Results 1 to 7 of 7

Math Help - Rational Numbers vs Integers

  1. #1
    Newbie
    Joined
    Jan 2010
    Posts
    9

    Rational Numbers vs Integers

    Hi,
    I can't seem to wrap my head around this proof of rational numbers and integers.

    The statement says:
    For all rational numbers a, there is a rational number b, such that a + b is not an integer.
    I say this statement is true.

    I would suppose there is a rational number a, where a = p/q for some integers p and q.
    But I'm not sure what to let b equal, in order for it to be a rational every time. I've tried letting b equal some manipulation of a, but nothing seem to work. For example, I tried letting b = 1/a or q/p. But it p was 0, then b is undefined and wouldn't work.

    Can I get help on this please?

    Thanks in advance!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Aug 2009
    From
    Israel
    Posts
    976
    Separate it to 2 cases:

    1) a is an integer;
    2) a is not an integer.

    Do you see what to do now?

    Oh, and welcome to the forums
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jan 2010
    Posts
    9
    Thanks for the quick reply!

    So for the first case, where a is an integer, would it be okay if I wrote:

    Case1: Suppose p divides q, and thus a is an integer.
    Let b = x/y; for some integers x and y, and y cannot equal 0
    (a) + (x/y) = (ay + x)/y; where (ay + x) is an integer
    Therefore a + b is not an integer

    Case2: Suppose p does not divide q, and thus a is not an integer
    Let b = s/t; for some integers s and t, and t cannot equal 0
    (a) + (s/t) = (at + s)/t; where (at + s) is an integer
    Therefore a + b is not an integer

    I really feel like I'm not being clear enough here, because (ay + x) could still be divisible by y and (at+s) could still be divisible by t, but did I fix that by saying for some x/y, and s/t?

    And thank you again for the welcome!

    Edit: After reading that through, I'm almost positive I did that wrong, and now I'm more lost than ever.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Aug 2009
    From
    Israel
    Posts
    976
    You're welcome.

    You're not being clear enough - that's correct:
    What you need to do is choose a number b, such that a+b is not an integer.

    The purpose of separating the cases is that for each case, you can easily find a suitable b.
    I'll help you with the easier one (well, they're both easy once you understand what you need to do ): If a is not an integer, you can choose b=0. Then a+b = a, and a is not integer, hence a+b is not an integer. Alternatively, for this case, you can choose any integer to be b, and that would work as well.

    Now, what b must you choose, so that a+b is not an integer, when a is an integer?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Jan 2010
    Posts
    9
    Oh I think I see now.
    So if a were an integer, I could let b = 1/2.
    But now can I just say, "integers adding with non-integers will result in a non-integer"? Or is there an easier way to state this fact?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member
    Joined
    Aug 2009
    From
    Israel
    Posts
    976
    Well, that mostly depends on the sort of class you're taking..
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Jan 2010
    Posts
    9
    Okay, the important thing is that I finally understand how to do this.
    Thanks again!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Proof: All rational numbers are algebraic numbers
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: September 5th 2010, 11:26 AM
  2. Replies: 7
    Last Post: August 3rd 2010, 02:31 PM
  3. Integers numbers
    Posted in the Number Theory Forum
    Replies: 3
    Last Post: July 2nd 2009, 11:44 AM
  4. Replies: 8
    Last Post: September 15th 2008, 05:33 PM
  5. integers & rational numbers
    Posted in the Algebra Forum
    Replies: 2
    Last Post: May 13th 2007, 07:37 AM

Search Tags


/mathhelpforum @mathhelpforum