- #1
cragar
- 2,552
- 3
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: let's 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 can't 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.
and I want to show that this is irrational.
Proof: let's 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 can't 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.