# (Ir)rational Behavior of Calculation

We had the following set of exercises with a group of secondary school students (in Ressun lukio).

Assume that a and b are positive real numbers. Then assume one of the following:

(i) a and b are both rational
(ii) a and b are both irrational
(iii) a is rational and b is irrational.

Then

(a) can a + b be rational/irrational?
(b) can ab be rational/irrational?
(c) can $$a^b$$ be rational/irrational? In case (iii) consider also $$b^a$$.

This makes 20 exercises total.

For details see this
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### 1 Response to (Ir)rational Behavior of Calculation

1. Christian says:

(i) a and b are both rational:

then a+b=rational, ab=rational, $latex a^{b}$= irrational. For the last one, such an example are the numbers a=5, b= $latex \frac{1}{2}.$

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(ii) a and b are both irrational

then a+b= both rational and irrational, i.e. a=$latex 1+\sqrt{2}$ and $latex 1-\sqrt{2}$

ab= both rational and irrational, i.e. a=$latex \sqrt{2}$ and $latex \sqrt{2}$ are irrationals, however the product is rational.

$latex a^{b}$=both rational and irrational again. We consider $latex a =\sqrt{2}^{\sqrt{2}}$ and $latex b=\sqrt{2}$\$. Then the product is rational.