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Homework Help: Help with proof for rational number problem

  1. Apr 25, 2010 #1
    1. The problem statement, all variables and given/known data
    For all a in the set of real numbers, if a is rational, [tex]a + \sqrt{2}[/tex] is irrational.
    You may use that [itex]\sqrt{2}[/itex] is irrational and the sum and difference of rational numbers is rational.

    2. Relevant equations

    3. The attempt at a solution

    My proof seems way too simple, I don't trust it. Can anyone see anything wrong with this?


    Let a be an arbitrary rational number. Therefore a = m/n for some integers m and n (n is not 0).

    Suppose [itex]a + \sqrt{2}[/itex] is rational. Therefore [tex]\frac{m}{n} + \sqrt{2} = \frac{q}{r}[/tex] where q and r are integers (r is not 0).

    [tex] \sqrt{2} = \frac{q}{r} - \frac{m}{n}[/tex]

    BUT this is a contradiction because we know that [tex]\sqrt{2}[/tex] is an irrational number. But since [tex]\frac{q}{r}[/tex] is rational and [tex]\frac{m}{n}[/tex] is rational, we know [tex]\frac{q}{r} - \frac{m}{n}[/tex] is rational because the difference of two rationals is rational.

    Therefore [tex]\frac{m}{n} + \sqrt{2}[/tex] is irrational.

  2. jcsd
  3. Apr 25, 2010 #2
    yep, that's right. generally, a rational plus an irrational is an irrational. Proven in the exact same way
  4. Apr 25, 2010 #3
    Thank you very much :D
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