Question

prove that √5 is irrational ​

Answer 1

Answer:

Let 5 be a rational number.

then it must be in form of qp where, q=0 ( p and q are co-prime)

p2 is divisible by 5.

So, p is divisible by 5.

So, q is divisible by 5.

Thus p and q have a common factor of 5.

We have assumed p and q are co-prime but here they a common factor of 5.

Step-by-step explanation:

Answer 2

$\underline{ \mathfrak{ \: To\:Prove:- \: }}$

• √5 is an irrational number.

$\underline{ \mathfrak{ \: Proof:- \: }}$

Let √5 be a rational number, which can be written in the form of p/q, where p and q are integers and q ≠ 0

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

$\underline{\:\textsf{ Squaring on both sides :}}$

$\sf{\hookrightarrow (\sqrt{5})^2 = \bigg(\dfrac{p}{q} \bigg)^2 }$

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

$\bf{\hookrightarrow 5q^2 = p^2....1) }$

______________

• 5 divides p²

• 5divides p

______________

Let p = 5x

Put the value of 'p' in equation (1)

$\sf{\hookrightarrow 5q^2 = (5x)^2 }$

$\sf{\hookrightarrow 5q^2 = 25x^2 }$

$\sf{\hookrightarrow q^2 = 5x^2 }$

______________

• 5 divides q²

• 5 divides q

______________

Thus, 5 divides p and q.

• It means 5 is a common factor of p and q. This contradicts the assumption as there is no common factor of p and q.

$\therefore\;{\underline{\sf{ \bf{ \sqrt{5}}\;is\; irrational.}}}$

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━