Question

Prove that √3 is an irrational number.

No Spam Answers.

Answer 1

This can be answered by the method of contradiction.

At first, let us assume that _/3 is a rational number.

Then, it can be written in the form of p/q, where p&q are integers and co primes and also q is not equal to zero.

Then,

_/3=p/q

(_/3) square= (p/q)square

3=p sq/q sq

3.q sq= p sq……….. (1)

So, 3 divides p sq

Therefore,3 divides p[since the Fundamental Theorem Of Arithmetic]

Let p=3r

Then ,after substituting in the equation (1), we get,

3.q sq=(3r) sq

3.q sq=9 r sq

q sq=9 r sq/3

q sq=3.r sq

So,3 divides q sq

Therefore 3 divides q. [ since the Fundamental Theorem Of Arithmetic]

Already it is proved that 3 divides p and now it is also proved that 3 divides q. That means 3 is the common factor of p and q. So, p and q are not co primes. But our assumption is that p and q are co primes. This contradiction has arisen because of our wrong assumption .i.e. _/3 is a rational number. So our assumption is false. Hence _/3 is an irrational number.

Hence ,proved.

Answer 2

hey dear



here is your answer



we have to prove √3 is irrational number




So



Let us assume that the opposite that is √3 is rational



Hence √3 can be written in the form of a /b





where a and b ( b unequal to zero) are Co prime ( no common factors other than 1 )



hence √3 = a / b



√3 b = a



squaring on both the sides


(√3b) ^2 = ( a) ^2


3b^2 = a^2



hence 3 divides a^2



a^2 / 3 = b^2




( By theorem if P is a prime number and p divides a^2 then p divides a where a is the positive number)


So 3 divide a also


hence we can say that


a / 3 = c ( where c is some integer)


So, a. = 3C




Now we know that



3b^2 = a^2


putting a = 3c



3b^2. = 3 c^2




3b^2 = 9 c^2



b^2 = 1 / 3 * 9 c^2



b^2 = 3c^2



b^2 /3 = c^2



hence 3 divide b^2




By (1) & ( 2)




3 divides both a and b



hence 3 is a factor of a and b


so a & b have factor 3




Therefore a & b are not Co prime



hence our assumptions is wrong


by contradiction


√3 is irrational



hence proved



hope it helps


thank you