Question

Prove that root 5 is irrational brainly

Answer 1

$\huge{\underline{\underline{\mathcal{\pink{ANSWER}}}}}$

Step-by-step explanation:

don't mind my writing please

ur answer is in attachment.

MARK AS BRAINLIEST

Answer 2

$\rule{400}{4}$

Question :-

Prove that √5 is an irrational number ?

$\rule{400}{4}$

Answer :-

Required to prove :-

• √5 is an irrational number

$\rule{400}{4}$

Solution :-

Given to prove that √5 is an irrational number .

So,

Let's assume on the contradicton that √5 is a rational number .

Here, √5 is equal p by q ( where p and q are integers , q ≠ 0 and p,q are co - primes ) .

So,

$\tt{ \sqrt{5} = \dfrac{p}{q}}$

Now perform cross multiplication

Hence,

5q = p

Now do squaring on both sides

$\tt{ {(5q)}^{2} = {(p)}^{2}}$

5q² = p²

From the fundamental theorem of arithmetic ,

we know that ,

If a divides p²

Then , a divides p also .

So,

5 divides p²

Similarly,

5 divides p also .

However;

Let's consider the value of p as 5k

( where k is any positive integer )

So,

5q = 5k

Now do squaring on both sides

$\tt{ {(\sqrt{5}q)}^{2} = {(5k)}^{2}}$

5q² = 25k²

q² = 5k²

Now interchange the position of the terms

Hence,

5k² = q²

Again recall the fundamental theorem of arithmetic .

So,

from the above it is clear that ,

5 divides q²

so, 5 divides q .

Here,

From the above we can conclude that ,

5 is common factor of both p and q .

But ,

According to the rational numbers properties ,

p and q are co-primes which means that they have only common factor as 1 .

Hence, this contradicton is due to wrong assumption .

That ,

√5 is a rational number .

Therefore,

Our assumption is wrong .❌

So,

√5 is an irrational number .

$\rule{400}{4}$

✅ Hence proved .

$\rule{400}{4}$