Answer
Claim: If a and b are rational numbers, b ≠ 0, and r is an irrational number, then a + br is irrational.
Negation: If a and b are rational numbers, b ≠ 0, and r is an irrational number, then a + br is rational.
Proof: Suppose that a and b are rational numbers, b ≠ 0, and r is an irrational number such that a + br is rational.
By definition of rational, a = c/d and b = e/f.
By substitution, a + br = c/d + er/f.
By laws of algebra, c/d + er/f = (cf + der)/(df).
Since c, d, e, f, r are all integers and products and differences of integers are integers, (cf + der) and (df) are both integers. By the zero sum property, (df) ≠ 0.
Let s = (cf + der)/(df).
Therefore, s is a quotient of two integers (cf + der) and (df) with (df) ≠ 0.
By definition of rational, s is rational, which contradicts the supposition that s is irrational.
In conclusion, the supposition is false and a + br is irrational.
Work Step by Step
Claim: If a and b are rational numbers, b ≠ 0, and r is an irrational number, then a + br is irrational.
Negation: If a and b are rational numbers, b ≠ 0, and r is an irrational number, then a + br is rational.
Proof: Suppose that a and b are rational numbers, b ≠ 0, and r is an irrational number such that a + br is rational.
By definition of rational, a = c/d and b = e/f.
By substitution, a + br = c/d + er/f.
By laws of algebra, c/d + er/f = (cf + der)/(df).
Since c, d, e, f, r are all integers and products and differences of integers are integers, (cf + der) and (df) are both integers. By the zero sum property, (df) ≠ 0.
Let s = (cf + der)/(df).
Therefore, s is a quotient of two integers (cf + der) and (df) with (df) ≠ 0.
By definition of rational, s is rational, which contradicts the supposition that s is irrational.
In conclusion, the supposition is false and a + br is irrational.
