Problem 88 - Calculus with Concepts in Calculus

Question

Prove that \(\sqrt{2}\) is irrational. (Hint: Assume that \(\sqrt{2}=\) \(p / q\), where \(p\) and \(q\) are integers such that at most one of them is divisible by \(2 .\) It can be shown that a square integer is divisible by 2 only if it is also divisible by 4 . Use this fact to show first that \(p\) is divisible by 2 and then that \(q\) is also divisible by \(2 .\) This contradicts the assumption.)

Step-by-Step Solution

Answer

\(\sqrt{2}\) is irrational because assuming it is rational leads to a contradiction (both \(p\) and \(q\) being divisible by 2).

Step 1: Assume Rationality

Assume that \(\sqrt{2}\) is rational, meaning it can be written as the fraction \(\frac{p}{q}\) where \(p\) and \(q\) are integers with no common factors other than 1 and not both divisible by 2.

Step 2: Square the Equation

Square both sides of the equation \(\sqrt{2} = \frac{p}{q}\) to obtain \(2 = \frac{p^2}{q^2}\), which can be rearranged to \(2q^2 = p^2\).

Step 3: Analyze Divisibility of \(p\)

From \(2q^2 = p^2\), we see that \(p^2\) is even (since it's divisible by 2). Hence, \(p\) must also be even (because an odd number squared is odd). Let \(p = 2k\) for some integer \(k\).

Step 4: Substitute and Simplify

By substituting \(p = 2k\) into the equation \(2q^2 = p^2\), we get \(2q^2 = (2k)^2 = 4k^2\), leading to \(q^2 = 2k^2\). Thus, \(q^2\) is also even, indicating \(q\) is even.

Step 5: Reach a Contradiction

Both \(p\) and \(q\) are even, meaning they have a common factor of 2, which contradicts our initial assumption that \(p\) and \(q\) have no common factors other than 1. Hence, \(\sqrt{2}\) is irrational.

Key Concepts

Proof by Contradiction

When tackling the question of the irrationality of \( \sqrt{2} \), we often employ a proof technique called "proof by contradiction." This approach starts by assuming the opposite of what we intend to prove and then showing that this assumption leads to a logical inconsistency.
Here's how it works in our case:

• We assume that \( \sqrt{2} \) is rational, which means it can be expressed as a fraction \( \frac{p}{q} \), where \( p \) and \( q \) are integers.
• These integers should have no common factors other than 1, meaning they are coprime, and at most, one of them should be divisible by 2.

By proceeding with our assumption, we eventually uncover that both \( p \) and \( q \) must be divisible by 2. This shattered our initial assumption of coprime integers and leads to a contradiction. Therefore, the initial assumption must be false, proving that \( \sqrt{2} \) cannot be rational and is indeed irrational. This technique is powerful because it uses logical consistency to rigorously establish truths.

Rational Numbers

Rational numbers are numbers that can be expressed as the quotient of two integers. In simpler terms, these are numbers we can write as fractions such as \( \frac{a}{b} \), where \( a \) and \( b \) are integers, and \( b eq 0 \).
Some key points here are:

• Integers themselves are rational numbers because any integer \( n \) can be written as \( \frac{n}{1} \).
• Rational numbers include both positive and negative values, as well as zero.
• Rational numbers are dense, meaning between any two rational numbers, there are infinitely many other rational numbers.

In proving \( \sqrt{2} \) as irrational, we began by assuming it could be expressed as a rational number. However, our logical journey brought us to an inconsistency, leading us to the fact that not all numbers fit neatly into the rational category, illustrating the existence of irrational numbers.

Divisibility

The concept of divisibility is key when dealing with numbers, particularly in this proof of irrationality.
Divisibility means that one integer can be divided by another without leaving a remainder. In our exercise, we explored:

• An integer is divisible by 2 if it is even. Such numbers can be expressed in the form \( 2k \), where \( k \) is an integer.
• If \( p^2 \) is divisible by 2 (as found by \( 2q^2 = p^2 \)), then \( p \) itself must be divisible by 2 because the square of an odd number cannot be even.
• We substitute \( p = 2k \) which eventually led to \( q^2 = 2k^2 \), making \( q \) also divisible by 2.

This insight into divisibility establishes the very contradiction we found—both \( p \) and \( q \) being divisible by 2 strips them of their status as coprime, confirming the irrational nature of \( \sqrt{2} \). Understanding divisibility is crucial for exploring integers' properties and how they relate to other number types.