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Proving that (p)^{1/n} is Irrational when p is Prime and n>1Problem: x^{n}
- p = 0
or, equivalently, x = (p)^{1/n}. and specify that n > 1.^{11} Prove that x is an irrational number.
Solution: Let x be rational; i.e., let x = a/b where a and b are integers. Then: p =
x^{n} = a^{n}/b^{n} = a rational number.
Since p is prime, then p is an integer. Thus, either: b^{n}
= 1 or b^{n} = a^{m} where m < n ^{12}
Special case: If m = n - 1, then p = a, and b^{n} = p^{n - 1}, or p = b^{(n/[n - 1])} = b. But if this latter statement is true, then p = 1, and we violate the assumption that p is any prime. Since the assumption that x is a rational number leads to contradictions in all possible cases, we must conclude that x is irrational.
^{11}
Notice that, for n = 1, x = p, and p = p/1 = p/
= /,
which is a rational number. The proof for irrationality is only valid
when n > 1.
^{12} If n = m, then p = 1, and the assumption that p is any prime is violated. ^{13} Under this condition, p = a...a (taken n times), and a is a factor [divisor] n times over. |
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