Question

Prove that √5 is irrational​

Answer 1

s root 5 is irrational because rational numbers if we written in p/q amd q is not equal to zero so root 5 we cant written in the form p/q. hence proved.plz follow me and mark as brain list

Answer 2

$answer \: -$

$we \: have \: to \: prove \: \sqrt{5 } \: is \: irrational \:$

Let us assume the opposite ,

$i.e \sqrt{5} \: is \: rational$

$hence \: it \: can \: be \: written \: as \: \frac{a}{b}$

$here a and b (b is not equal to 0) are co - prime{no common factor other than 1$

$hence \: \sqrt{5} = \frac{a}{b}$

$\sqrt{5b } = a$

squaring both side :-

$({ \sqrt{5b} })^{2} = {a}^{2}$

${5b}^{2} = {a}^{2}$

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

Hence,

$5 \: divides \: {a}^{2}$

${ by theorem if p is a prime no. and p divides {a}^{2} then p divides a where a is a positive number$

So, 5 shall divide a Also.................(1)

Hence We can say ,

$\frac{a}{5} = c \: \: where \: c \: is \: some \: integer$

So,

$a = 5c$

Now we know that,

$5 {b}^{2} = {a}^{2}$

Putting a = 5c

$5 {b}^{2} = ({5c})^{2}$

${5b}^{2} = {25c}^{2}$

${b}^{2} = \frac{1}{5 } \times {25c}^{2}$

${b}^{2} = {5c}^{2}$

$\frac{ {b}^{2} }{5} = {c}^{2}$

$Hence \: 5 \: Divides \: {b}^{2}$

Theorem :-

${if p is a prime no. and p divides {a }^{2} then p divides a where p is positive no.$

So, 5 divides a also...............(2)

By (1) and (2)

5 divides both a and b.

Hence , 5 is factor of Both a and b.

so, a and b both have 5 as factor

then,

they both ( a and b) are not co-prime no.

Hence, Our assumption is wrong

Therefore By Contradiction ,

$\sqrt{5} \: is \: irrational$


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