Math, asked by nc5977068, 5 months ago

prove that √5 is an irrational numbers​

Answers

Answered by Nid45
0

ANSWER

We have to show that

5

is irrational.

We will prove this via the method of contradiction.

So let's assume

5

is rational.

Hence, we can write

5

in the form

b

a

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

=0.

5

=

b

a

squaring both sides we have

⇒5=

b

2

a

2

⇒5b

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

⇒5b

2

=a

2

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

⇒5b

2

=25c

2

⇒b

2

=5c

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

5

is irrational.

Hence, solved.

Answered by Anonymous
3

Step-by-step explanation:

Let \:  \sqrt{5}  \: is \: rational \\ ∴ \:  \sqrt{5}  =  \frac{a}{b} , \:  where  \: a  \: and  \: b  \: are   \:  \\  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  integers  \: and \:  b ≠ 0 \\  =  > 5 =  \frac{a {}^{2} }{b {}^{2} }  \\ and \:   \: a {}^{2}  = 5b {}^{2}  \\  =  > a {}^{2}  \: is \: divisible \: by \: 5 \: and \: so \: \:  a \: \\  is \: also \: divisible \: by \: 5. \\ Let \: \:  \:  \:  \:  \:   a = 5c \\ ∴ \: \:  \:  \:  \:  \:  \:  \:  a {}^{2}  = 5b {}^{2}  \\  =  > 25c {}^{2}  = 5b {}^{2}  \\ and, \:  \:  \:  b {}^{2}  = 5c {}^{2}  \\  =  > b {}^{2}  \: is \: divisible \: by \: 5 \: and \: so \:  \: b \\ is \: divisible \: by \: 5 \\ ∵a \:  \: is \: divisible \: by \: 5 \:  and \: b \:  \: is \:  \\ also \: divisible \: by \: 5 \\  =  >  \frac{a}{b }  \:  \: is \: not \: rational \\  =  >  \sqrt{5}  \:  \: is \: not \: rational \\ and, so \:  \sqrt{5}  \:  \: is \: irrational.

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