Question

prove that √5 is a irrational number.
Only genius answer my questions​

Answer 1

$\text{ \huge{ \underline{ \purple{ \underline{✿Solution!}}}}}$

$\sf{We \: have \: to \: prove \: that  \sqrt{5} \:  is \: irrational \: number}$

$\sf{let's assume  \sqrt{5}  is rational.}$

$\sf{Hence, \: we \: can \: write  \sqrt{5} \:  in \: the \: form  \frac{a}{b} ,}$

$\sf{where \:  a  \: and  \: b \:  are \: co-prime \: numbers \: such \: that}$

$\sf{a, b, { \in{R \: and \: b≠0}}}$

$\sf{ \therefore{ \sqrt{5} = \frac{a}{b} }}$

$\sf{squaring \: both \: sides \: we \: have }$

$\sf{ = > 5 = \frac{a {}^{2} }{ {b}^{2} } }$

$\sf{ = > 5b { }^{2} = a {}^{2} }$

$\sf{ = > \frac{a {}^{2} }{5} = {b}^{2} }$

$\sf{Hence, \: 5 \: divides \:   {a}^{2} }$

Now, a theorem tells that if 'P' is a prime number and P divides a^{2} then P should divide 'a', where a is a positive number.

$\sf{Hence \: 5 \: devides \: {a}^{2} - - - - - - - - - - - - - (1)}$

$\sf{ \therefore{We \: can \: say \: = > \frac{a}{5} = c - - - - - - - - - - - - (2)}}$

$\sf{we \: already \: know \: that \: ⇒5b {}^{2} =a {}^{2} }$

From (2), we know a=5c substituting that in the above equation we get,

$\sf{5b {}^{2} = 25 {c}^{2} } \\ \sf{ = > {b}^{2} = 5 {c}^{2} } \\ \sf{ = > \frac{ {b}^{2} }{5} = {c}^{2} }$

Hence, 5 divides b^{2} And by the above mentioned theorem we can say that 5 divides b as well.

$\sf{hence \: 5 \: devides \: {b}^{2} - - - - - - - - - - - - (3)}$

So from (2) and (3) we can see that both a and b have a common factor 5. Therefore a and b are no co-prime. Hence our assumption is wrong.

$\sf{ \blue{ \boxed{ \therefore{ By \: contradiction,   \: \sqrt{5}  is irrational. }}}}$

Hence,Proved!

Answer 2

Let

5

be a rational number.

then it must be in form of

q

p

where, q



=0 ( p and q are co-prime)

5

=

q

p

5

×q=p

Suaring on both sides,

5q

2

=p

2

--------------(1)

p

2

is divisible by 5.

So, p is divisible by 5.

p=5c

Suaring on both sides,

p

2

=25c

2

--------------(2)

Put p

2

in eqn.(1)

5q

2

=25(c)

2

q

2

=5c

2

So, q is divisible by 5.

.

Thus p and q have a common factor of 5.

So, there is a contradiction as per our assumption.

We have assumed p and q are co-prime but here they a common factor of 5.

The above statement contradicts our assumption.

Therefore,

5

is an irrational number.