Question

prove that
$\sqrt{5}$
is an irrational number​

Answer 1

Answer:

I'm too tired and sooooooo small

Answer 2

HERE'S YOUR ANSWER MATE....

We have to show that

.

is irrational.

We will prove this via the method of contradiction.

So let's assume

.

is rational.

Hence, we can write

.

in the form

.

, where a and b are co-prime numbers such that a,b,∈R and b



=0.

∴

.

squaring both sides we have

⇒

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

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

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

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

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.

Hence, 5 divides a ......(1)

∴ we can say that

${5}^{a} = c$

we already know that

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

..............(2)

From (2), we know

$a = {5}^{c}$

substituting that in the above equation we get,

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

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

⇒

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

$hence \: 5 \: divides \: {b}^{2}$

. And by the above mentioned theorem we can say that 5 divides b as well.

hence, 5 divides b .........(3)

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

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

Hence, solved.

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AKSHITA