Question

√5 is an irrational number​

Answer 1

Answer:

yes root 5 is irrational number

what i have to do

Answer 2

Correct Question:

Prove that $\sf\sqrt{5}$ is an irrational number.

AnswEr:

Let us consider that $\sf\sqrt{5}$ is an rational number.

So, it can be written in the form of $\sf\dfrac{p}{q}.$

where p & q are co - prime and q ≠ 0.

Such that,

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

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

$\dag\bold{\underline{\sf{Squaring\; Both\; Sides}}}$

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

$:\implies\sf\; 5q^2 = p^2$

$\dag\bold{\underline{\sf{Here\; p \; is\; divisible\; by \; 5}}}$

$\rule{150}3$

Let p = 5k where k is an integer.

$\dag\bold{\underline{\sf{Again\; Squaring\; Both\; Sides}}}$

$:\implies\sf\; p^2 = 25 k^2$

Substituting 5q² = p².

$:\implies\sf\; 5q^2 = 25 k^2$

$:\implies\sf\; q^2 = 5k^2$

Here, we can see that q is also divisible by 5.

Now we can say that both p and q have atleast 5 as it's common factor.

So, it arises contradiction because of our wrong assumption.

$\bold{\underline{\sf{\blue{Hence,\; \sqrt{5}\; is \; an \; irrational\; number.}}}}$