Question

$prove \: that \: \sqrt{3 \: is \: irrational}$
full solution needed !


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Answer 1

Refer to the attachment above for the answer ⬆️☑️

Hope this helps you ❤️

Answer 2

Let us on the contrary that √3 is rational number.

Then, there exist positive integer a and b

Such that,

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

Where, a and b are co-prime i.e.

there HCF is 1

Now,

$\sqrt{3} = \frac{a}{b} \\ →3 = \frac{a {}^{2} }{b {}^{2} } \\ →3b {}^{2} = a {}^{2} \\ →3 \: divides \: a {}^{2} (3 \: divides \: 3b {}^{2} ) \\ →3 \: divides \: a \: ...(1) \\ →a = 3c for \: sum \: integer \: c \: \\ →a {}^{2} = 9c {}^{2} \\ →3b {}^{2} = 9c {}^{2} ...(a {}^{2} = 3b {}^{2} ) \\ →b {}^{2} = 3c {}^{2} \\ →3 \: divides \: b {}^{2} \\ →3 \: divides \: b \: ...(2)$

From (1) and (2), we observe that a and b have at least 3 as a common factor. But, this contradicts the fact that a and b are co prime. This means that our assumption is not correct.

Hence, √3 is an irrational number.