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Irrationaltiy of sqrt(2)

  1. Mar 5, 2005 #1

    mathwonk

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    Have you seen this argument? If sqrt(2) = p/q is in lowest terms, then also 2/1 = p^2/q^2 is in lowest terms. Since lowest terms is unique, p^2 = 2 and q^2 = 1. Thus sqrt(2) is the integer p. But 1^2 is too small, 2^2 is too big and all the resta re even bigger, so this is false. So sqrt(2) is not rational.

    does anyone have a shorter one?
     
  2. jcsd
  3. Mar 5, 2005 #2

    dextercioby

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    Nope.I've known this proof from 6-th grade.Learnt it in school.It's much nicer (though roughly equivalent) than the one with prime factors...

    Daniel.
     
  4. Mar 6, 2005 #3

    mathwonk

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    boy you had a good 6th grade teacher!
     
  5. Mar 6, 2005 #4

    matt grime

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    A better proof, perhaps, and one that works for all non-squares.

    let sqrt(x) = p/q, then xq^2=p^2, counting primes with multiplicity, unless x is a perfect square there will be an odd number of prime factors of the left hand side and an even number of the right handside
     
  6. Mar 6, 2005 #5

    mathwonk

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    thats a nice proof, but when i first saw that as a student i didn't understand it, so i came up with the one above as using fewer tools, or at least apparently so.

    moreover, doesn't the first proof above also work on all non squares? I.e. if p/q = sqrt(n) is in lowest terms, then so also is p^2/q^2 = n/1, so p^2 = n, and hence n is a perfect square.

    of course both proofs rest on exactly the same fact, uniqueness of factorization into primes.
     
  7. Mar 6, 2005 #6

    matt grime

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    I just don't like using fractions unnecessarily.
     
  8. Mar 6, 2005 #7

    mathwonk

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    chaque a son gout. but you do not mind using prime factorization.

    To me it seems rather hard to avoid fractions in discussing rational numbers, but i agree the correct first step in a problem involving fractions is normally to remove them.
     
    Last edited: Mar 6, 2005
  9. Mar 7, 2005 #8
    This may be the same argument as you're using, don't know really. Assume that sqrt(2) is rational, then we have p and q such as sqrt(2) = p/q. This gived 2 = p^2/q^2. But since gcd(p,q)=1 then p^2/q^2 isn't an integer unless q=1. This gives the equation 2 = p^2 with integers, no integer can solve this equation, and hence sqrt(2) cannot be rational.
     
  10. Mar 11, 2005 #9
    would it not be:

    p^2 = nq^2

    ?
     
  11. Mar 11, 2005 #10

    mathwonk

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    well if p^2 /q^2 = n/1, and the left side is known to be in lowest terms, and if the lowest terms form of a fraction is unique, then since the right side n/1 is obviously also in lowest terms, then the two fractions must be equal, i.e;. they must have the same top and same bottom.

    Since the tops are equal, p^2 = n.

    so of course you are right, that p^2 = nq^2, but in fact also q^2 = 1.
     
  12. Mar 12, 2005 #11
    I've heard that the nth root of 2 (n an integer, greater than or equal to 2) is irrational. How would one prove that?
     
  13. Mar 12, 2005 #12

    Hurkyl

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    Have you tried applying the same technique?
     
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