Question

Prove that √3 is an irrational number

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Answer 1

Solution –

Let us assume that √3 is irrational

Then, there exist a positive number

a and b

Now,

$\sqrt{3 \: } = \frac{a}{b} \\ \\ squaring \: on \: both \: sides \: \\ \\ 3 = \frac{ {a}^{2} }{ {b}^{2} } \\ \\ 3b^{2} = {a}^{2}.........(1)$

So,

If 3 is the factor of $a^{2}$

then 3 is also a factor of a ...........(2)

Now,

let a = 3c (where c is any integer)

again, squaring on both sides

${a}^{2} = (3c)^{2} \\ \\ {a}^{2} = 9c^{2} \\ \\ putting \: the \: value \: of \: a^{2} \: in \: equation \: 1 \\ \\ 3b^{2} = {a}^{2} \\ \\ 3b^{2} = 9c^{2} \: (a^{2} = 9c^{2} ) \\ \\ b^{2} = \frac{9 {c}^{2} }{3} \\ \\ b^{2} = 3 {c}^{2} \\ \\ so \: if \: 3 \: is \: the \: factor \: of \: b^{2}$

then 3 is also a factor of b

since,

we observed that 3 is the factor of

a and b

but the contradicts the fact that a and b are coprime.

This means that our assumption is not correct.

Hence √3 is an irrational number.

Answer 2

Answer:

The number √3 is irrational ,it cannot be expressed as a ratio of integers a and b. To prove that this statement is true, let us Assume that it is rational and then prove it isn't (Contradiction).

Explanation: