1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    jbunniii

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

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

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

    Then

    [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

    jbunniii

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Looks good.
     
  6. Feb 2, 2012 #5
    sweet thanks for the help
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook