divisor

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