Is my proof correct? if c>0 prove that (ac,bc)=c(a,b)

Question

This is how I did my proof although Im doubting weather or not I can do this one thing which I will mention explicitly, also (a,b) is the gcd or greatest common divisor, and every lower case letter is an integer.

Proof. Suppose (a,b)= d, then we know that d=ax+by for some x and y (This is a well-known theorem that states that the gcd of a and b is the least element of the set of ax+by so x and y should be numbers such that d is the least element of the set)

$d\cdot c$ = (ac)x+(bc)y

This is where I'm having an issue, I want to say that (ac,bc)=dc by the aforementioned theorem, but for that to be true dc would have to be the least element of the set of acx+bcy, so does multiplying this equation by c preserve this property or can we not know that?

Does this answer your question? Prove that $(ma, mb) = |m|(a, b)\ [GCD & LCM Distributive Law]$ - found through using an Approach0 search. Note there are quite a few other duplicates, e.g., $\gcd (ca, cb) = \gcd (a, b)c$ if $c > 0$, How to prove that $z\gcd(a,b)=\gcd(za,zb)$, etc. Also, there's closely related ones, e.g., about integral domains, such as ... — John Omielan, Nov 24 '22 at 02:51
(cont.) Integral domain( not merely $\mathbb{Z}$ ). Prove that $\gcd(ca, cb) = c\gcd(a, b)$, In an Integral Domain is it true that $\gcd(ac,ab) = a\gcd(c,b)$?, etc. — John Omielan, Nov 24 '22 at 02:58
See Andre's answer in the linked dupe for this method, and see my answer for a generalization that works in any gcd domain (where Bezout equations for the gcd need not exist). Please search for answers before posting questions to help avoid duplication. — Bill Dubuque, Nov 24 '22 at 14:02

score -1 · Answer 1 · answered Nov 24 '22 at 02:18

-1

Ok I think I found a satisfying answer, if we imagine the table of with x,y and d column and then multiply c to everything the acx+bcy for x and y such that d is the lowest element of the set ax+by, then in our new table with xc, yc and dc then acx+bcy would be the least element of the set.

so dc=ac(x)=bc(y) implies dc=(ac,bc)=c(a,b).

answered Nov 24 '22 at 02:18

cheesewiz

81

Please strive not to post more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Nov 25 '22 at 09:58

Is my proof correct? if c>0 prove that (ac,bc)=c(a,b)

1 Answers1