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Homework Help: Rational and Irrational Numbers

  1. Jan 25, 2005 #1
    I need to show that a rational + irrational number is irrational. I am trying to do a proof by contradiction.

    So far I have:

    Suppose a rational, b irrational.
    Then a = p/q for p, q in Z.
    Then a + b = p/q + b = (p + qb) / q
    But I don't know where to go from here because I still have a rational plus an irrational and that is what I am trying to show.

    Also, would a similar proof work to show that an irrational + irrational = irrational?
  2. jcsd
  3. Jan 25, 2005 #2


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    When you're doing a contradiction, you should assume the opposite of the premise and show that it leads to an absurd conclusion.

    For this problem, try this :

    Let [itex]a = \frac{p}{q}[/itex] where p and q are coprime integers.

    b is irrational.

    Now let us say their sum is a rational number, which can be expressed as [itex]\frac{s}{t}[/tex], where s and t are coprime integers.


    [tex]a + b = \frac{s}{t}[/tex]

    [tex]b = \frac{s}{t} - a = \frac{s}{t} - \frac{p}{q} = \frac{sq - pt}{qt}[/tex]

    We have just shown that b can be expressed as the ratio of two integers. But b is irrational.

    This is a contradiction. Hence the assumption is false and the original premise is true.

    This isn't even true in general. What can you say about the sum of [tex]\sqrt{2}[/tex] and [tex](5 - \sqrt{2})[/tex] ?
  4. Jan 25, 2005 #3


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    The proof is one line. :smile:

    So, you're doing proof by contradiction, and have assumed that there are rational numbers p and q, and an irrational number z, such that:

    p + z = q


    It is also fairly easy to construct counterexamples... but the method might be easier to find once you've done the first problem.
  5. Sep 2, 2010 #4
    I need to show that a rational - irrational number is irrational. I am trying to do a proof by contradiction.
    Plz send me the related answer as I need it....
    Last edited: Sep 2, 2010
  6. Sep 2, 2010 #5


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    Since this clearly has nothing to do with physics, I am moving it to "precalculus homework".
  7. Sep 2, 2010 #6


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    Homework Helper

    Anjuyogi, take a look at the previous posts?
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