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Proof by contradiction, i think i got it, its long but i think it works, suggestions

  1. Sep 26, 2006 #1
    Hello everyone! THe directions are the following:
    Carefully formulate the negations of each of the statements. Then prove each statement by contradiction.

    Here is the problem:
    If a and b are rational numbers, b!=0, and r is an irrational number, then a+br is irrational.

    I made it into a universal statement:
    [tex]\forall[/tex] real numbers a and b and r, if a nd b are rational such that b != 0, and r is irrational, then a + br is rational.

    I then took the negation:
    [tex]\exists[/tex] rational numbers a and b, b != 0 and irrational number r such that a + br is rational.

    Suppose not. That is, suppose there are rational numbers a and b, with b != 0 and r is an irrational number such that a + br is rational. By definition of rational a + br = e/f and a = m/n and b = x/y for some integers e, f, m, n, x, and y with n,y,x, and f != 0. By substitution

    a + br = e/f
    a = m/n;
    r = irrational
    b = x/y

    m/n + (x/y)*r = e/f

    r = y/x(e/f – m/n)

    r = (yen-fmy)/fnx

    now yen-fmy and fnx are both integers, and fnx != 0. Hence r is a quotient of the integers yen-fmy and fnx with fnx != 0. Thus, by definition of rational, r is rational, which contradicts the supposition that a+br is irrational.

    I know there are alot of variables, but i wasn't sure how else to show that r is infact a quotient of integers.

    Any comments on what i did if its correct or flawed? Thank you!
  2. jcsd
  3. Sep 27, 2006 #2


    User Avatar
    Staff Emeritus
    Science Advisor

    What you've done is perfectly correct but you are right- it's hard to read with all those letters.

    You might just use the fact that the set of rational numbers is closed under addition and multiplication- and every non-zero rational number has a rational multiplicative inverse.

    If a+ br= c then r= (c-a)(1/b). If a, b, c, b not 0, are all rational then c-a is rational (because the rational numbers are closed under addition), 1/b is rational (b is not 0 so has a rational multiplicative inverse), (c-a)(1/b) is rational (the rational numbers are closed under multiplication).
  4. Sep 27, 2006 #3
    That does look alot nicer, thanks for the help! :biggrin:
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