Question

Prove the following statements using any method from Chapters 4,5 or 6 . The number \(\log _{2} 3\) is irrational.

Short answer

Given the contradiction we've arrived at, we must conclude that \(log_{2} 3\) is an irrational number.

Step-by-step solution

Assumption of Rationality

Begin by assuming that \(log_{2} 3\) is rational. Therefore, it can be written as a fraction of two integers, say \(p/q\). So, we can write the equation as follows: \(log_{2} 3 = p/q\).

Reframe the Equation

Next, we convert the logarithmic equation into an exponential one, which leads to the equation \(2^{p/q} = 3\). This would mean that \(2^p = 3^q\). Here we see that the left side of the equation is an even number (since any power of 2 results in an even number), while the right side is an odd number (since any power of 3 results in an odd number).

Contradiction

We've reached a contradiction; an even integer cannot be equal to an odd integer. Thus, our initial assumption that \(log_{2} 3\) could be expressed as a ratio of two integers \(p/q\) was incorrect.