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Sums, differences, products and quotients of an irrational and a rational are irrational.

Prove that if with , and , then are all irrational, i.e., are all in .

Proof. Since , we know there are integers and such that . Now we consider each of the elements we wish to show are irrational.

1. Suppose otherwise, that . Then, there exist such that This contradicts our assumption that is irrational.

2. Since is irrational, so is (by part (a), the sum of a rational and an irrational cannot be rational and since is rational, cannot have rational). But then, and by part (a) this sum must be irrational.
3. Suppose otherwise, that . Then there exist such that . Further, since , we know exists. Contradicting our assumption that is irrational.

4. First, we since is irrational, we have , and thus exists. Further, since is rational, and is irrational, by (c) we must have irrational as well. Then by (c), since , we must have irrational.
5. By (d) we know is irrational, and since is rational, by (c), we must have irrational.

One comment

1. Anonymous says:

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