Question

prove √3 is irrational number​

Answer 1

Answer:

Let us assume to the contrary that √3 is a rational number.It can be expressed in the form of p/qwhere p and q are co-primes and q≠ 0.⇒ √3 = p/q⇒ 3 = p²/q² (Squaring on both the sides)⇒ 3q2 = p²………………………………..(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 = 3rwhere r is some integer.⇒ p² = 9r²………………………………..(2)from equation (1) and (2)⇒ 3q²= 9r²⇒ q²= 3r²Where q2²is multiply of 3 and also q is multiple of 3.Then p, q have a common factor of 3. This runs contrary to their being co-primes. Consequently, p / q is not a rational number. This demonstrates that √3 is an irrational number.

Answer 2

Answer:

ASSUMPTION,

$let \: \sqrt{3} \: be \: rational$

$then$

$\sqrt{3} = \frac{a}{b} (where \: a \: and \: b \: are \: co - prime \: integers \: and \: b \: is \: not \: equal \: to \: 0)$

$\sqrt{3} b = a$

on squaring both sides,

$( \sqrt{3} {b})^{2} = ( {a})^{2}$

or,

$3 {b}^{2} = {a}^{2} ....(1)$

this means that a² is divisible by 3 or a is divisible by 3

so we can also write,

$a = 3c(for \: some \: integer \: c)$

on squaring both the sides.

${a}^{2} = ({3c})^{2}$

or,

${a}^{2} = {9c}^{2} ...(2)$

On substituting equation (1) and equation (2)

${3b}^{2} = {9c}^{2} \\ {b}^{2} = {3c}^{2}$

which also means that b is divisible by 3 pr b is divisible by 3

if this is so, it will mean that a and b are divisible to each other and are not CO-PRIME integers.

CONTRADICTION TO MY ASSUMPTION

Conclusion - UNDER ROOT 3 IS AN IRRATIONAL NUMBER

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