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Is this a valid proof?

  1. Oct 6, 2013 #1
    Is this a valid proof? Also is this way of doing it valid?

    Statement : There are infinitely many natural numbers n where the square root of n is rational.

    Proof:
    sqrt of n = x (where x is natural)
    n= x squared

    And n can be any natural number(x) squared ,and there are infinitely many natural numbers (x)
    therefore there are infinitely many n which has a natural square root. Since natural numbers are rational , there are infinitely many natural numbers n where the square root of n is rational.
     
  2. jcsd
  3. Oct 6, 2013 #2

    Ray Vickson

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    Your argument seems fine. There is one small point, though, that you should consider: how do you know that different integers x yield different integers x^2? (You need to argue that in order to know that squaring does not cause everything to collapse into a small, finite set of values.)
     
  4. Oct 6, 2013 #3
    Try going like this...
    There are two possibilities:
    1.There are infinitely many natural numbers n where the square root of n is rational.
    2.There are finite natural number whose roots are rational.
    If things were go as 1. then the proof is complete. If things were to go as 2. then we have to prove it wrong..
    If second condition is satisfied then we will come up with a maximum number nmax (say) whose root is rational. Prove that there exist another number greater than nmax whose root is rational.
    So the 2. possibility is ruled out, hence we are left with only one possibility which is 1.
    Regards
     
  5. Oct 6, 2013 #4
    Thanks for responding.

    Isn't the square of an integer always different?(except 1 or 0 i guess)?

    And I dont understand the "(You need to argue that in order to know that squaring does not cause everything to collapse into a small, finite set of values.)" part (specifically what are you referring to that is collapsing?).
     
  6. Oct 6, 2013 #5

    Yes i know of proof by contradiction, but i used a more straightforward method. Though im not entirely sure my proof works.
     
  7. Oct 6, 2013 #6

    Dick

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    It's perfectly valid. The set of all n^2 for n any natural number i.e. {1^2,2^2,3^2,...} has an infinite number (since there are no duplicates, that's what Ray was saying about 'collapsing') of elements all of which are all natural numbers. Their square roots are all natural numbers, i.e. {1,2,3....} hence rational. That's what you are saying, right? In fact, those are the ONLY natural numbers that have rational square roots, but you don't need to show that.
     
  8. Oct 7, 2013 #7

    HallsofIvy

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    No, that is not true. (2)^2= (-2)^2. However, YOU are working with natural numbers, not integers (do you understand the difference?) so it is true. But the point is that, unless you are "given" that, you need to prove it. Suppose m and n are different natural numbers. Then [itex]n- m\ne 0[/itex] because they are different. Further [itex]m+n\ne 0[/itex] (why?). Therefore [itex]m^2- n^2= (m- n)(m+ n)[/itex] is not 0 so [itex]m^2[/itex] is not equal to [itex]n^2[/itex].

    For example, when squaring, the set {-3, -2, 2, 3} "collapses" to the smaller set {4, 9}. The proof above shows that cannot happen for the natural numbers.
     
    Last edited: Oct 7, 2013
  9. Oct 7, 2013 #8
    Ah right i confused naturals(whole numbers larger than zero) with integers (integers include less than zero whole number right). Thanks for the explanation.
     
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