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Help with proving irrationality

  1. Sep 27, 2009 #1
    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.


    2. Relevant 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.

    1. The problem statement, all variables and given/known data

    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.
     
  2. jcsd
  3. Sep 28, 2009 #2
    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. =><=
     
  4. Sep 28, 2009 #3

    Dick

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    Homework Helper

    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?
     
  5. Oct 1, 2009 #4
    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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