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Chapter 4: Problem 5

Zeigen Sie, dass \(\log _{2} 3\) irrational ist.

Short Answer

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Question: Prove that \(\log_{2}3\) is irrational. Answer: \(\log_{2}3\) is irrational because assuming it to be rational leads to a contradiction in the equation \(2^a = 3^b\), where the left side is an even integer and the right side is an odd integer, which cannot be equal.

Step by step solution

01

Proof by Contradiction

Assume that \(\log_{2}3\) is a rational number. That is, it can be expressed as a fraction \(\frac{a}{b}\) where \(a\) and \(b\) are integers and \(b \neq 0\). Then, we have $$\log_{2}3 = \frac{a}{b}$$
02

Convert logarithm to exponent

Now, let's convert this logarithm equation to an exponential equation with base 2 and the given fraction. This gives: $$2^{\frac{a}{b}} = 3$$
03

Manipulate the equation

To isolate the equality, raise both sides to the power of \(b\). This gives: $$(2^{\frac{a}{b}})^b = 3^b$$ Which simplifies to: $$2^a = 3^b$$
04

Analyze the parity

Look at the equation \(2^a = 3^b\). The left side of the equation is a power of 2, which means that it is an even integer. In contrast, the right side of the equation is a power of 3, making it an odd integer.
05

Find the contradiction

Since an even integer cannot be equal to an odd integer, the equation \(2^a = 3^b\) demonstrates a contradiction to our original assumption that \(\log_{2}3\) is a rational number. Therefore, the assumption that \(\log_{2}3\) is rational is false, which means that \(\log_{2}3\) must be irrational.

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