Prove that if 2|(a-b) and 3|(b-a), then 6|(a-b)
\(2 | (a-b) \implies a-b = 2k\) \(3 | (b-a) \implies b-a = 3m\)
since \(\gcd(2, 3) = 1\) we can write \(2x + 3y = 1\) for some integers \(x\) and \(y\)
\(a-b = (a-b)\times 1 = (a-b)(2x+3y) = (a-b)2x + (a-b)3y\) \(= -3m*2x + 2k * 3y = 6(-mx+ky )\) so, \(6 | a-b\)
let me knw if smthng doesnt make sense
Thank you very much. Why does gcd(2,3)=1 imply that 2x + 3y =1 ?
good question :) gcd of any two integers is the LOWEST possible positive integer that can be written as linear combination of those two integers
thats mouthful definition
you need to prove it upfront, before pursuing this current problem
gcd(a, b) = smallest possible positive integer than can be represented as "ax + by"
since gcd(2, 3) = 1 we can find x and y such that 2x + 3y = 1
That's very helpful, thank you!
np.. u wlc :) in general : if \(a|c\) ,\(b|c\), and \(\gcd(a, b) = 1\), then \(ab | c\)
I miss number theory, this stuff is so fun.:)
yes :) and after seeing the kind of problems u post... i think NT is relatively easy compared to abstract algebra or analysis lol
I love the abstract, and despise the analysis. Don't tell eliassaab :)
well actually I dont mind it if its new, but when we look at the same thing through 20 different ways I get .....
I'm sure NT gets nutty at some point, I have only taken a 300 level intro class.
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
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)
looks much simpler ! thnks @eliassaab :)
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:)
YW. Glad you got an A @zzr0ck3r
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