Question

prove that root 5 is irrational number​

Answer 1

Answer:

"√5 is an “irrational number”.

Given:

√5

To prove:

√5 is a rational number

Solution:

Let us consider that √5 is a “rational number”.

We were told that the rational numbers will be in the “form” of form Where “p, q” are integers.

So,

we know that 'p' is a “rational number”. So 5 \times q should be normal as it is equal to p

But it did not happens with √5 because it is “not an integer”

Therefore, p ≠ √5q

This denies that √5 is an “irrational number”

So, our consideration is false and √5 is an “irrational number”."

Answer 2

Let us assume that √5 is rational such that they can be written in the form a/b where a and b are coprime integers

Now, √5 = a/b (b≠ 0)

Squaring both side:

=> 5 = a²/b²

=> 5b² = a² ------- (i)

=> 5 divides a² [since, 5 divides 5b²]

=> 5 divides a

let a = 5c for some integer c

putting a = 5c in (i), we get:

=> 5b² = 25c²

=> b² = 5c²

=> 5 divides b² [since, 5 divides 5c²]

=> 5 divides b

Thus, 5 is a common factor of a and b

But it contradicts that a and b have no common factor other than 1 (coprime)

This contradiction arose due to our wrong assumption.

Therefore, √5 is irrational

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