Abstract algebra gcd proof.

[musicalwatch]

Question: Prove that if m|a and m|b, then gcd(a/m,b/m)=gcd(a,b)/m, provided that m>0.

Since m|a and m|b, a can be written as a=mt for some integer t. Likewise, b=mq for some integer q.

Therefore, we can rewrite the left part of gcd equation as gcd(mt/m,mq/m)=gcd(t,q)=tr+qs, where r and s are some integers.

The right part of the equation will become: gcd(a,b)/m = gcd(mt, mq)/m= mtr/m+mqs/m=tr+qs, where r and s are some integers

tr+qs=tr+qs..... I have a feeling that this proof isn't correct. After all, r, s, t, q are "some" integers so some of my statements are most likely wrong.

I also tried working with m|gcd(a,b) and gcd(a,b)=mx+my for all integers x and y but couldn't develop this thought.

 

[mathmari]

Question: Prove that if m|a and m|b, then gcd(a/m,b/m)=gcd(a,b)/m, provided that m>0.

Since m|a and m|b, a can be written as a=mt for some integer t. Likewise, b=mq for some integer q.

Therefore, we can rewrite the left part of gcd equation as gcd(mt/m,mq/m)=gcd(t,q)=tr+qs, where r and s are some integers.

The right part of the equation will become: gcd(a,b)/m = gcd(mt, mq)/m= mtr/m+mqs/m=tr+qs, where r and s are some integers

tr+qs=tr+qs..... I have a feeling that this proof isn't correct. After all, r, s, t, q are "some" integers so some of my statements are most likely wrong.

I also tried working with m|gcd(a,b) and gcd(a,b)=mx+my for all integers x and y but couldn't develop this thought.

m|a => a=mt
m|b => b=mq
So, gcd(a/m, b/m)=gcd(t,q)=tr+qs=C

gcd(a,b)=gcd(mt,mq)=mtx+mqy=D

We want to show that C=\(\displaystyle \frac{D}{m}\) => D=mC
We have that mC=mtr+mqs, that implies D|mC
We have D=mtx+mqy, while mC|mt & mC|mq, which implies that mC|D.
So, D|mC & mC|D => D=mC.

 

[lookagain]

Question: Prove that if m|a and m|b, then gcd(a/m,b/m)=gcd(a,b)/m, provided that m>0.

Let a, b, m, p, r, and t be positive integers.



\(\displaystyle If \ \ m\bigg|a \ \ and \ \ m\bigg|b, \ \ then \ \ gcd\bigg(\dfrac{a}{m}, \ \dfrac{b}{m}\bigg) \ = \ \dfrac{gcd(a, \ b)}{m}, \ \ provided \ \ that \ \ m > 0.\)



m is a common denominator of a and b. \(\displaystyle \ \ \) Let p be a divisor, such that mp = gcd(a, b).


\(\displaystyle (mp) \bigg|a \ \ \implies a = mpr\)



\(\displaystyle (mp)\bigg|b \ \ \implies b = mpt\)



And then the gcd(r, t) = 1.



--------------------------------------------------------------


\(\displaystyle gcd\bigg(\dfrac{a}{m}, \ \dfrac{b}{m}\bigg) \ = \)



\(\displaystyle gcd\bigg(\dfrac{mpr}{m}, \ \dfrac{mpt}{m}\bigg) \ = \)



\(\displaystyle gcd(pr, \ pt) \ = \)



\(\displaystyle p*gcd(r, \ t) \ = \)


\(\displaystyle p \ \)*


-----------------------------------------------------



\(\displaystyle \dfrac{gcd(a, \ b)}{m} \ =\)



\(\displaystyle \dfrac{gcd(mpr, \ mpt)}{m} \ = \)



\(\displaystyle \dfrac{mp*gcd(r, \ t)}{m} \ = \)


\(\displaystyle \dfrac{mp}{m} \ = \)


\(\displaystyle p\) **



* \(\displaystyle \ = \ \) ** \(\displaystyle \ = \ \) p




musicalwatch, do you know if this type of proof would be accepted?