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Chapter 3: Problem 74

Show that \(\log 2\) is irrational.

Short Answer

Expert verified
Assuming \(\log 2\) is rational, we can express it as \(\frac{m}{n}\), where \(m\) and \(n\) are integers. Converting to exponential form, we get \(10^m = 2^n\). Examining the prime factors, the left side has both 2 and 5 as factors, while the right side has only 2. This contradiction shows our original assumption must be false, proving that \(\log 2\) is irrational.

Step by step solution

01

Assume \(\log 2\) is rational

For this exercise, we will use the properties of logarithms and assume that \(\log 2\) can be expressed as a ratio of two integers \(m\) and \(n\), where \(n \neq 0\): \[\log 2 = \frac{m}{n}\]
02

Convert the equation to exponential form

Now, we'll convert the equation from logarithmic form to exponential form by using the base of logarithm, which is 10. If \(\log_a b = c\), then \(a^c = b\). In our case, \(a = 10\), \(b = 2\), and \(c = \frac{m}{n}\). Thus: \[10^{\frac{m}{n}} = 2\]
03

Rearrange the equation to isolate n

Let's rearrange the equation so that we have \(n\) on one side by raising both sides to the power of \(n\): \[\begin{aligned} \left(10^{\frac{m}{n}}\right)^n &= 2^n \\ 10^m &= 2^n \end{aligned}\] Now the exponent on the left side has been simplified to \(m\), and on the right side, it remains \(n\).
04

Examine the prime factors of each side

Let's look at the prime factorization of both sides of the equation. On the left side, the prime factors of the base 10 are 2 and 5. That makes the prime factorization of \(10^m\) consist of a pile of 2s and 5s multiplied together. It will look like this: \((2^m \cdot 5^m)\). Now, let's look at the prime factorization of the right side of the equation. The prime factorization of \(2^n\) is simply a pile of 2s multiplied together. It can be expressed as \((2^n)\).
05

Show the contradictions

We can clearly see that the left side of the equation, \(10^m\), has both 2 and 5 as factors, while the right side, \(2^n\), has only 2 as a factor. This means that the left side must have a factor of 5, but the right side does not have any factor of 5. This is a contradiction, as we reached an impossible situation regarding the factors arising from our assumption that \(\log 2 = \frac{m}{n}\).
06

Conclude the proof

Since our assumption that \(\log 2\) is rational led to a contradiction, we can conclude that \(\log 2\) is an irrational number.

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