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Proof of irrationality of sqrt(3)

Assume sqrt(3) = p/q in lowest terms. After squaring, 3q^2 = p^2, so p is divisible by 3. That then forces q to be divisible by 3, which contradicts lowest terms.

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Student-friendly explanation

The proof uses contradiction just like sqrt(2), but the prime number is 3 here. If both numerator and denominator become divisible by 3, the original fraction cannot have been in lowest terms, so the assumption fails.

How to write this in exams

  1. 1

    Start with the exact idea

    Assume sqrt(3) = p/q in lowest terms. After squaring, 3q^2 = p^2, so p is divisible by 3. That then forces q to be divisible by 3, which contradicts lowest terms.

  2. 2

    Then show how to use it

    1. Assume sqrt(3) = p/q in lowest terms. 2. Square both sides. 3. Show p is divisible by 3. 4. Substitute p = 3k. 5. Show q is divisible by 3. 6. Reach the contradiction.

  3. 3

    Add one concrete example

    Write p = 3k after showing p is divisible by 3, then continue until q is also shown to be divisible by 3.

  4. 4

    Avoid this incomplete answer

    Claiming that only p being divisible by 3 is enough is wrong; the proof needs the contradiction that both p and q share the factor 3.

Definition

Assume sqrt(3) = p/q in lowest terms. After squaring, 3q^2 = p^2, so p is divisible by 3. That then forces q to be divisible by 3, which contradicts lowest terms.

Example

Write p = 3k after showing p is divisible by 3, then continue until q is also shown to be divisible by 3.

Rule to remember

Assume sqrt(3) = p/q in lowest terms, then 3q^2 = p^2.

Memory hook

For sqrt(3), the prime 3 repeats its way into both sides.

Examples and method

Worked example

Let sqrt(3) = p/q in lowest terms. Then p^2 = 3q^2, so p is divisible by 3. Write p = 3k. Substituting gives 9k^2 = 3q^2, so q^2 = 3k^2 and q is also divisible by 3. That contradicts lowest terms.

Method to apply

1. Assume sqrt(3) = p/q in lowest terms. 2. Square both sides. 3. Show p is divisible by 3. 4. Substitute p = 3k. 5. Show q is divisible by 3. 6. Reach the contradiction.

Diagram support

Use the same proof ladder as sqrt(2), but replace 2 with 3 in every divisibility step.

How CBSE asks it

Prove sqrt(3) is irrational using contradiction, or identify the divisibility step in the proof.

Avoid common mistakes

Common confusion

Students repeat the sqrt(2) proof without changing the prime from 2 to 3.

Common wrong answer

Claiming that only p being divisible by 3 is enough is wrong; the proof needs the contradiction that both p and q share the factor 3.

Exam tip

Mention the prime 3 clearly at the divisibility step so the logic stays correct.

Quick check

What contradiction appears when proving sqrt(3) irrational by the fraction method?

Both p and q turn out divisible by 3, so the fraction cannot have been in lowest terms.

Answer writing and exam use

1-mark answer

Assume sqrt(3) = p/q in lowest terms. After squaring, 3q^2 = p^2, so p is divisible by 3. That then forces q to be divisible by 3, which contradicts lowest terms.

2-mark answer

Assume sqrt(3) = p/q in lowest terms. After squaring, 3q^2 = p^2, so p is divisible by 3. That then forces q to be divisible by 3, which contradicts lowest terms. Assume sqrt(3) = p/q in lowest terms, then 3q^2 = p^2. Write p = 3k after showing p is divisible by 3, then continue until q is also shown to be divisible by 3.

3-mark answer

The proof uses contradiction just like sqrt(2), but the prime number is 3 here. If both numerator and denominator become divisible by 3, the original fraction cannot have been in lowest terms, so the assumption fails. Assume sqrt(3) = p/q in lowest terms, then 3q^2 = p^2. Let sqrt(3) = p/q in lowest terms. Then p^2 = 3q^2, so p is divisible by 3. Write p = 3k. Substituting gives 9k^2 = 3q^2, so q^2 = 3k^2 and q is also divisible by 3. That contradicts lowest terms. Prove sqrt(3) is irrational using contradiction, or identify the divisibility step in the proof. Claiming that only p being divisible by 3 is enough is wrong; the proof needs the contradiction that both p and q share the factor 3.
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