Question

prove that √3 is irrational.​

Answer 1

Answer:

Let us assume to the contrary that √3 is a rational number. where p and q are co-primes and q≠ 0. ... This demonstrates that √3 is an irrational number.

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Answer 2

Answer:

${\huge{\boxed{\red{\mathscr{Answer}}}}}$

Let us assume that √3 is a rational number

So it can be written as

$\frac{a}{b} = \sqrt{3}$

$\green{\boxed{\sf \frac{a}{b} = \sqrt{3} }}$

(where a and b are co- prime numbers)

→Suppose that a and b have a common factor other than one ,then we can divide by the common factor and assume a and b are coprime.

So,.

b√3=a

SQUARING BOTH SIDES WE GET:

a²=3b²----------(1)

Therefore, a² is divisible by 3

and. a is also divisible by 3

( by the theorem that let P be a prime number if P divides a square then P divides a when a is positive integer)

So,..

we can write as

a=3c. (for some integer c) .

Put the value of 'a' in this from eq,(1)

3b²=9c²

and

b²=3c²

This means that b² is divisible by 3 and b is also divisble by 3

Therefore a and b have atleast(3) as common factor

And it contradicts the fact that a and b are co- prime

This arise due to our wrong assumption

So

$\huge\red{\boxed{\sf √3 \: is\: an\:irrational\:No.}}$