1. Not finding help here? Sign up for a free 30min 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* Sum of Rational and Irrational Numbers

  1. Jan 27, 2009 #1
    1. The problem statement, all variables and given/known data
    Prove by contradiction: If a and b are rational numbers and b != 0, and r is an irrational number, then a+br is irrational.

    In addition, I am to use only properties of integers, the definitions of rational and irrational numbers, and algebra.

    You guys should also know that I am new to proofs, so if I'm breaking convention in any blatant way, please lemme know.


    3. The attempt at a solution

    First, I figure I need to find the negation of this statement:

    Negation: There exists rational numbers a and b, b!=0, and irrational number r, such that a+br is rational.

    I'm pretty sure this negation is correct, but I've been wrong before

    Anyway, my proof would start like this:

    Proof: To prove that for any rational numbers a,b,b!=0, and irrational number r, a+br is irrational, let's suppose not. Suppose there exists rational numbers a,b,b!=0 and irrational number r such that a+br is rational.
    Let c,d,e,f be integers, by definition of rational, a+br can be rewritten as:

    [tex]\frac{c}{d}[/tex] + [tex]\frac{e}{f}[/tex]*r is rational.

    [tex]\frac{e}{f}[/tex]*r = -[tex]\frac{c}{d}[/tex]

    r = -[tex]\frac{cf}{de}[/tex]

    [tex]\frac{cf}{de}[/tex] can be written as a quotient of integers, so it is rational, therefore, our negation is false, therefore, our theorem is true?

    I think I'm on the right track, but I know its messy and I might have made some logic mistakes...Also, my professor said that if we didn't have to explicitly say why b!=0, we probably weren't doing it right, and I never really did, so I have no idea...any help would be greatly appreciated.

    edit-I guess if b was 0, that would make the quantity a+(br) 0, which leaves you with a, which is just a rational number
     
  2. jcsd
  3. Jan 27, 2009 #2

    Mark44

    Staff: Mentor

    In a proof by contradiction, you assume that the original hypothesis is true and that the conclusion is false.

    Your original hypothesis is "a and b are rational numbers and b != 0, and r is an irrational number". Your original conclusion is "a + br is irrational".

    The negative of your original conclusion is that a + br is not irrational, or in other words that a + br is rational.

    Now start by assuming original hypothesis is true and that a + br is rational (the negated original conclusion). You should arrive at a contradiction, from which you can conclude that when the original hypothesis is true, the original conclusion is true, as well.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?