Question

Prove that √5 is irrational.​

Answer 1

Answer:

This Is your Answer

Step-by-step explanation:

: √5

We need to prove that √5 is irrational

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

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

√5 is an irrational number

Hence proved

Answer 2

Step-by-step explanation:

Let us prove √5 by contradiction.

Let us suppose that √5 is a rational.It means that we have co-prime Integers 'a' and 'b' (b is not equal to 0)

such that,

$\sqrt{5} = \sf \frac{a}{b} \\ \sf{b \sqrt{5}} = a \: \: \: equation \: 1$

Squaring on both sides, we get

$\sf5 {b}^{2} = {a}^{2}$

It means that '5' is factor of 'a²'

Hence,5 is also factor of 'a' by theorem2

If 5 is a factor of 'a', it means that we can write a=5c for some integer 'c'.Substituting value of 'a' in equation 1

$\sf{5 {b}^{2}} = 25 {c}^{2} = {b}^{2} = 5 {c}^{2}$

It means that '5' is factor of 'b²'.

Hence, '5' is also factor of 'b' by theorem 3

From '2' and '3',we can say that '5' is factor of both 'a' and 'b'

But 'a' and 'b' are co-primes.

Therefore,our assumption was wrong,√5 cannot be irrational.

$\bold \green{ \underline{</strong><strong>P</strong><strong>lease \: mark \: it \: as \: brainlist \: answer}}$