Question

prove that √5 is irrational

Answer 1

Proof :

For now,
Let us assume that √5 is a Rational Number which can be written in the form of x/y where x and y are the prime numbers.

So,
$\sqrt{5} = \frac{x}{y}$

Squaring both the sides we get,

${( \sqrt{5} )}^{2} = {( \frac{x}{y} )}^{2} \\ \\ 5 = \frac{ {x}^{2} }{ {y}^{2} } \\ \\ 5 {y}^{2} = {x}^{2}$

5a is divisible by x where a is some integer

And 5a is also divisible by x^2

x = 5a

$5 {y}^{2} = {(5a)}^{2} \\ \\ 5 {y}^{2} = 25 {a}^{2} \\ \\ {y}^{2} = 5 {a}^{2}$

5b is divisible by c where b is some integer
5b is also divisible by c^2

But, above we said that x and y are prime numbers and don't have any common factor.

This contradicts √5 a irrational number.

This contradiction is because of our wrong assumption that √5 is a Rational number.

Hence proved :)

Answer 2

HI

Proof:

Let us assume to the contrary that √5 is rational

Therefore, √5 = a/b , where a and b are coprime integers and b ≠ 0

Squaring on both sides,

5 = a²/b²

5b² = a² ......(1)

Since, a² is divisible by 5,

Therefore, a is divisible by 5

Let a = 5c and substitute in eq(1)

5b² = (5c)²

5b² = 25c²

b² = 5c²

Since, b² is divisible by 5,

Therefore, b is divisible by 5

➡️ a and b have at least one common factor i.e. 5

This contradicts the fact that a and b are coprime.

➡️ This contradiction has arisen due to our incorrect assumption that √5 is rational.

Therefore, we conclude that √5 is irrational.

Hence Proved !

Hope it proved to be beneficial....