Problem 22 - Discrete Mathematics with Applications

Question

Prove by contradiction, where \(p\) is a prime number. \(\log _{10} 2\) is an irrational number.

Step-by-Step Solution

Answer

Assume that \(\log_{10}2\) is a rational number. Then, there exist coprime integers \(a\) and \(b\) such that \(\log_{10}2 = \frac{a}{b}\). Rewrite the expression as \(10^{\frac{a}{b}} = 2\) and raise both sides to the power of \(b\), resulting in \(10^a = 2^b\). This leads to a contradiction, as the left side is a multiple of 10 (and thus also a multiple of 5), while the right side is a multiple of 2 with no factor of 5. Therefore, the assumption is false, and \(\log_{10}2\) must be an irrational number.

Step 1: Assumption

Assume that \(\log_{10}2\) is a rational number. So, there exist coprime integers \(a\) and \(b\) such that: \(\log_{10}2 = \frac{a}{b}\)

Step 2: Rewrite the expression

Using the definition of logarithms, rewrite the expression as: \(10^{\frac{a}{b}} = 2\)

Step 3: Take both sides to the power of \(b\)

Raise both sides of the equation to the power of \(b\): \((10^{\frac{a}{b}})^b = 2^b\) This simplifies to: \(10^a = 2^b\)

Step 4: Identify contradiction

In the equation \(10^a = 2^b\), we see that the left-hand side of the equation is a multiple of 10 (since \(10^a\)), while the right-hand side is a multiple of 2 (since \(2^b\)). Notice that any multiple of 10 is also a multiple of 2 (even number) and 5, so these factors must be in the prime factorization of \(10^a\). But the prime factorization of \(2^b\) only contains the prime number \(2\), which means that there is no multiple of 5 on the right-hand side. This is a contradiction because we have proved that both sides of the equation cannot be equal, as one side contains a factor of 5 while the other side does not.

Step 5: Conclusion

Since we reached a contradiction under the assumption that \(\log_{10}2\) is a rational number, we can conclude that the assumption is false. Therefore, \(\log_{10}2\) must be an irrational number.

Key Concepts

Proof by Contradiction

Proof by contradiction is a powerful method of mathematical proof, particularly useful for establishing the truth of a statement by showing that its opposite leads to a logical inconsistency. The process starts by assuming the opposite of what you want to prove. If you can deduce a contradiction from this assumption, you can conclude that it must be false, which means the original statement is true.

In the context of showing that \(\log_{10}2\) is an irrational number, we start by assuming the opposite: that \(\log_{10}2\) is a rational number. This assumption implies that it can be expressed as a ratio of two integers. Our goal will be to follow this assumption to its logical end and find the contradiction which, in turn, proves the initial statement.

Prime Factorization

Prime factorization is the process of breaking down a composite number into the prime numbers that, when multiplied together, give the original number. Mathematically, every integer greater than 1 either is a prime number itself or can be uniquely represented as a product of prime numbers, which is known as its prime factorization. For example, the prime factorization of 10 is \(2 \times 5\).

In proving the irrationality of \(\log_{10}2\), the concept of prime factorization is key. Once \(10^a = 2^b\) has been established through the assumed rational expression of \(\log_{10}2\), we examine the prime factors on both sides. Since 10 is composed of the prime factors 2 and 5, \(10^a\) will necessarily include the factor 5, which \(2^b\) lacks. This discrepancy highlights the inherent contradiction and is the crux in demonstrating the irrationality of \(\log_{10}2\).

Logarithms

Logarithms are an essential part of mathematics, especially when dealing with multiplication and division of very large or very small numbers. A logarithm answers the question: 'To what power must we raise one number to obtain another number?' For example, the base 10 logarithm of 1000 answers the question: 'To what power must we raise 10 to get 1000?' The answer is 3, because \(10^3 = 1000\).

In our exercise, we're dealing with \(\log_{10}2\), the power to which we must raise 10 to get 2. Our derivation starts from assuming this logarithm is rational, and that opens the door for manipulation using properties of logarithms and exponentiation, eventually leading to the realization that the assumption does not hold. By navigating through these mathematical terrains, students hone their understanding of how logarithms interact with other mathematical principles, such as exponents and prime factorization.