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Square root of 2 divided by 0 is rational?

  1. Oct 17, 2014 #1
    Dear All,
    Please help me understand how ## \sqrt{2} ## divide by 0 is rational as stated in the excerpt from alan F beardon's book?
     

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  3. Oct 17, 2014 #2

    phinds

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    I don't follow his logic, but in any case it is actually undefined, and thus it is not meaningful to talk about it being rational or irrational. I think probably what he is saying is more complex than that "(sqrt 2) / 0 is rational"
     
  4. Oct 18, 2014 #3
    He is saying that the multiplicative inverse of the element ##x=a+b\sqrt{2}## in the field ##\mathbb{Q}(\sqrt{2})## is the element ##x^{-1}=\frac{a}{a^2-2b^2}+\frac{-b}{a^2-2b^2}\sqrt{2}##.

    There part where he mentions that ##a^2-2b^2\neq 0## is so that we can rest assured that the formula that he's given us for ##x^{-1}## is defined for all pairs of rationals ##a## and ##b##.
     
  5. Oct 18, 2014 #4

    PeroK

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    You've misunderstood him. If ##a^2 - 2b^2 = 0## then ##\sqrt{2} = \frac{a}{b}## is rational. So, you know that ##a^2 - 2b^2 \ne 0##
     
  6. Oct 18, 2014 #5
    Hey PeroK, I think this is what you mean??

    ##
    a^2 - 2b^2 = 0
    ##
    can be written as
    ##
    (a + \sqrt{2}b)(a - \sqrt{2}b) = 0
    ##
    here note that a and b are rational numbers ## {a + b√2 : a, b ∈ Q}, ##
    so either ## (a + \sqrt{2}b)## or ## (a - \sqrt{2}b) ## must equal 0.
    lets say ## (a - \sqrt{2}b) = 0 ## then ## \sqrt{2} = a/b ## which is a contradiction so ## (a - \sqrt{2}b) = 0 ## cannot happen.
     
  7. Oct 18, 2014 #6

    PeroK

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    That's not quite the way I'd do it! I'd say:

    ##a^2 - 2b^2 = 0 \ \ \Rightarrow \ \ a^2 = 2b^2 \ \Rightarrow \ \ \frac{a^2}{b^2} = 2 \ \Rightarrow \ \ (\frac{a}{b})^2 = 2##
     
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