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P^2 = 2, proof question

  1. Oct 7, 2014 #1
    I am reading this paragraph in little Rudin, right at the beginning.

    Let A be the set of all positive rationals, p such that p2 < 2 and let B consist of all rationals p such that p2 > 2. We shall now show that A contains no largest number, and B...

    To do this we associate with each rational p > 0 the number

    q = p - ((p2 - 2) / (p + 2))

    Now, I can't see where this is going. We want to pick a q that is rational, and always bigger than p but less than root(2) or always smaller than p but greater than root(2) depending on whether we are considering the set A or B.

    Because we want q to be rational we can't pick the number directly in between p and root(2), that is (p + root(2) /2).

    I guess my question is, how would one come up with that calculation for q? I can't for the life of me wrap my head around where that formula for q would come from if I was trying to pick q on my own..
  2. jcsd
  3. Oct 7, 2014 #2


    Staff: Mentor

    From above, and given that p > 0, $$q = p - \frac{p^2 - 2}{p + 2} = \frac{p^2 + 2p - p^2 + 2}{p + 2} = 2\frac{p+1}{p+2} < 2$$
    Look at the two cases separately, in one of which p ##\in## A, and the other in which p ##\in## B.

    If p < ##\sqrt{2}## (p ##\in## A), then the formulation of q that you gave shows that q > p. Can you see why? Similarly, if p > ##\sqrt{2}## , (p ##\in## B), q < p. Both of these hinge on whether we're subtracting a negative quantity from p or subtracting a positive quantity.

    I'm not sure what led to the formulation of q, but it might be helpful to draw and label a part of the number line for each of the two cases, showing the relative positions of p, q, and ##\sqrt{2}##.
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