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Proof of irrationality of sqrt(2)

Assume sqrt(2) = p/q in lowest terms. After squaring, 2q^2 = p^2, so p is even. That leads to q also being even, which contradicts the fraction being in lowest terms.

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Student-friendly explanation

This is a proof by contradiction. If both p and q become divisible by 2, then the fraction was not in lowest terms after all. That contradiction shows sqrt(2) cannot be rational.

How to write this in exams

  1. 1

    Start with the exact idea

    Assume sqrt(2) = p/q in lowest terms. After squaring, 2q^2 = p^2, so p is even. That leads to q also being even, which contradicts the fraction being in lowest terms.

  2. 2

    Then show how to use it

    1. Assume sqrt(2) = p/q in lowest terms. 2. Square both sides. 3. Show p is even. 4. Substitute p = 2k. 5. Show q is even. 6. Contradict lowest terms.

  3. 3

    Add one concrete example

    If p is even, write p = 2k. Then 4k^2 = 2q^2, so q^2 = 2k^2 and q is also even.

  4. 4

    Avoid this incomplete answer

    Saying p is even, so sqrt(2) is rational, is wrong because the same argument forces q to be even too.

Definition

Assume sqrt(2) = p/q in lowest terms. After squaring, 2q^2 = p^2, so p is even. That leads to q also being even, which contradicts the fraction being in lowest terms.

Example

If p is even, write p = 2k. Then 4k^2 = 2q^2, so q^2 = 2k^2 and q is also even.

Rule to remember

Assume sqrt(2) = p/q in lowest terms, then 2q^2 = p^2.

Memory hook

Even numerator leads to even denominator; lowest terms cannot survive.

Examples and method

Worked example

Let sqrt(2) = p/q in lowest terms. Then p^2 = 2q^2, so p is even. Write p = 2k. Substituting gives 4k^2 = 2q^2, so q^2 = 2k^2 and q is even. Both being even is impossible in lowest terms.

Method to apply

1. Assume sqrt(2) = p/q in lowest terms. 2. Square both sides. 3. Show p is even. 4. Substitute p = 2k. 5. Show q is even. 6. Contradict lowest terms.

Diagram support

A proof ladder with the steps assumption, square, evenness, contradiction, and conclusion works well.

How CBSE asks it

Prove sqrt(2) is irrational, or identify the contradiction in the proof.

Avoid common mistakes

Common confusion

Students stop after proving p is even and forget to show that q is even too.

Common wrong answer

Saying p is even, so sqrt(2) is rational, is wrong because the same argument forces q to be even too.

Exam tip

In a proof by contradiction, always end with the words 'this contradicts lowest terms'.

Quick check

Why does the proof of sqrt(2) fail once both p and q are even?

Because a fraction with both numerator and denominator even is not in lowest terms, so the starting assumption must be wrong.

Answer writing and exam use

1-mark answer

Assume sqrt(2) = p/q in lowest terms. After squaring, 2q^2 = p^2, so p is even. That leads to q also being even, which contradicts the fraction being in lowest terms.

2-mark answer

Assume sqrt(2) = p/q in lowest terms. After squaring, 2q^2 = p^2, so p is even. That leads to q also being even, which contradicts the fraction being in lowest terms. Assume sqrt(2) = p/q in lowest terms, then 2q^2 = p^2. If p is even, write p = 2k. Then 4k^2 = 2q^2, so q^2 = 2k^2 and q is also even.

3-mark answer

This is a proof by contradiction. If both p and q become divisible by 2, then the fraction was not in lowest terms after all. That contradiction shows sqrt(2) cannot be rational. Assume sqrt(2) = p/q in lowest terms, then 2q^2 = p^2. Let sqrt(2) = p/q in lowest terms. Then p^2 = 2q^2, so p is even. Write p = 2k. Substituting gives 4k^2 = 2q^2, so q^2 = 2k^2 and q is even. Both being even is impossible in lowest terms. Prove sqrt(2) is irrational, or identify the contradiction in the proof. Saying p is even, so sqrt(2) is rational, is wrong because the same argument forces q to be even too.
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