Question

prove that root 5 is irrational​

Answer 1

Solution :

Let us find the square root of 5 by Long Division Method

[Refer to the attachment]

$\implies \sf \sqrt{5} = 2.2360679..$

We observe that the decimal representation of √5 is neither terminating nor repeating.

Hence,

$\star \: \: \sf \red{ \sqrt{5} \: is \: an \: irrational \: number. }$

Aliter :

We shall prove this by the method of contradiction. If possible,let us assume √5 is rational number. Then,

$\longrightarrow \sf \: \sqrt{5} = \dfrac{p}{q}$

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

Squaring both sides

$\longrightarrow \sf \: 5 = \dfrac{ {p}^{2} }{ {q}^{2} }$

$\longrightarrow \sf \: {p}^{2} = 5 {q}^{2} ....(i)$

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

[p²= (2m+1)² = 4m² + 4m + 1,which is odd. This is a contradiction to the fact that p² is even]

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

$\longrightarrow \sf \: {p} = 5m ,\: where \: m \: is \: an \: integer \:$

$\longrightarrow \sf \: {p}^{2} = {25m}^{2}$

Using equation i)

$\longrightarrow \sf \: 5 {q}^{2} =25 {m}^{2}$

$\longrightarrow \sf \: {q}^{2} =5 {m}^{2}$

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

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

5 divides both the integers p and q. Hence,it is the factor of p and q. Therefore ,p and q aren't co-primes.

Our assumption is wrong

Hence,

$\star \: \: \sf \red{ \sqrt{5} \: is \: an \: irrational \: number. }$

Answer 2

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QUESTION :

prove that root 5 is irrational.

SOLUTION :

Let us assume that √5 is rational.

Here, ( p , q ) = 1 , q not equal to 0, P and Q are natural numbers.

So,

$\sqrt{5} = \dfrac{p}{q} \: \\ \\ (p. \: q \: ) = 1 \\ \\ \: squareing \\ \\ 5 = \dfrac{ {q}^{2} }{ {p}^{2} }$

So,

$5 { p } ^ 2 = { q } ^ 2$

So ,

${ p } ^ 2 = \dfrac{ { q } ^ 2 }{5}$

So,

Q ^ 2 is a factor of 5

=> Q is a factor of 5

So, we have showed that Q is a factor of 5

Let Q be 5 k

$=> 5 = \dfrac{25 { k } ^ 2 }{{p} ^ 2 }$

$\dfrac{{ p} ^ 2}{5} = { k } ^ 2$

So, P ^ 2 is a factor of 5

=> P is a factor of 5

So, both p and q are factors of 5.

But P and Q have no common factors as mentioned above.

So , √5 is a irrational number.

Hence proved By Contradiction.

ADDITIONAL INFORMATION -

√5 = 2.360679975.............