Problem 75 - Calculus: One and Several Variables

Question

Prove that \(\sqrt{2}\) is irrational. HINT: Assume that \(\sqrt{2}=p / q\) with the fraction written in lowest terms. Square both sides of this equation and argue that both \(p\) and \(q\) must be divisible by 2.

Step-by-Step Solution

Answer

The proof involves a contradiction. By assuming that \( \sqrt{2} \) is rational, squaring this equation leads us to conclude that both \( p \) and \( q \) must be divisible by 2, contradicting our initial assumption. Therefore, \( \sqrt{2} \) is proved to be irrational.

Step 1: Assumption

Assume the converse of what's to be proven. Thus, assuming \( \sqrt{2} \) is a rational number, then it can be expressed as a fraction in the lowest terms, \( \sqrt{2} = \frac{p}{q} \) where \( p \) and \( q \) are integers and have no common factors other than 1.

Step 2: Squaring both sides

Square both sides of the equation, which results in \( 2 = \frac{p^2}{q^2} \), or equivalently \( p^2 = 2q^2 \). This implies that \( p^2 \) is an even number (as it's 2 times something), and therefore \( p \) must also be even (since only even numbers have even squares). If \( p \) is even, then \( p = 2k \) for some integer \( k \).

Step 3: Substitution and simplification

Substitute \( p = 2k \) into \( p^2 = 2q^2 \) gives us \( (2k)^2 = 2q^2 \) or \( 4k^2 = 2q^2 \), which simplifies to \( q^2 = 2k^2 \). Similar to \( p \), this shows that \( q^2 \), and therefore \( q \), is also even.

Step 4: Showing the contradiction

From Steps 2 and 3, it is concluded that both \( p \) and \( q \) are even, meaning they share a common factor of 2. This contradicts our initial assumption that \( p \) and \( q \) have no common factors other than 1. Therefore, the assumption that \( \sqrt{2} \) is rational must be incorrect, proving \( \sqrt{2} \) is irrational.

Key Concepts

Rational and Irrational Numbers

Understanding the difference between rational and irrational numbers lays the foundation for comprehending why numbers like \(\sqrt{2}\) are irrational. In mathematics, rational numbers are those that can be expressed as the quotient or fraction of two integers. The denominator in this fraction is non-zero. These numbers can also be represented as either repeating or terminating decimals. Examples include \(\frac{1}{2}\), \(0.75\), or \(\frac{5}{4}\).

In contrast, irrational numbers cannot be accurately represented as fractions of integers. They have non-repeating, non-terminating decimal expansions. This means that no matter how much you try to write them out as a decimal, you'll never find a repeating pattern or an end. \(\pi\) and \(\sqrt{2}\) are typical examples of irrational numbers. The concept of irrationality was quite unsettling when it was discovered by the ancient Greeks, as it challenged the notion that all quantities could be expressed as ratios of whole numbers.

So, when working with exercises like proving that \(\sqrt{2}\) is irrational, it's essential to recall that if \(\sqrt{2}\) were rational, it would be possible to express it as a fraction in simplest form. But, as the exercise solution demonstrates, this leads to a logical contradiction, reinforcing its place among the irrational numbers.

Proof by Contradiction

The method of proof by contradiction is a powerful and often-used tool in mathematics. This type of argument establishes the truth of a statement by showing that the assumption of its falsehood inevitably leads to a contradiction. The key steps in a proof by contradiction involve:

• Assuming the opposite of what you're trying to prove.
• Making logical deductions from this assumption.
• Finding a contradiction — a situation where two outcomes cannot possibly coexist.
• Concluding that since the assumption leads to an absurdity, it must be false, and therefore, the original statement is proven true.

These steps are elegantly used in establishing the irrationality of \(\sqrt{2}\) by first assuming it is rational and then logically deducing that this assumption leads to an impossibility. Proof by contradiction does not provide a 'direct' proof but instead confirms the validity of a claim by showing the impossibility or the absurdity of the contrary. It's a testament to the axiom 'by excluding the impossible, whatever remains, however improbable, must be the truth.'

Square Roots

The concept of a square root is central to many mathematical disciplines, including algebra, geometry, and calculus. A square root of a number 'x' is a value that, when multiplied by itself, gives 'x'. Symbolically, if \(y^2 = x\), then \(y\) is the square root of \(x\). For positive real numbers, there are always two square roots: one positive and one negative. For instance, the positive square root of 4 is 2, because \(2^2 = 4\), but -2 is also a square root because \( (-2)^2 = 4\) as well.

Properties of Square Roots

Understanding the properties of square roots is crucial when addressing problems involving irrationality. One key property is that the square root of a positive integer is either an integer or an irrational number. It is never a non-integer rational number. This is particularly relevant to the example at hand, \(\sqrt{2}\), which can't be an integer because no integer squared gives 2. This property leads to the exploration of whether \(\sqrt{2}\) could be a non-integer rational number. The steps to prove \(\sqrt{2}\) is irrational involve dealing with the peculiarities of square roots and their relation to both rational and irrational numbers, showcasing how square roots often serve as the first meaningful exposure to the world of irrational numbers for many students.