SOLUTION: To show that is an irrational number unless n is a perfect square, explain how the assumption that is rational leads to a contradiction of the fundamental theorem of arithmetic by
Algebra ->
Square-cubic-other-roots
-> SOLUTION: To show that is an irrational number unless n is a perfect square, explain how the assumption that is rational leads to a contradiction of the fundamental theorem of arithmetic by
Log On
Question 1041791: To show that is an irrational number unless n is a perfect square, explain how the assumption that is rational leads to a contradiction of the fundamental theorem of arithmetic by the following steps:
(A) Assume that n is not a perfect square, that is, does not belong to the sequence 1, 4, 9, 16, 25, . . . . Explain why some prime number p appears an odd number of times as a factor in the prime factorization of n.
(B) Suppose that where a and b are positive integers, Explain why
(C) Explain why the prime number p appears an even number of times (possibly 0 times) as a factor in the prime factorization of
(D) Explain why the prime number p appears an odd number of times as a factor in the prime factorization of
(E) Explain why parts (C) and (D) contradict the fundamental theorem of arithmetic. Answer by solver91311(24713) (Show Source):
Fundamental Theorem of Arithmetic: Every integer greater than 1 is either prime itself or the product of primes.
Beyond this, I can't help you much because you have left stuff out of the questions. I suspect you cut and paste from an on-line course and some of the information was graphical that did not copy and render. Next time READ what you have in the dialog box BEFORE you hit send.
John
My calculator said it, I believe it, that settles it
