Question

Show that 4-root3 is irrational

Answer 1

Answer:

Yes because 4 root 3 doesn't has square value

Answer 2

Answer:

$4 - \sqrt{3}$

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

Hope it helps you...

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