Solution for Section 3.6 Question 3 (by contradiction)

3. We are asked to prove, using the method of contradiction, the statement: "rQ with r non-zero, if (s is irrational), then (r·s is irrational).

The negation of the given statement is:  $ r Q with r non-zero such that (s is irrational) and (r·s is rational).

We now assume that the negation is true and we need to show that this leads to a contradiction.

Proof Suppose that r and r·s are rational numbers with r non-zero and that s is irrational.

Since r is a non-zero rational number, there exist integers a and b such that r = a / b where b0, a0.
Since r·s is a rational number, there exist integers c and d such that  r·s = c / d where d0.

Since a, b, c and d are all integers, and a, b and d are non-zero,   c·b  and a·d   are also integers, and a·d   is non-zero. Hence s can be written as a quotient (fraction) of integers. Therefore s is rational.

We have now discovered that if the negation of this statement were true, then s would be both rational and irrational. This is clearly impossible and hence the original statement must be true. Therefore "rQ with r non-zero, if (s is irrational), then (r·s is irrational).

Back to Section 3.6