Exercise 1.2.2.

Show that there is no rational number \(r\) satisfying \(2^r = 3\).

Solution:

Suppose that, for contradiction, there exists a rational number such that \(2^r = 3\). Let \(r = p/q\) where \(\gcd(p,q) = 1\) for \(p,q \in \mathbb{Z}\). We will consider three cases:

1. \(p/q > 0\)
2. \(p/q < 0\)
3. \(p/q = 0\)

Case 1:

Suppose that \(p/q > 0 \). Then \[ 2^{p/q}=3 \implies 2^p=3^q \] Since \(2^p\) is even and \(3^q\) is odd for all \(p,q \in \mathbb{N}\), this is a contradiction and no such \(r > 0\) exists.

Case 2:

Suppose that \(p/q < 0\). Without loss of generality, assume that \(p<0\) and \(q>0\). Then \[ 2^{-{p/q}}=\frac{1}{2^{p/q}}=3 \implies \frac{1}{2^p}=3^q\] Then \(\frac{1}{2^p} < 1\) while \(3^q> 1\). This is a contradiction and no such \(r < 0 \) exists.

Case 3:

Suppose that \(p/q=0\). Then \[2^0=3 \implies 1=3\] This is a contradiction and no such \(r=0\) exists.

Hence, there is no rational number \(r\) satisfying \(2^r = 3\).

Q.E.D.