Question

prove that √5 is an irrational number. ​

Answer 1

Solution:-

Let us assume that $\sf{{\sqrt {5}}}$ is a rational number.

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

$\sf\dfrac{p}{q}$ are integers and have no common factor and $\sf{{q }}$≠$\sf{{0}}$

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

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

Case I:

Let $\sf{{q }}$=$\sf{{1}}$

$\implies \sf{{5\times 1 = }}$$\sf\dfrac{p^{2}}{1}$

$\implies \sf{{5 = }}$$\sf\dfrac{p^{2}}{1}$

$\implies \sf{{5 = p^{2}}}$

There is no integer whose square is $\sf{{5}}$

∴This case is not possible

Case II:

Let $\sf{{q }}$≠$\sf{{1}}$

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

The answer will be a non integer.

As Integer ≠ Non Integer

∴ This case is also not possible.

So, $\sf{{\sqrt {5}}}$ is not a rational number.

∴ √5 is an irrational number.

Answer 2

Answer:

Given: √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