Question

prove that √2 is a irrational no.

Answer 1

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Let's suppose √2 is a rational number. Then we can write it √2=a/b where a,b are whole numbers.[b is not equals 0]

We additionally assume that this A/b is simplified to lowest terms, since that can obviously be done with any fraction. Notice that in order for a/b to be in simplest terms, both of a and b cannot be even. One or both must be odd. Otherwise, we could simplify a/b further.

so,

if, √2=a/b

then, 2=a^2/b^2

If we substitute a=2k

then,

2=(2k)^2/b^2

=>2=4k^2/b^2

That means, b^2 from which follows again that b itself is even. And that is a contradiction.

Thus our original assumption =√2 is not correct. Therefore √2 cannot be rational. [Proved]

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Answer 2

Heya

Your answer is given below :

Solution :

=> let us assume on the contrary that √2 is a rational number.Then, there exists positive integers a and b such that

$= > \sqrt{2} = \frac{a}{b} \: \: \: \: (where \: a \: \: \: and \: b \: \: are \: coprime ) \\ = > { (\sqrt{2} )}^{2} = { (\frac{a}{b} )}^{2} \\ = > 2 {b}^{2} = {a}^{2} \\ = > \frac{2}{ {a}^{2} } \\ by \: using \: theorem \\ = > \frac{2}{a} \: \: \: (i) \\ = > a = 2c \: \: \: \: \: \: for \: some \: interger \: c$
$= > {a}^{2} = 4 {c}^{2} \\ = > 2 {b }^{2} = 4 {c}^{2} \\ = > {b}^{2} = 2 {c}^{2} \\ = > \frac{2}{ {b}^{2} } \\ by \: using \: theorem \\ = > \frac{2}{b} \: \: \: \: \: \: (ii)$
from (i) and (ii) we obtained that 2 is a common factor of a and b but this contradicts the fact that a and b have no common factor other than 1 this means that our supposition is wrong

hence √2 is an irrational number.

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