**Proofs of Irrationality**1. **Prove $ \sqrt{5} $ is irrational** To show that $ \sqrt{5} $ is irrational, assume the contrary: that $ \sqrt{5} $ is rational. Then it can be expressed as a fraction $ \frac{a}{b} $, where $ a $ and $ b $ are integers with no common factors other than 1, and $ b \neq 0 $. \[ \sqrt{5} = \frac{a}{b} \implies 5 = \frac{a^2}{b^2} \implies a^2 = 5b^2 \] This implies that $ a^2 $ is divisible by 5, hence $ a $ itself must be divisible by 5. Let $ a = 5k $ for some integer $ k $. \[ (5k)^2 = 5b^2 \implies 25k^2 = 5b^2 \implies 5k^2 = b^2 \] So $ b^2 $ is also divisible by 5, which means $ b $ is divisible by 5. This contradicts our assumption that $ a $ and $ b $ have no common factors other than 1. Therefore, $ \sqrt{5} $ is irrational.2. **Prove $ 2 - 3\sqrt{5} $ is irrational** Assume for contradiction that $ 2 - 3\sqrt{5} $ is rational. Let it be expressed as a fraction $ \frac{m}{n} $, where $ m $ and $ n $ are integers, and $ n \neq 0 $. Rearrange the equation: \[ 2 - \frac{m}{n} = 3\sqrt{5} \implies 3\sqrt{5} = 2 - \frac{m}{n} \] Solving for $ \sqrt{5} $: \[ \sqrt{5} = \frac{2n - m}{3n} \] Since both $ \frac{m}{n} $ and $ 2 $ are rational, their difference

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Description: **Proofs of Irrationality**1. **Prove $ \sqrt{5} $ is irrational** To show that $ \sqrt{5} $ is irrational, assume the contrary: that $ \sqrt{5} $ is rational. Then it can be expressed as a fraction $ \frac{a}{b} $, where $ a $ and \( b

Proof of 5 is irrational: Let us assume that 5 is rational. 5=xy, where x,y are rational and y≠0.…

Answered: Prove v5 is irrational Prove 2– 3/5 is…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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**Proofs of Irrationality**

1. **Prove $ \sqrt{5} $ is irrational**

To show that $ \sqrt{5} $ is irrational, assume the contrary: that $ \sqrt{5} $ is rational. Then it can be expressed as a fraction $ \frac{a}{b} $, where $ a $ and $ b $ are integers with no common factors other than 1, and $ b \neq 0 $.

\[
\sqrt{5} = \frac{a}{b} \implies 5 = \frac{a^2}{b^2} \implies a^2 = 5b^2
\]

This implies that $ a^2 $ is divisible by 5, hence $ a $ itself must be divisible by 5. Let $ a = 5k $ for some integer $ k $.

\[
(5k)^2 = 5b^2 \implies 25k^2 = 5b^2 \implies 5k^2 = b^2
\]

So $ b^2 $ is also divisible by 5, which means $ b $ is divisible by 5. This contradicts our assumption that $ a $ and $ b $ have no common factors other than 1. Therefore, $ \sqrt{5} $ is irrational.

2. **Prove $ 2 - 3\sqrt{5} $ is irrational**

Assume for contradiction that $ 2 - 3\sqrt{5} $ is rational. Let it be expressed as a fraction $ \frac{m}{n} $, where $ m $ and $ n $ are integers, and $ n \neq 0 $.

Rearrange the equation:

\[
2 - \frac{m}{n} = 3\sqrt{5} \implies 3\sqrt{5} = 2 - \frac{m}{n}
\]

Solving for $ \sqrt{5} $:

\[
\sqrt{5} = \frac{2n - m}{3n}
\]

Since both $ \frac{m}{n} $ and $ 2 $ are rational, their difference

Expert Solution

Step 1

Proof of $\sqrt{5}$ is irrational:

Let us assume that $\sqrt{5}$ is rational.

$\sqrt{5} = \frac{x}{y}, where x, y are rational and y \neq 0 .$

Square both sides of the equation as follows.

$5 = \frac{x^{2}}{y^{2}}$

Simplify further as follows.

$5 y^{2} = x^{2} 5 y^{2} and x^{2} must have the same number of prime factors . x^{2}, y^{2} have even number of prime factors But 5 y^{2} have odd number of prime numbers . a contradiction since a number cannot have both odd number and even number of prime factors .$

The assumption that $\sqrt{5}$ is rational is wrong. Therefore, $\sqrt{5}$ is irrational.

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