Prove that the fifth of root of 15 is irrational

Question

anonymous · Answer

same proof by contradiction as $\sqrt2$ is irrational
suppose it is, and derive a contradiction 
$$\sqrt[5]{15}=\frac{a}{b}\15=\frac{a^5}{b^5}\15b^5=a^5$$ and contradiction like with prime factorization on both sides of the equal sign

anonymous · Answer

Can you go into more detail about last part?

anonymous · Answer

I follow up until the last part. I've done the proof of 2^1/2, but I dont understand how to compare it to any other root

anonymous · Answer

yes i can be in more detail but i will say it in english you can write the math

anonymous · Answer

on the right hand side you have $a^5$ 
however $a$ factors as the product of primes, when you raise it to the 5th power, each prime will have a power that is a multiple of 5

anonymous · Answer

on the left hand side you have $b^5$ and the same argument holds, each prime factor will be raised to a multiple of 5 for that
but the left hand side is $15b^5=3\times 5\times b^5$ so the prime $3$ will have an exponent that is NOT a multiple of 5, as will the prime $5$
since there is only one way to factor a number as the product of primes (fundamental theorem of arithmetic) that is a contradiction

anonymous · Answer

okay, i don't really understand that, but it works. Thank you