Question

prove that 2√3 is an irrational number​

Answer 1

Answer:

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Answer 2

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Let us assume that 6+2√3 is rational,

so,

$2 \sqrt{3} = \frac{a }{b}$

where a and b are co prime

$\sqrt{3} = \frac{a}{2b}$

we know that √3 is rational but it is shown in the form of

$\frac{p}{q}$

so, this contradiction is arissen because of our wrong assumption,

so,

2√3 is irrational

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