Q1) prove that if d is a positive integer, d\a and d\b, then (a,b) = d...if and only if (d\a, d\b) = 1.
Q2) Prove that if P is a prime and a is an integer, and p does not divide a then (a,p) = 1.
yeah I am sorry...that form is d divides a and d divides b...I messed up with my signs!
I don't think so we have studied this before! I think we have to use the division algorithm for that purpose...
but I tried solving it...please approve if that makes sense or help me out!
when we say the common divisor of a and b is d it means this:
a = d * x
b = d * y
so when we divide a on d ( a/d) 'x' would remain.
in other words x = a / d
and in the same way when we divide a on d (a/d) y would remain. in other words y = b/ d
note that (x, y)= 1 because if it wasn't 1 the greatest divisor of a and b would be more than d ( for example if (x,y)= s we would have (a, b) = (dx, dy) = d*(x,y)= d*s)
so (x,y)=1
and (a/d , b/d)= ( x , y)=1...
Thanks for offering help!
It looks confusing to me... though it looks correct ^^
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Known fact : d is a common divisor of a and b.
We want to prove , where x and y are defined like you did
Proving
It corresponds to what you've written here :
The thing is that it's not very good to sa "it's that because if it was not..."note that (x, y)= 1 because if it wasn't 1 the greatest divisor of a and b would be more than d ( for example if (x,y)= s we would have (a, b) = (dx, dy) = d*(x,y)= d*s)
so (x,y)=1
and (a/d , b/d)= ( x , y)=1...
What you've done is sort of proving the contrapositive.
Suppose . Then
So we have proved that , which is equivalent to .
Yes, it looks like it's a lot, but I'm just trying to explain as much as possible the reasoning
Proving
If , then ... The rest follows