How to prove that $\sqrt{2}$ is irrational by contradiction?

Question

I'm preparing for IIT-JEE and I need to understand the classic proof that $\sqrt{2}$ is irrational. I've seen the proof but I have some doubts about the logic.

The proof assumes $\sqrt{2} = p/q$ in lowest terms, then squares both sides to get $2q^2 = p^2$ . Then it argues that $p$ must be even, so $p = 2k$ , and substituting gives $q^2 = 2k^2$ , so $q$ is also even, contradicting the assumption that $p/q$ is in lowest terms.

My question: Why must $p$ be even if $p^2$ is even? And why does this proof not work for $\sqrt{4}$ (which is rational)?

Irbah Pro · Accepted Answer

Let me give you a thorough breakdown of this problem, because there are actually several layers to understand.

First principles:

The definition you're working with is correct, but there's a subtlety most textbooks gloss over. The limit has to exist AND be the same regardless of the path you take to approach the point. In one dimension, there are only two paths (from left and right), so checking those is enough. In higher dimensions, there are infinitely many paths, which is why multivariable limits are harder.

For your specific function:

Try approaching along the line y = mx. You'll find that the limit depends on m — different slopes give different values. This immediately tells you the limit doesn't exist. This is the standard technique for proving a multivariable limit doesn't exist.

A deeper insight:

The reason this happens is that the function is not continuous at the origin. The limit depends on the angle of approach, not just the distance. Functions that are well-behaved at a point have limits that are independent of the approach path. This is equivalent to the function being continuous at that point.

Practical advice:

When you're faced with a multivariable limit, first try approaching along a few simple paths (x-axis, y-axis, y = x). If you get different values, you're done — the limit doesn't exist. If they all agree, you still need to prove the limit exists using the epsilon-delta definition or by bounding the function.

There's no shortcut for the existence proof, but the squeeze theorem (sandwich theorem) often works well. Find upper and lower bounds that both approach the same value, and your function is squeezed between them.

Ethan Caldwell · Answer

These are excellent questions that get to the heart of the proof. Why p 2 even ⇒ p even? This relies on the fundamental property of integers: if an integer is odd, its square is odd. Let p = 2 k + 1, then p 2 = 4 k 2 + 4 k + 1 = 2 ( 2 k 2 + 2 k ) + 1, which is odd. By contrapositive: if p 2 is even, then p cannot be odd, so p is even. Why doesn't this work for 4? If we try the same proof for 4 = p / q in lowest terms: 4 q 2 = p 2 Here p 2 is divisible by 4, which implies p is even, say p = 2 k. Then: 4 q 2 = 4 k 2 ⇒ q 2 = k 2 So q = k, meaning p / q = 2 k / k = 2, which is already in lowest terms when k = 1. The proof does not force a contradiction because the factor of 4 cancels completely. The critical difference is that 2 is not a perfect square, so the prime factorisation argument forces p and q to share a factor of 2. For 4 = 2, the equation 4 q 2 = p 2 is satisfied by p = 2 q without contradiction.

How to prove that √2 is irrational by contradiction?

2 Answers

How to prove that √2 is irrational by contradiction?

2 Answers