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Mathematics 15 Online
OpenStudy (anonymous):

Prove that if 2|(a-b) and 3|(b-a), then 6|(a-b)

ganeshie8 (ganeshie8):

\(2 | (a-b) \implies a-b = 2k\) \(3 | (b-a) \implies b-a = 3m\)

ganeshie8 (ganeshie8):

since \(\gcd(2, 3) = 1\) we can write \(2x + 3y = 1\) for some integers \(x\) and \(y\)

ganeshie8 (ganeshie8):

\(a-b = (a-b)\times 1 = (a-b)(2x+3y) = (a-b)2x + (a-b)3y\) \(= -3m*2x + 2k * 3y = 6(-mx+ky )\) so, \(6 | a-b\)

ganeshie8 (ganeshie8):

let me knw if smthng doesnt make sense

OpenStudy (anonymous):

Thank you very much. Why does gcd(2,3)=1 imply that 2x + 3y =1 ?

ganeshie8 (ganeshie8):

good question :) gcd of any two integers is the LOWEST possible positive integer that can be written as linear combination of those two integers

ganeshie8 (ganeshie8):

thats mouthful definition

ganeshie8 (ganeshie8):

you need to prove it upfront, before pursuing this current problem

ganeshie8 (ganeshie8):

gcd(a, b) = smallest possible positive integer than can be represented as "ax + by"

ganeshie8 (ganeshie8):

since gcd(2, 3) = 1 we can find x and y such that 2x + 3y = 1

OpenStudy (anonymous):

That's very helpful, thank you!

ganeshie8 (ganeshie8):

np.. u wlc :) in general : if \(a|c\) ,\(b|c\), and \(\gcd(a, b) = 1\), then \(ab | c\)

OpenStudy (zzr0ck3r):

I miss number theory, this stuff is so fun.:)

ganeshie8 (ganeshie8):

yes :) and after seeing the kind of problems u post... i think NT is relatively easy compared to abstract algebra or analysis lol

OpenStudy (zzr0ck3r):

I love the abstract, and despise the analysis. Don't tell eliassaab :)

OpenStudy (zzr0ck3r):

well actually I dont mind it if its new, but when we look at the same thing through 20 different ways I get .....

OpenStudy (zzr0ck3r):

I'm sure NT gets nutty at some point, I have only taken a 300 level intro class.

ganeshie8 (ganeshie8):

ahh i never tried abstract/analysis lol @eliassaab elementary NT is fun and easy to understand... advanced nt is harder than abstract and everything i feel... i had a look once and realized soon enough that it was not for meh

OpenStudy (anonymous):

You can also argue this way. 2 occurs in the prime factorization of ( a-b) 3 occurs in the prime factorization of ( a-b) then\[ (a-b)=2^n 3^n q \, \text { where } n\ge 1; m\ge 1 \] Hence 6 divides (a-b)

ganeshie8 (ganeshie8):

looks much simpler ! thnks @eliassaab :)

OpenStudy (zzr0ck3r):

ps @eliassaab I got a 93% on the final and an A in the class thanks for all the help this term in measure theory:)

OpenStudy (anonymous):

YW. Glad you got an A @zzr0ck3r

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