Can someone explain why B^2 * P definitely has an odd number of prime factors?
-
-
-
This Tweet is unavailable.
New conversation -
-
-
Serious question, is there an integer for which the square root is rational but not an integer?
-
No because a non-integer rational number is comprised of two numbers that don't share prime factors. Squaring those numbers means the results continue to not share prime factors.
End of conversation
New conversation -
-
-
isn't that easily generalisable to any integer that is not a square, i.e., whose prime factors do not all have even powers?
-
No because if P =15 then P has an even number of factors. Of course it is irrational, but this particular proof doesn't work.
- 3 more replies
New conversation -
-
-
You can make a more general argument that all integers that are not perfect nth powers have irrational nth roots by proving 2 things: 1: That the integer is between two consecutive nth powers. 2: That the nth power of any non-integer rational is a non-integer rational.
-
Or in the spirit of the given proof, given an integer x that isn't an n'th power, let p be a prime that doesn't appear a multiple of n times in the prime factorization of x. Then x * b^n = a^n, and p appears non-zero times (mod n) on the left, and zero times (mod n) on the right.
End of conversation
New conversation -
-
-
Not sure of the assumption that A^2 or B^2 will have even number of prime factors. For example, both 7 and 7^2=49, have the same number of prime factors
-
7*7 -> 2 factors which are prime. And yes, equal.
End of conversation
New conversation -
Loading seems to be taking a while.
Twitter may be over capacity or experiencing a momentary hiccup. Try again or visit Twitter Status for more information.