Question

$✠\ \textless \ br /\ \textgreater \ QUESTIØN\ \textless \ br /\ \textgreater \ QUESTIØN✠$

$Prove that \sf\sqrt{3} \ \textless \ br /\ \textgreater \ 3\ \textless \ br /\ \textgreater \ ​$


is an Irrational Number.

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Answer 1

Answer:

Hello ✌

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proof :-

Let us assume to the contrary that √3 is a rational number.

It can be expressed in the form of p/q

where p and q are co-primes and q ≠ 0.

$\sqrt{3} = \frac{p}{q}$

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

$3 {q}^{2} = {p}^{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ..1$

It means that 3 divides p2and also 3 divides p because each factor should appear two times for the square to exist.

So we have p = 3r

where r is some integer.

$⇒ \: {p}^{2} = 9 {r}^{2}$

$⇒ {q}^{2} = 3 {r}^{2}$

We note that the lefthand side of this equation is even, while the righthand side of this equation is odd, which is a contradiction. Therefore there exists no rational number r such that

${r }^{2} = 3$

Hence the root of 3 is an irrational number.

.: Hence Proved..

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Answer 2

for answer check out the above image

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