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Sloppy Proof in Spivak's Calculus?

  1. Jun 11, 2012 #1
    So I was reading Spivak's Calculus for fun and I found on the bottom of page 25, 4th edition:
    Suppose that [itex]\sqrt{2}[/itex] were rational; that is, suppose there were natural numbers p and q such that [itex]\frac{p^{2}}{q^{2}} = 2[/itex].
    We can assume that p and q have no common divisor (since all common divisors could be divided out to begin with). Now we have
    p[itex]^{2}[/itex] = 2q[itex]^{2}[/itex].
    This shows that p[itex]^{2}[/itex] is even, and consequently p must be even; that is p = 2k for some natural number k. Then
    p[itex]^{2}[/itex] = 4k[itex]^{2}[/itex]=2q[itex]^{2}[/itex],
    so 2k[itex]^{2}[/itex] = q[itex]^{2}[/itex].
    This shows that q[itex]^{2}[/itex] is even, and consequently q is even. Thus both p and q are even, contradicting the fact that p and q have no common divisor. This contradiction completes the proof.

    Am I missing something or is this proof not complete and sloppy?! Spivak didn't even consider the possibility that q is odd and p is even; in which case, we can write p as 2k and q as 2m+1 and THEN check the equivalency. In this case we find that two sides won't be equivalent when we substitute 2k and 2m+1 in the given equation and that would complete the proof.
     
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  3. Jun 11, 2012 #2
    The line [itex] 2k^2=q^2 [/itex] implies that [itex]q^2[/itex] is even which in turn implies that [itex]q[/itex] is even. To recap the whole proof, assume [itex]\sqrt{2}=p/q[/itex], a reduced fraction. [itex]p^2/q^2=2 [/itex] implies that p must be even which then implies that q must be even which implies that the fraction is not reduced which is a contradiction.
     
  4. Jun 11, 2012 #3

    AlephZero

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    He did consider that case. After he divides out any common factors, there are 3 possible cases left
    p is odd and q is even
    p is even and q is odd
    p is odd and q is odd

    He then proves that actually both p and q must be even, which means they still have a common factor.

    You made a different (and correct) proof of one of the three cases, but Spivak's proof handles all three cases at once by showing they are all impossible
     
  5. Jun 11, 2012 #4
    Ahh I see how that completes the proof! The line I was looking at was p[itex]^{2}[/itex]=2q[itex]^{2}[/itex]. This line suggests that p is even but q can be even or odd. If q is even we have contradiction as we assumed that common divisors were taken care of. So q must be odd and then we can move on to show LHS [itex]\neq[/itex] RHS after substitution of p as 2k and q as 2m+1 in the equation p[itex]^{2}[/itex]=2q[itex]^{2}[/itex]
    But writing p as some number 2k is definitely more efficient.

    Thanks a lot!
     
  6. Jun 11, 2012 #5
    Yes! I see it now. Thanks a lot :)
     
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