Question

solve that root 3 is an irrational number​

Answer 1

Let $\sqrt{3}$ be rational

Then$\sqrt{3}$=p/q(where p and q are co-prime)

Squaring Both Sides,

3=$p^{2}/q^{2}$

$3q^{2}=p^{2}$

This Means That 'p' has prime factor 3      ..........(1)

Now,let p=3k

Therefore,

$3q^{2}=(3k)^{2} \\\\3q^{2}=9k^{2} \\\\q^{2}=3k^{2}$

This Means That 'q' has prime factor 3     .........(2)

By (1) and(2)

we get that both 'p' and 'q' has prime factors as 3

=> 'p' and 'q' are not co-prime

=>$\sqrt{3}$ is irrational

This is contradiction because  of our wrong assumption

Hence,$\sqrt{3}$ is Irrational

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Answer 2

Answer:

Yes....

Step-by-step explanation:

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