Help with proving irrationality

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The discussion revolves around proving the irrationality of square roots and their generalizations using properties of prime factorization. The user expresses confusion about how to apply the fundamental theorem of integers to show that sqrt(n) is irrational unless n is a perfect square. They consider a proof by contradiction, assuming sqrt(n) is rational and leading to a contradiction by examining the implications of prime factors. The conversation emphasizes the importance of unique prime factorization in establishing the irrationality of roots of non-square integers. Overall, the thread seeks clarity on how to effectively structure the proof while leveraging the definitions and theorems provided in Spivak's text.
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From Spivak's 4th edition

I'm having some difficulties knowing how to prove these things I need to prove. If someone could help me out, I would appreciate it.


Homework Equations



The book defines a prime number as this: A natural number p is called a prime number if it is impossible to write p=ab for natural numbers a and b unless one of these is p, and the other 1. If n>1 is not a prime, then n=ab, with a and b both < n, if either a or b is not a prime it can be factored similarly.

Homework Statement



I got a fine but then

A fundamental theorem about integers states that this factorization (was talking about factoring stuff down to primes in previous problem) is unique except for the order of the factors. Thus, for example, 28 can never be written as a product of primes one of which is 3, nor can it be written in a way that involves 2 only once.

b. Using this fact, prove that sqrt(n) is irrational unless n = m2 for some natural number m.

c. Prove more generally that ksqrt(n) is irrational unless n = mk


I am so confused on what to do, and how that fact helps me? I was thinking of trying to do a proof by contradition, and attempt to show that sqrt(n) is rational? But I'm not sure how to do this because then I could just prove n=m2. i have no idea how to show that if n =/= m2, sqrt(n) is irrational. Please help.
 
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Does this mean anything at all for b?

Proof by contradiction. Assume sqrt(n) is rational and n=/=m2 for any rational m.

sqrt(n) = p/q where p,q are rational numbers.

if n =/= m2, p/q =/= m. A rational number over a rational number has to rational. =><=
 
You've got the right start. Assume sqrt(n) rational. So n=(a/b)^2. You can also assume a and b have no common prime factors (otherwise you could just divide them out and get a/b in 'lowest terms'). Now that means b^2*n=a^2. Suppose a prime p divides b. Then it also divides a^2 which means it divides a. But we assumed it doesn't. So b isn't divisible by any primes. What is b?
 
there's a part in the book above that question about proving the irrationality of cube root, square root, 5-root.. et c and then it tells you to think about how that wouldn't work if it was something like 4- root for example. I think that part could help you
 
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