proof about m/nth root of a prime.

cragar

Lets take a prime number and raise it to m/n where m and n are coprime. x,y are coprime
and I want to show that this is irrational.
Proof: lets assume for the sake of contradiction that
[itex] P^{\frac{m}{n}}=\frac{x}{y} [/itex]
P is prime and m,n,x,y are integers.
no we take both sides to the nth power and then multiply the y term over.
[itex] y^n P^m=x^n [/itex]
now we factor y and x into their prime factorization.
[itex] {p^a.....{P_{t}}^b}^n P^m={{P_{q}}^c........}^n [/itex]
okay so if the left side is equal to the right side.
on the left side we know that P has at least m factors, but if P is also contained
in y then it has m+n factors. but if x and y are coprime then x has no common factors with y so y cant have factors of P in it. so now if x contains multiples of P , in order to have the same amount of factors it must be true that ne=m where e is the number of factors of P in x.
but this would imply that m and n are not coprime therefore this is a contradiction and our original number is irrational.

chiro

Hey cragar.

The thing I see is that when you get P^m = xⁿ/yⁿ, then it means that (x/y) have to be a prime since primes have no factors other than 1 or itself. But the only way for this to happen is if x and y are not co-prime and also relevant powers of that prime.

Thus you have a contradiction and you have proved the result that there exists no rational form given x and y are co-prime. Also because we are dealing with a prime, we know that the RHS of the above must be an integer and also that it has a particular decomposition since primes only have factors of 1 and itself.

Edit: Correction should have been / instead of *

cragar

how did you get [itex] p^m=x^ny^n [/itex]

chiro

Sorry, I've changed it: should have been / instead of *

cragar

okay I see, ya that is also a contradiction too and right off the bat.