Question

prove that root 3 is an irrational number​

Answer 1

Answer:

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Answer 2

$\large \bold{ \sf \: Solution : }$

To prove that √3 is an irrational number,We have to find the square root of √3 by Long Division Method.

[Refer to the attachment.]

$\therefore \sf \: \sqrt{3} = 1.732050807...$

We observe that the decimal representation of √3 is neither terminating nor repeating.

$\red {\sf \: Hence \: \sqrt{3} \: is \: an \: irrational \: number} \:$

$\large \bold{ \sf \: Aliter : }$

We shall prove this by the method of contradiction. If possible, let us assume that √3 is a rational number. Then,

$\sf \: \sqrt{3} = \dfrac{p}{q}$

Where p and q are integers having no common factor and q is not equal to 0.

$\rightarrow \sf \: 3 = \dfrac{ {p}^{2} }{ {q}^{2} } \: ( { Squaring \: both \:sides })$

$\rightarrow \sf \: {p}^{2} = 3 {q}^{2} ...(i)$

$\rightarrow \sf \: {p}^{2} \: is \: an \: even \: integer$

$\rightarrow \sf \: {p} \: \: is \: an \: even \: integer$

$\rightarrow \sf \: {p} = 3m ,where \: m \: is \: integer$

$\sf \rightarrow \: {p}^{2} = 9 {m}^{2}$

$\sf \rightarrow \: 3 {q}^{2} = 9 {m}^{2} \: \: ({Using \: eq.i})$

$\sf \rightarrow \: {q}^{2} =3 {m}^{2}$

$\sf \rightarrow \: {q}^{2} \: is \: an \: even \: integer$

$\sf \rightarrow \: {q} \: is \: an \: even \: integer$

3 divides both the integers p and q. Hence,It is the factor of p and q. Therefore,p and q aren't co-prime.

Hence,Our assumption is wrong.

$\red {\sf \sqrt{3} \: is \: an \: irrational \: number}$