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!