Question

show that 2-root3 is irrational no
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Answer 1

$\large{\underline{\bf{\pink{Correct\:Question:-}}}}$

prove that 2+ √3 is irrational.

$\large{\underline{\bf{\purple{To\:prove:-}}}}$

• ✦ 2 + √3 is irrational.

$\huge{\underline{\bf{\red{Proof:-}}}}$

Let us assume, that 2 + √3 is rational.

Then,

There exist co - primes a and b (b≠ 0)

So,

➝ (2 + √3) = a/b

$\mapsto \rm\:( 2+ \sqrt{3} ) = \frac{a}{b} \: \\ \\\mapsto \rm\: \sqrt{3} = \frac{a}{b} - 2 \\ \\ \mapsto \rm\: \sqrt{3} = \frac{a - 2b}{b} \\ \\ \rm\:since \: \bf \: a \rm\: and \: \bf \: b \rm \: are \: intigers \: \\ \rm \: so \:, \: \frac{a - 2b}{b} \: \: is \: rational. \\ \\ \rm \: so ,\: \sqrt{3} \: is \: also \: rational.$

But , we know that √3 is irrational.

So, this contradiction is arissen because of our wrong assumption.

Hence, (2 + √3 ) is irrational.

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Answer 2

Answer:

$2 - \sqrt{3}$

Let us assume to the contrary that root3 is rational

$\sqrt{3 = \frac{a}{b} }$

Where a and b are integers, a and b are coprimes, b is not equal to 0.

Squaring on both sides, we get

$3 = \frac{a {}^{2} }{b {}^{2} } \\ 3b {}^{2} = a {}^{2} \\ a {}^{2} = 3b {}^{2}(multiple \: of \: 3) \\ a {}^{2} \: is \: divisible \: by \: 3 \\ a \: is \: also \: divisible \: by \: 3(1)$

Let a=3c [c is an integer]

Squaring on both sides, we get

$a {}^{2} = 9c {}^{2} \\ 3b {}^{2} = 9c {}^{2} (from \: 1) \\ b {}^{2} = 3c {}^{2} (multiple \: of \: 3) \\ b {}^{2} \: is \: divisible \: by \: 3 \\ b \: is \: also \: divisible \: by \: 3(2)$

From (1) and (2),a and b have a common factor 3 other than 1.

Therefore, a and b are not coprimes

This contradiction arisen because of our wrong assumption that root3 is rational

Therefore, conclude root3 is irrational

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