Question

prove that √5 is irrational.​

Answer 1

To proof :

• √5 is irrational

Prove :

Let us consider √5 be a rational number.

$: \implies \sf{ \sqrt{5} = \dfrac{p}{q} }$

Where, p and q are co prime number and q ≠ 0.

Squaring both the sides,

$: \implies \sf{5 = \dfrac{ {p}^{2} }{ {q}^{2} } }$

$: \implies \sf{5 {q}^{2} = {p}^{2} }....(i)$

5 is a factor of p²

°.° 5 is a factor of p.

Let, p = 5k

Putting the value of p in equation (i)

➝ 5q² = (5k)²

➝ 5q² = 25k²

➝ q² = 5k²

5 is a factor of q².

°.° 5 is a factor of q.

So, 5 is a factor of both p and q which contradicts the statement.

.°. √5 is an irrational number.

Answer 2

To prove:-

√5 is irrational

Proof:-

Let's assume that √5 is irrational.

• So we can write this. as p/q form.

p and q are co-integers and q≠0

Thus

$\\ \sf{:}\longrightarrow \sqrt{5}=\dfrac{p}{q}$

• On squaring both of the sides.

$\\ \sf{:}\longrightarrow (\sqrt{5})^2=\left(\dfrac{p}{q}\right)^2$

$\\ \sf{:}\longrightarrow 5=\dfrac{p^2}{q^2}$

• Using cross multiplication

$\\ \sf{:}\longrightarrow 5q^2=p^2\dots(1)$

$\\ \sf{:}\longrightarrow \dfrac{p^2}{5}=q^2$

• As 5 divides p so p is a multiple of 5

$\\ \sf{:}\longrightarrow p=5m$

• By squaring

$\\ \sf{:}\longrightarrow p^2=(5m)^2=25m^2\dots(2)$

• From eq(1) and eq(2)

$\\ \sf{:}\longrightarrow 5q^2=25m^2$

$\\ \sf{:}\longrightarrow q^2=5m^2$

• q² is a multiple of 5
• q is a multiple of 5

Conclusion:-

p,q have a common factor 5.

• so they are co-primes.
• Therefore, p/q is not a rational number

Hence

√5 is not a rational number

• √5 is irrational

Proved