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Homework Help: Irrationality Proofs

  1. Jun 23, 2010 #1
    1. The problem statement, all variables and given/known data

    a) Prove that [tex]\sqrt{3}, \sqrt{5}, \sqrt{6}[/tex] are irrational. Hint: To treat [tex]\sqrt{3}[/tex], for example, use the fact that every integer is of the form [tex]3n[/tex] or [tex]3n + 1[/tex] or [tex]3n + 2[/tex]. Why doesn't this proof work for [tex]\sqrt{4}[/tex]?

    b) Prove that [tex]2 ^ 1/3[/tex] and [tex]3 ^ 1/3[/tex] are irrational.

    2. Relevant equations

    Can't think of any, really.

    3. The attempt at a solution

    a) I futzed around with fractions involving [tex]3n[/tex] and [tex]3n + 1[/tex] and so forth but I never really got anywhere. I figure that the proof must have something to do with 3, or [tex]\sqrt{3}^2[/tex], but I can't tell exactly what.

    b) I actually got something here: I reasoned that [tex]\sqrt[3]{2} = (\sqrt[3]{2}^2)\sqrt{2}[/tex]. Since we know that [tex]\sqrt{2}[/tex] is irrational, and that (rational) x (irrational) is irrational unless one of the terms is [tex]0[/tex], and that (rational) x (rational) is rational, we can conclude that it is impossible for [tex]\sqrt[3]{2}[/tex] to be rational because if it was, then a rational number ([tex]\sqrt[3]{2}^2[/tex]) multiplied by an irrational number ([tex]\sqrt{2}[/tex]) would equal a rational number, which here cannot be possible. Is this a proof by contradiction?
     
    Last edited: Jun 23, 2010
  2. jcsd
  3. Jun 23, 2010 #2

    Mark44

    Staff: Mentor

    What you meant to say but your LaTeX script wasn't quite right was this:
    Prove that [tex]2 ^ {1/3}[/tex] and [tex]3 ^ {1/3}[/tex] are irrational.

    If your exponent inside a [ tex] tag takes two or more characters, surround them with braces - { }.

    You can also use cube roots, using \sqrt[3]{ ... }, like this:
    Prove that [tex]\sqrt[3]{2}[/tex] and [tex]\sqrt[3]{3}[/tex] are irrational.


     
  4. Jun 23, 2010 #3
    Sorry about that. I went back and fixed the [tex]\sqrt[3]{2}[/tex] in the second half but forgot about the first!
     
  5. Jun 23, 2010 #4

    Mark44

    Staff: Mentor

    To prove that, say, sqrt(2) is irrational, assume the opposite: that it is rational and so can be written as a/b, with no factors in common in a and b.

    sqrt(2) = a/b ==> 2 = a2/b2

    You should arrive at a contradiction, which means that your assumption that sqrt(2) is rational must not be true.
     
  6. Jun 23, 2010 #5
    Oh yes, I know that (Spivak actually provides what is essentially the same proof). I'm just looking to see if my reasoning is correct [b) in "The attempt at a solution"]
     
  7. Jun 23, 2010 #6

    Mark44

    Staff: Mentor

    In b, you have [tex]\sqrt[3]{2} = (\sqrt[3]{2}^2)\sqrt{2}[/tex]
    Why would you think this? The left side is about 1.4422; the right side is about 2.2449.
    I don't get what you're trying to do here.
     
  8. Jun 23, 2010 #7
    Oh dear. It's an arithmetic error that I overlooked (I multiplied two of the exponents when I should have added them).

    I guess there's a reason Spivak's proof looked so much longer than mine!

    Should probably turn attention back to a), then.
     
  9. Jun 24, 2010 #8

    Gib Z

    User Avatar
    Homework Helper

    This doesn't use the hint, but you could generalize the proof in Spivak for sqrt 2 to any non-perfect square. Keep an eye out for which step its crucial that its not a perfect square.
     
  10. Jun 24, 2010 #9
    The step that most sticks out to me is the one in which Spivak asserts that [tex]p^2 = 2q^2[/tex]. If 2 were some perfect square, then the equation would look like [tex]p^2 = (aq)^2[/tex] where [tex]a[/tex] is some integer, which would allow one to solve for [tex]p/q[/tex]. But I'm puzzled as to how I could generalize that proof, since it seems to rest on the fact that the numbers involved are even, which wouldn't work for 3.

    In other words, I'd end up with [tex]p^2 = 3q^2[/tex]. All this seems to tell me is that [tex]p[/tex] and [tex]q[/tex] are both even or both odd. But whereas Spivak's proof for 2 relied on [tex]p[/tex] and [tex]q[/tex] being even, and so having a common factor of 2, that doesn't seem to be possible with [tex]p[/tex] and [tex]q[/tex] being odd.
     
  11. Jun 24, 2010 #10
    You can also look at the factorization in prime numbers. If you have:

    sqrt[p] = r/s

    where p is a prime number and r and s are integers that have no divisirs in common, then you have:

    p = r^2/s^2 ---------->

    r^2 = p s^2


    If you were to factor both sides and count (by multiplicity) how many prime factors there are on each side, what would you find?
     
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