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Homework Help: Proof about rational numbers.

  1. Feb 2, 2012 #1
    1. The problem statement, all variables and given/known data
    Show that if [itex] a\in\mathbb{Q} [/itex] and [itex] t\in\mathbb{I} [/itex]
    then [itex] a+t\in\mathbb{I} [/itex] and [itex] at\in\mathbb{I} [/itex]
    as long as a≠0

    3. The attempt at a solution
    Let [itex] a=\frac{x}{y} [/itex] where x and y are integers. and t is an irrational number
    If I have a+t . since t cannot be written as a fraction, there's no way an integer times an irrational number will be an integer so this number will be irrational. and also at and a+t would be irrational.
  2. jcsd
  3. Feb 2, 2012 #2


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    I didn't understand your argument. I suggest starting by assuming the contrary. Suppose

    [tex]a + t \not\in \mathbb{I}[/tex]


    [tex]a + t \in \mathbb{Q}[/tex]

    So there is some rational [itex]r[/itex] such that [itex]a + t = r[/itex]. Can you explain why this is impossible?
  4. Feb 2, 2012 #3
    ok i see. So we assume that a+t is a rational number. let a=x/y
    and let a+t=L/M=x/y+t=L/M
    and when we subtract x/y from both sides and we get a common denominator and simplify the right hand side we get that t is a rational number. which is a contradiction, therefore a+t is an irrational number. does this work
  5. Feb 2, 2012 #4


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    Looks good.
  6. Feb 2, 2012 #5
    sweet thanks for the help
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