Question

prove that √3 is an irrational
​

Answer 1

ANSWER:-


Let us assume to the contrary that √3 is a rational number.

It can be expressed in the form of p/q

where p and q are co-primes and q≠ 0.

⇒ √3 = p/q

⇒ 3 = p2/q2 (Squaring on both the sides)

⇒ 3q2 = p2………………………………..(1)

It means that 3 divides p2 and also 3 divides p because each factor should appear two times for the square to exist.

So we have p = 3r

where r is some integer.

⇒ p2 = 9r2………………………………..(2)

from equation (1) and (2)

⇒ 3q2 = 9r2

⇒ q2 = 3r2

Where q2 is multiply of 3 and also q is multiple of 3.

Then p, q have a common factor of 3. This runs contrary to their being co-primes. Consequently, p / q is not a rational number. This demonstrates that √3 is an irrational number.

THANKS :)

Answer 2

Answer:

Let us assume on the contrary that

3

is a rational number.

Then, there exist positive integers a and b such tthat

3= ba

where, a and b, are co-prime i.e. their HCF is 1

Now,

3

=

b

a

⇒3=

b

2

a

2

⇒3b

2

=a

2

⇒3 divides a

2

[∵3 divides 3b

2

]

⇒3 divides a...(i)

⇒a=3c for some integer c

⇒a

2

=9c

2

⇒3b

2

=9c

2

[∵a

2

=3b

2

]

⇒b

2=3c 2

⇒3 divides b

2

[∵3 divides 3c

2

]

⇒3 divides b...(ii)

From (i) and (ii), we observe that a and b have at least 3 as a common factor. But, this contradicts the fact that a and b are co-prime. This means that our assumption is not correct.

Hence, 3 is an irrational number.

Step-by-step explanation:

hope it will help u...