Question

Prove that √3 is an irrational number. pls show how you slove this question​

Answer 1

Solution:-

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When the square root of a number is a whole number, this number is called a perfect square. Many square roots are irrational numbers, meaning there is no rational number equivalent.

Answer 2

$\large\bf\underline\red{Answer \: :-}$

Let us assume on the contrary √3 is an rational number.

Then, there exist positive integer a and b

such that

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

where, a and b, are co prime i.e. their HCF is 1.

Now,

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

$\sf\implies{3 = \frac{a {}^{2} }{b {}^{2} } } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \sf\implies{3b {}^{2} = a {}^{2} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \sf\implies{3 \: divides \: a {}^{2} \: \: \: [ \: \because \: 3 \: divides \: 3b {}^{2} \: ] } \\ \\\sf\implies \purple{3 \: divides \: a \: \: \: ...(1)} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\\sf\implies{a = 3c \: for \: some \: integer \: c} \: \: \: \: \: \: \: \: \: \: \\ \\\sf\implies{a {}^{2} = 9c {}^{2} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\\sf\implies{3b {}^{2} = 9c {}^{2} \: \: \: } [ \: \because \: a {}^{2} \: = \: 3b {}^{2} \: ] \: \: \: \: \: \: \: \\ \\ \sf\implies{b {}^{2} = 3c {}^{2} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \sf\implies{3 \: divides \: b {}^{2} \: \: \: [ \: \because \: 3 \: = \: 3c {}^{2} \: ]} \: \: \: \: \: \: \: \\ \\ \sf\implies \purple{3 \: divides \: b \: \: \: ...(2)} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

From (1) and (2), we observe that a and b have at least 3 as 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.

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