Question

explain why √5 is irrational​

Answer 1

Step-by-step explanation:

Let's prove this by the method of contradiction-

Say, √5 is a rational number. ∴ It can be expressed in the form p/q where p,q are co-prime integers.

⇒√5=p/q

⇒5=p²/q²  {Squaring both the sides}

⇒5q²=p²  (1)

⇒p² is a multiple of 5. {Euclid's Division Lemma}

⇒p is also a multiple of 5. {Fundamental Theorm of arithmetic}

⇒p=5m

⇒p²=25m²   (2)

From equations (1) and (2), we get,

5q²=25m²

⇒q²=5m²

⇒q² is a multiple of 5. {Euclid's Division Lemma}

⇒q is a multiple of 5.{Fundamental Theorm of Arithmetic}

Hence, p,q have a common factor 5. this contradicts that they are co-primes. Therefore, p/q is not a rational number. This proves that √5 is an irrational number. 

For the second query, as we've proved √5 irrational. Therefore 2-√5 is also irrational because difference of a rational and an irrational number is always an irrational number.

Answer 2

Answer:

let

$\sqrt{5}$

is rational

√5=a/b where a and b are integers and b is not equal to zero.

5=a^2/b^2

a^2=5b^2

# a^2 is divisible by 5 and so a is also divisible by 5

Let a= 5c

a^2=5b^2

25c^2=5b^2

b^2 is divisible by 5 and so b is also divisible by 5

since a is divisible by 5 and b is also divisible by 5

a/b is not rational

√5 is not rational

so √5 is irrational.

Step-by-step explanation:

This method is contradiction method in which we assume the opposite of that required by us.

If you want then can also use division method.