We use the method of proof by contradiction. We begin by assuming that √5 is a rational number, meaning it can be expressed as a quotient of two integers.

Since √5 is assumed to be rational, we can write it as (a)/(b), where a and b are co-prime integers with no common factor other than 1, and b ≠ 0.

Squaring both sides eliminates the radical, giving 5 = (a²)/(b²). Multiplying both sides by b² yields a² = 5b², which shows that a² is divisible by 5.

By the fundamental theorem of arithmetic, if a prime p divides a², then p divides a. Since 5 is prime and divides a², it follows that 5 divides a. Hence, we can write a = 5c for some integer c.

Substituting a = 5c into a² = 5b² yields 25c² = 5b². Dividing both sides by 5, we get b² = 5c². This implies 5 divides b², and therefore 5 divides b.

Prove that' √5 is irrational

Answer: Hence, by contradiction, $\sqrt{5}$ is an irrational number.

Step-by-step solution

Step 1: Assume the Contrary

We use the method of proof by contradiction. We begin by assuming that $\sqrt{5}$ is a rational number, meaning it can be expressed as a quotient of two integers.

Step 2: Express as Co-prime Integers

Since $\sqrt{5}$ is assumed to be rational, we can write it as $\frac{a}{b}$ , where $a$ and $b$ are co-prime integers with no common factor other than $1$ , and $b \neq 0$ .

Step 3: Square Both Sides and Rearrange

Squaring both sides eliminates the radical, giving $5 = \frac{a^2}{b^2}$ . Multiplying both sides by $b^2$ yields $a^2 = 5b^2$ , which shows that $a^2$ is divisible by $5$ .

Step 4: Deduce that 5 Divides a

By the fundamental theorem of arithmetic, if a prime $p$ divides $a^2$ , then $p$ divides $a$ . Since $5$ is prime and divides $a^2$ , it follows that $5$ divides $a$ . Hence, we can write $a = 5c$ for some integer $c$ .

Step 5: Substitute and Deduce that 5 Divides b

Substituting $a = 5c$ into $a^2 = 5b^2$ yields $25c^2 = 5b^2$ . Dividing both sides by $5$ , we get $b^2 = 5c^2$ . This implies $5$ divides $b^2$ , and therefore $5$ divides $b$ .

Step 6: Conclude the Contradiction

We have shown that both $a$ and $b$ have $5$ as a common factor. This contradicts our initial assumption that $a$ and $b$ are co-prime. The contradiction arose because of our incorrect assumption that $\sqrt{5}$ is rational. Hence, $\sqrt{5}$ is irrational.

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