SOLUTION TO PROBLEM 14

Since $\sqrt{2}$ is our first and easiest example of irrational number, what about

$$N=\sqrt{2}^{\sqrt{2}}?$$

If N were rational, then the answer to the question above would be YES. But what if it's irrational? Well, in that case consider

$$M=N^{\sqrt{2}}=2$$

which is obviously rational, so again the answer is YES. Since Nis a real number it must be either rational or irrational hence the answer to this problem is YES.
Note: Observe that we solved the problem using the number $\sqrt{2}^{\sqrt{2}}$ but without really deciding if it is rational or not.

Questions and/or comments should be directed to Judy Downey or Griff Elder

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Last modified:   Sat Dec 8 12:56:27 CST 2001