Hint 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新 is irrational).

The negation of the given statement is:  $ r Q with r non-zero such that (s is irrational) and (r新 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新 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新 is rational, there exist integers c and d such that  r新 = c / d where d0.

Now try to write s in terms of a, b, c and d. You would like to contradict the fact that s is irrational.

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