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# A simple proof?

Given:$a=b\times q+r,0\leq r < b$.Yes,that is the division equation.Try to prove that,$GCD(a,b)=GCD(b,r)$.Try not to use any theorem for solving this.Just use simple Algebra.This is from my class $$10^{th}$$ textbook,it was just written as a statement and no proof was given.I came up with one which I will post afterwards.

2 years, 9 months ago

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To prove :- If $$a=bq+r$$, then $$(a,b)=(b,r)$$, where $$(x,y)$$ denotes GCD of x and y. Proof:- Let $$(a,b)=d$$ and $$(b,r)=e$$. Then, (a,b) =$$d \Rightarrow d|a$$ and $$d|b \Rightarrow d|a$$ and $$d|qb \Rightarrow d|(a-qb) \Rightarrow d|r$$. Hence, $$d|r$$ and $$d|b \Rightarrow \boxed{d|(b,r)}$$. Similarly, (b,r) = $$e \Rightarrow e|b$$ and $$e|r \Rightarrow e|bq$$ and $$e|r \Rightarrow e|(bq-r) \Rightarrow e|a$$. Hence, $$e|a$$ and $$e|b \Rightarrow \boxed{e|(a,b)}$$. We see from the boxed expressions that $$(a,b)|(b,r)$$ and $$(b,r)|(a,b) \Rightarrow (a,b) = (b,r)$$. Q.E.D.

- 2 years, 9 months ago

I completed the proof in the same way - I personally remember the proof from "Elementary Number Theory " by Jones - It's a great book :)

- 2 years, 9 months ago

Thanks a lot for your suggestion sir 👍 ! The book is really awesome for beginners 💎 !

- 2 years, 9 months ago

Very nice proof!But how did you come up with it so fast?Have you seen similar problems before?

- 2 years, 9 months ago

Yeah, I am pretty familiar with these kind of problems ! I came across problems involving proofs of gcd while doing number theory...

- 2 years, 9 months ago

So,is this your own?Do u have any NT books?

- 2 years, 9 months ago

Not my own bro, I just remember seeing this proof in some book ! Btw this is really important one, it is the basis of the Euclids Algorithm ! For good books, check out David A. Santos books, they are available in pdf format freely, and they are really good books to follow 🎓 📱 !

- 2 years, 9 months ago

Yes,yes this is really an important one!It is the basis of Euclid's Algorithm,that is how I came across this.It must be in your book also.Check it!

- 2 years, 9 months ago

Oh!Thanx for the recommendations!Thanx a lot!!Grateful!

- 2 years, 9 months ago