1. ## Addition of relatively prime numbers

RTP: If $\displaystyle gcd(a,b)=1$, then $\displaystyle gcd(a, (a+b))=1$, where a,b are positive integers

Proof
Note $\displaystyle 1=as+bt$
Assume $\displaystyle gcd(a, (a+b))=k$
Then
$\displaystyle am+(a+b)n=a(m+n)+bn=k*1$
$\displaystyle a(m+n)+bn=aks+bkt$
Thus
$\displaystyle m+n=ks,n=kt$
So
$\displaystyle m+kt=ks,m=ks-kt$
So
$\displaystyle a(ks-kt)+bkt=aks+bkt$
$\displaystyle -akt=0$
This is impossible, so our assumption is wrong
Thus our statement is true

Questions
1. Is this proof valid?
2. Can I generalize to :If gcd(a,b)=k, then gcd(a, (a+b))=k
3. Is there a simpler non-contradiction, non-FTA proof?

2. This thread is a repost. Sorry. Marking it as solved and requesting that it be removed please

3. Originally Posted by I-Think
RTP: If $\displaystyle gcd(a,b)=1$, then $\displaystyle gcd(a, (a+b))=1$, where a,b are positive integers

Proof
Note $\displaystyle 1=as+bt$
Assume $\displaystyle gcd(a, (a+b))=k$
Then
$\displaystyle am+(a+b)n=a(m+n)+bn=k*1$
$\displaystyle a(m+n)+bn=aks+bkt$
Thus
$\displaystyle m+n=ks,n=kt$
So
$\displaystyle m+kt=ks,m=ks-kt$
So
$\displaystyle a(ks-kt)+bkt=aks+bkt$
$\displaystyle -akt=0$
This is impossible, so our assumption is wrong
Thus our statement is true

Questions
1. Is this proof valid?
2. Can I generalize to :If gcd(a,b)=k, then gcd(a, (a+b))=k
3. Is there a simpler non-contradiction, non-FTA proof?
Originally Posted by I-Think
This thread is a repost. Sorry. Marking it as solved and requesting that it be removed please
I can't find your other thread. Your proof looks fine from what I see. Yes, you're theorem generalizes. One merely needs note that if $\displaystyle \text{gcd}(a,a+b)\mid \text{gcd}(a,b)$ since $\displaystyle z\mid a\text{ and }z\mid a+b\implies z\mid a\text{ and }z\mid b$. Also, $\displaystyle \text{gcd}(a,b)\mid \text{gcd}(a,b)$ since $\displaystyle z\mid a\text{ and }z\mid b\implies z\mid a\text{ and }z\mid a+b$
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