Question

Show that 2 root 5 is irrational

Answer 1

SOLUTION :

Let we suppose 2√5 is a rational number.

$\mapsto\sf{2\sqrt{5} =\dfrac{p}{q}\:\:\:\:\:\bigg[where\:p\:and\:q\:are\:co-prime\:\&\:q\neq 0\bigg]}$

So, multiplying by 1/2 both the sides.

$\mapsto\sf{\cancel{2}\sqrt{5} \times \dfrac{1}{\cancel{2}} =\dfrac{p}{q} \times \dfrac{1}{2} }\\\\\\\mapsto\sf{\sqrt{5} =\dfrac{p}{2q} }$

∴ 2, p & q are numbers.

∴ p/2q is rational number.

So, √5 is rational number.

But our contradiction fact that √5 is an irrational number.

∴ √5 ≠ p/2q.

Thus;

2√5 is an irrational number .

Answer 2

Please mark it as the brainliest

Step-by-step explanation:

Let $\sqrt{5}$ be is a rational number.

$\sqrt{5} =\frac{a}{b}$                                 where a & b are co-prime integers and b $\neq$ 0

squaring both the sides

$5 = \frac{a^{2} }{b^{2} }$

$b^{2} * 5 =a^{2}$                                  ------------ equation (1)

=> 5 is a factor of $a^{2}$

which means 5 is also a factor of a

Let m be any integer

a = 5m

putting the value of a in equation (1)

$\sqrt{5b^{2} } = 5m\\5b^{2} = 25m^{2}\\b^{2} = 5 * m^{2}$

=> 5 is a factor of $b^{2}$

which means 5 is also a factor of b

=>  5 is the common factor of a & b

This contradicts the condition that a & b are co-prime number.

Therefore $\sqrt{5}$ is an irrational number

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Let $2\sqrt{5}$ be a rational number

$2\sqrt{5} = \frac{a}{b}$                                   where a & b are co-prime integers and b $\neq$ 0

$\sqrt{5} = \frac{a}{2b}$

Since a & b are integers then $\frac{a}{2b}$ is also a rational number

but we know that $\sqrt{5}$ is irrational                (as proved above)

Therefore our assumption is wrong.

$2\sqrt{5}$  is irrational