Question

Show that \(\sqrt{2}\) is irrational. HINT: Show that if \(\sqrt{2}=m / n,\) where \(m\) and \(n\) are integers, then both \(\mathrm{m}\) and \(n\) must be even. Obtain a contradiction from this.

Short answer

Question: Prove that \(\sqrt{2}\) is irrational. Answer: Using the contradiction method, we assumed that \(\sqrt{2}\) is rational, which means there exist integers \(m\) and \(n\) such that \(\sqrt{2} = \frac{m}{n}\), and they are coprime (have no common factors other than 1). We then proceeded to show that both \(m\) and \(n\) must be even, which contradicts the initial assumption that they have no common factors other than 1. Therefore, \(\sqrt{2}\) must be irrational.

Step-by-step solution

Define the assumption that \(\sqrt{2}\) is rational.

Assume that \(\sqrt{2}\) is rational, which means there exist integers \(m\) and \(n\) such that \(\sqrt{2} = \frac{m}{n}\), where \(m\) and \(n\) have no common factors other than 1 (i.e., they are coprime).

Square both sides of the equation.

To get rid of the square root, square both sides of the equation: \((\sqrt{2})^2 = (\frac{m}{n})^2\) Which simplifies to: \(2 = \frac{m^2}{n^2}\)

Multiply both sides of the equation by \(n^2\) to get rid of the denominator.

Multiply both sides of the equation by \(n^2\) to get: \(2n^2 = m^2\)

Analyze parity of \(m^2\) given that \(2\) is a factor of \(2n^2\).

Since \(2n^2\) is an even number (because \(2\) is a factor of it), then \(m^2\) must also be an even number, as it is equal to \(2n^2\). So, \(m\) must be an even number, as the square of an odd number is odd and the square of an even number is even.

Rewrite \(m\) as \(2k\) for some integer \(k\).

Since \(m\) is an even number, there exists an integer \(k\) such that \(m = 2k\).

Substitute \(2k\) for \(m\) in the equation \(2n^2 = m^2\).

Replace \(m\) with \(2k\) in the equation \(2n^2 = m^2\): \(2n^2 = (2k)^2\)

Simplify the equation.

Simplify the equation: \(2n^2 = 4k^2\)

Divide both sides of the equation by \(2\) to get rid of the factor of \(2\) from the equation.

Divide both sides by \(2\) to get: \(n^2 = 2k^2\)

Analyze parity of \(n^2\) given that \(2\) is a factor of \(2k^2\).

Since \(2k^2\) is an even number (because \(2\) is a factor of it), then \(n^2\) must also be an even number, as it is equal to \(2k^2\). So, \(n\) must be an even number, as the square of an odd number is odd and the square of an even number is even.

State the contradiction and conclude that \(\sqrt{2}\) is irrational.

We have reached a contradiction. We initially assumed that \(m\) and \(n\) have no common factors other than 1, but we have shown that both \(m\) and \(n\) must be even. Since both are even, they share a common factor of \(2\), which contradicts our initial assumption. This means that our assumption that \(\sqrt{2}\) is rational must be incorrect. Therefore, \(\sqrt{2}\) must be irrational.

Key concepts

Rational Numbers

Rational numbers are numbers that can be expressed as a fraction, where both the numerator and the denominator are integers. Importantly, the denominator cannot be zero. For example, numbers like \( \frac{1}{2} \), \( 3 \), and \( -4 \) are rational because they can be written as fractions: \( \frac{1}{2} \), \( \frac{3}{1} \), and \( \frac{-4}{1} \), respectively. In any rational number \( \frac{m}{n} \), the integers \( m \) and \( n \) should be coprime, meaning they have no common factors other than 1. This ensures the fraction is in its simplest form.

Rational numbers contrast with irrational numbers, which cannot be expressed as such a fraction. Numbers like \( \sqrt{2} \) are irrational, as they have non-repeating, non-terminating decimal expansions, making them impossible to precisely write as a simple fraction of two integers. Understanding rational numbers is foundational for studying more complex numerical systems.

Proof by Contradiction

Proof by contradiction is a powerful mathematical technique used to establish the truth of a statement by assuming the opposite is true and then showing that this assumption leads to a contradiction. This approach works because if assuming the opposite leads to an impossible situation, the original statement must be true.

In the case of showing that \( \sqrt{2} \) is irrational, you start by assuming \( \sqrt{2} \) is rational. This means it can be expressed as \( \frac{m}{n} \), where \( m \) and \( n \) are integers with no common factors other than 1.

By manipulating this equation step-by-step, you eventually find that both \( m \) and \( n \) must be even, which contradicts the initial assumption that they have no common factors other than 1. Therefore, the assumption that \( \sqrt{2} \) is a rational number must be false, leading to the conclusion that \( \sqrt{2} \) is indeed irrational. This method is elegant because it reveals the truth by simply disproving the opposite.

Even and Odd Numbers

Even and odd numbers are fundamental concepts in mathematics, distinguishing numbers based on their divisibility by 2. An even number is any integer that can be divided by 2 without leaving a remainder, such as 2, 4, 6, and so on. Conversely, an odd number leaves a remainder of 1 when divided by 2, examples include 1, 3, and 5.

In our exploration of whether \( \sqrt{2} \) is irrational, even and odd numbers play a crucial role. When analyzing the equation \( 2n^2 = m^2 \), the conclusion follows that if \( m^2 \) is even, then \( m \) must also be even. This is because the square of any odd number remains odd, thus confirming that \( m \) cannot be odd.

Once \( m \) is determined to be even, we can express it as \( 2k \), leading further to the conclusion that \( n \) must also be even. This chain of reasoning shows how seemingly simple concepts like even and odd numbers can have significant implications in proving more complex mathematical ideas.