myntfalskj4

Answered

2022-07-01

MathJax(?): Can't find handler for document
MathJax(?): Can't find handler for document
I suggested the following problem to my friend: prove that there exist irrational numbers a and b such that ab is rational.

Now, his inital solution was like this: let's take a rational number r and an irrational number i. Let's assume $a={r}^{i}$ b=\frac{1}{i}$$

So we have

${a}^{b}=({r}^{i}{)}^{\frac{1}{i}}=r$

which is rational per initial supposition. b is obviously irrational if i is. My friend says that it is also obvious that if r is rational and i is irrational, then ${r}^{i}$ is irrational. I quickly objected saying that $r=1$ is an easy counterexample. To which my friend said, OK, for any positive rational number r, other than 1 and for any irrational number i ${r}^{i}$ is irrational. Is this true? If so, is it easily proved? If not, can someone come up with a counterexample?

Now, his inital solution was like this: let's take a rational number r and an irrational number i. Let's assume $a={r}^{i}$ b=\frac{1}{i}$$

So we have

${a}^{b}=({r}^{i}{)}^{\frac{1}{i}}=r$

which is rational per initial supposition. b is obviously irrational if i is. My friend says that it is also obvious that if r is rational and i is irrational, then ${r}^{i}$ is irrational. I quickly objected saying that $r=1$ is an easy counterexample. To which my friend said, OK, for any positive rational number r, other than 1 and for any irrational number i ${r}^{i}$ is irrational. Is this true? If so, is it easily proved? If not, can someone come up with a counterexample?

Answer & Explanation

soosenhc

Expert

2022-07-02Added 16 answers

Consider ${2}^{{\mathrm{log}}_{2}3}$

babyagelesszj

Expert

2022-07-03Added 7 answers

Let $r$ be a positive rational number and i a positive irrational number. If ${r}^{i}$ is rational, then ${r}^{i}=\frac{a}{b}$ for a,$a,b\in \mathbb{Z}$ such that$b\ne 0$. In particular, $i={\mathrm{log}}_{r}\left(\frac{a}{b}\right)$.

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