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Square root of two

  1. Aug 30, 2014 #1
    1. The problem statement, all variables and given/known data
    In this task we will show that √2 is irrational. Assume that √2 = a/b where both a and b are natural numbers. Let a = p1p2p3...pn, and b = q1q2q3...qm be the prime factors

    a) explain why 2q1q1q2q2q3q3...qmqm = p1p1p2p2p3p3...pnpn

    √2=a/b

    b√2=a

    √2q1q2q3...qm=p1p2p3...pn

    (√2(q1q2q3...qm))2=(p1p2p3...pn)2

    2q1q1q2q2q3q3...qmqm = p1p1p2p2p3p3...pnpn

    b) explain why there can't be the same amount of 2-factors on the left and the right

    Im stuck on this one...

    Is it because then a and b could be further shortened/divided?
     
  2. jcsd
  3. Aug 30, 2014 #2

    pasmith

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    Is there some reason why you didn't write that as [tex]
    2q_1^2 \dots q_m^2 = p_1^2 \dots p_n^2?[/tex]

    The left hand side is even. Therefore the right hand side is also. But the right hand side is the square of an integer, [itex]a[/itex]. Either [itex]a[/itex] is even or [itex]a[/itex] is odd. Only one of these is possible if [itex]a^2[/itex] must be even. What does that say about the minimum number of times [itex]a^2[/itex] can be divided by 2? What can you then say about the integer [itex]b^2 = a^2/2[/itex]?
     
  4. Aug 31, 2014 #3
    No, the book wrote it that way. So i did to.

    I don't know how to prove it, but i remember that if a2 is even, then a is even also(given that a2 is a square number which we know it is, because a was a whole number). So a is even and that would imply that i could write a=(2k) where k is a natural number 1, 2, 3, 4... If i square this i get that a2=(2k)2=4k2. I see from this that a2 could be divided by 2 two times and still be a natural number. If i know look at b2=a2/2 i can see that a2/2 can be written as 2k2.

    We get that:
    b2=2k2

    b2/k2=2

    b/k=√2

    This last result goes against what what we assumed. That a and b had no common factors? (the task doesn't really assume this, i assume this)

    (Or maybe we don't have to assume that a and b had no common factors. We can rather argue that since we originally had √2=a/b and got that this could be rewritten to √2=b/k, we could just repeat the process an infinite amount of times, and thus removing an infinite amount of 2-factors from the fraction. And this is absurd. So there can't be the same amount of 2-factors on the left and the right.)
     
    Last edited: Aug 31, 2014
  5. Aug 31, 2014 #4

    PeroK

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    You seem to be mixing up two different proofs. The simplest, which I think you've almost got above, is:

    Assume a, b have no common factors, then show that a and b are both even. You don't need prime factorisations for this.

    There is another proof using the unique prime factorisation of a and b. This can be used to show that if a and b have no common factors, then ##a^2## and ##b^2## have no common factors. Therefore, the square root of any whole number is either a whole number or irrational; it's never a/b with b ≠ 1.
     
  6. Aug 31, 2014 #5
    there is also a (last) task c)

    Use the fundamental theorem of arithmetic to derive a contradiction
    ----------------------------------------------------------------------------

    Wouldn't i be doing task c if i do what you say?

    I think I'm just supposed to - at least for now - explain why there can't be the same amount of 2-factors on the left and the right...
     
  7. Aug 31, 2014 #6

    pasmith

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    Is the number of factors of 2 on the left even or odd?
    Is the number of factors of 2 on the right even or odd?
     
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