Question

Suppose that a and b are relatively prime integers and that x is an integer. Show that if a divides the product bx , then a divides x
Use the existence of s, t such that sa + tb =1
Please, help

Answer 1

@kirbykirby

Answer 2

Please, check my stuff
a,b are relative prime --> there exist s, t in \(Z\) such that as + tb =1, times x both sides
asx + tbx = x
the first term divided by a, (since a there)
the second term divided by a ( given a divides bx)
therefore, a divides the left hand side --> a| x
Am I right?

Answer 3

yep :) it make sense

Answer 4

yeah, yeah !! thanks girl

Answer 5

np :P

Answer 6

Anyone is good at abstract?? Please, help.

Answer 7

Let n be an integer \(\geq 2\) , let \(\tau \in S_n\) and let X be a nonempty subset of {1,2,....,n}
Say that \(\tau\) fixes X if \(\tau (X)= X\), and say that \(\tau\) acts irreducibly provided the only nonempty subset of {1,2,....,n} fixed by \(\tau\) is {1,2,....,n} itself.
Prove \(\tau \) acts irreducibly if and only if \(\tau \) is an n-cycle.

Answer 8

^not sure I can help on that one :s

Answer 9

@zarkon