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Proof of irrationality of sqrt(5)

Assume sqrt(5) = p/q in lowest terms. After squaring, 5q^2 = p^2, so p is divisible by 5. That then forces q to be divisible by 5, which contradicts lowest terms.

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Student-friendly explanation

This proof follows the same contradiction pattern as the proofs for sqrt(2) and sqrt(3). The only change is the prime number 5. Once both p and q are shown to share factor 5, the assumption of lowest terms collapses.

How to write this in exams

  1. 1

    Start with the exact idea

    Assume sqrt(5) = p/q in lowest terms. After squaring, 5q^2 = p^2, so p is divisible by 5. That then forces q to be divisible by 5, which contradicts lowest terms.

  2. 2

    Then show how to use it

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

  3. 3

    Add one concrete example

    If p is divisible by 5, write p = 5k and continue until q is also shown to be divisible by 5.

  4. 4

    Avoid this incomplete answer

    Stopping after finding p divisible by 5 is incomplete because the contradiction comes from both numbers sharing the factor 5.

Definition

Assume sqrt(5) = p/q in lowest terms. After squaring, 5q^2 = p^2, so p is divisible by 5. That then forces q to be divisible by 5, which contradicts lowest terms.

Example

If p is divisible by 5, write p = 5k and continue until q is also shown to be divisible by 5.

Rule to remember

Assume sqrt(5) = p/q in lowest terms, then 5q^2 = p^2.

Memory hook

Prime 5 locks both ends of the fraction.

Examples and method

Worked example

Let sqrt(5) = p/q. Then p^2 = 5q^2, so p is a multiple of 5. Write p = 5k. Substituting gives 25k^2 = 5q^2, so q^2 = 5k^2. Therefore q is also a multiple of 5.

Method to apply

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

Diagram support

A short contradiction flowchart is enough: assumption, square, divisibility by 5, contradiction.

How CBSE asks it

Prove sqrt(5) is irrational, or identify the divisibility contradiction in the proof.

Avoid common mistakes

Common confusion

Students copy the earlier proofs but forget to replace the prime with 5 in every divisibility step.

Common wrong answer

Stopping after finding p divisible by 5 is incomplete because the contradiction comes from both numbers sharing the factor 5.

Exam tip

Keep the same proof steps and only change the prime factor to 5.

Quick check

What does the proof show when sqrt(5) = p/q is assumed in lowest terms?

It shows both p and q are divisible by 5, so the assumption cannot be correct.

Answer writing and exam use

1-mark answer

Assume sqrt(5) = p/q in lowest terms. After squaring, 5q^2 = p^2, so p is divisible by 5. That then forces q to be divisible by 5, which contradicts lowest terms.

2-mark answer

Assume sqrt(5) = p/q in lowest terms. After squaring, 5q^2 = p^2, so p is divisible by 5. That then forces q to be divisible by 5, which contradicts lowest terms. Assume sqrt(5) = p/q in lowest terms, then 5q^2 = p^2. If p is divisible by 5, write p = 5k and continue until q is also shown to be divisible by 5.

3-mark answer

This proof follows the same contradiction pattern as the proofs for sqrt(2) and sqrt(3). The only change is the prime number 5. Once both p and q are shown to share factor 5, the assumption of lowest terms collapses. Assume sqrt(5) = p/q in lowest terms, then 5q^2 = p^2. Let sqrt(5) = p/q. Then p^2 = 5q^2, so p is a multiple of 5. Write p = 5k. Substituting gives 25k^2 = 5q^2, so q^2 = 5k^2. Therefore q is also a multiple of 5. Prove sqrt(5) is irrational, or identify the divisibility contradiction in the proof. Stopping after finding p divisible by 5 is incomplete because the contradiction comes from both numbers sharing the factor 5.
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