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

mr_coffee

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.

Proof:
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!

HallsofIvy

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).

mr_coffee

That does look alot nicer, thanks for the help!