(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 [tex]a^b[/tex] be rational/irrational? In case (iii) consider also [tex]b^a[/tex].

This makes 20 exercises total.

About Vadim Kulikov

For details see this
This entry was posted in Algebra, Calculus, Education, Mathematics, Wednesday Problem. Bookmark the permalink.

One 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= $latex \frac{1}{2}.$

    =============================================================
    (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.

Leave a Reply

Your email address will not be published. Required fields are marked *