Question

Prove that √5 is irrational.​

Answer 1

Answer:

Step-by-step explanation:

Let us assume to the contrary that √5 is rational.

then it is of the form √5 = a/b ( where a and b are integers , b ≠ 0 and a and b are coprime)

b√5 = a

a = b√5

squaring on both sides

a² = 5b² -----(1)

Therefore a² is divisible by 5 and a is also divisible by 5

so , we can write a = 5c (for some integer c )

substitute a = 5c in eqn 1

5c = 5b²

This means b² is divisible by 5 and so b is also divisible by 5

therefore a and b have atleast 5 as a common factor

But this contradicts the fact that  a and b are co-prime.

This contradiction has arisen from our incorrect assumption that √5 is rational.

SO , WE CONCLUDE THAT √5 IS IRRATIONAL.

Answer 2

$\huge \red{\bf Answer}$

We need to prove that √5 is irrational

$\pink{\rm Proof:}$

Let us assume that √5 is a rational number.

Sp it t can be expressed in the form p/q where p,q are co-prime integers and q≠0

⇒√5=p/q

On squaring both the sides we get,

⇒5=p²/q²

⇒5q²=p² —————–(i)

p²/5= q²

So 5 divides p

ANSWER

p is a multiple of 5

⇒p=5m

⇒p²=25m² ————-(ii)

From equations (i) and (ii), we get,

5q²=25m²

⇒q²=5m²

⇒q² is a multiple of 5

⇒q is a multiple of 5

Hence, p,q have a common factor 5. This contradicts our assumption that they are co-primes. Therefore, p/q is not a rational number

$\green{ \bf \sqrt{5} \: is \: a \: irrational \: number}$