Question

prove that √5 is irrational number​

Answer 1

Answer: Hey buddy, hope this helps you! Please mark as the brainliest answer if you feel so!

Step-by-step explanation:

To Prove: $\sqrt{5}$ is irrational.

Proof: Let us assume that $\sqrt{5}$ is rational.

That means $\sqrt{5}$ = $\frac{p}{q}$, as a rational number can be represented as a fraction, where p and q are co-primes and rational integers.

Now, squaring on both sides, we get

5 = $\frac{p^{2} }{q^{2}}$

⇒ $q^{2}$ = $\frac{p^{2} }{5}$

Here, 5 divides both p and $p^{2}$.

Now, let p = 5c, for any integral value of c.

⇒ $(q)^{2}$ = $\frac{(5c)^{2} }{5}$

⇒ $(q)^{2}$ = $\frac{25c^{2} }{5}$

⇒$(q)^{2}$  = $5c^{2}$

⇒$c^{2}$ = $\frac{q^{2} }{5}$

Here, 5 divides both q and $(q)^{2}$.

So now, both p and q have a common factor as 5.

But this contradicts the fact that both p and q are co-primes.

This contradiction has arisen due to our wrong assumption that $\sqrt{5}$ is rational.

Therefore, $\sqrt{5}$ is irrational.

Hence, Proved.