Question

Prove that 3 is an irrational number.​

Answer 1

Answer:

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.

Answer 2

Answer:

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

Then, there exist positive integers a and b such that

Then, there exist positive integers a and b such that3=ba where, a and b, are co-prime i.e. their HCF is 1

Then, there exist positive integers a and b such that3=ba where, a and b, are co-prime i.e. their HCF is 1Now,

Then, there exist positive integers a and b such that3=ba where, a and b, are co-prime i.e. their HCF is 1Now,3=ba

Then, there exist positive integers a and b such that3=ba where, a and b, are co-prime i.e. their HCF is 1Now,3=ba⇒3=b2a2 

Then, there exist positive integers a and b such that3=ba where, a and b, are co-prime i.e. their HCF is 1Now,3=ba⇒3=b2a2 ⇒3b2=a2 

Then, there exist positive integers a and b such that3=ba where, a and b, are co-prime i.e. their HCF is 1Now,3=ba⇒3=b2a2 ⇒3b2=a2 ⇒3 divides a2[∵3 divides 3

⇒3 divides a...(i) 

⇒3 divides a...(i) ⇒a=3c for some integer c

⇒3 divides a...(i) ⇒a=3c for some integer c⇒a2=9c2 

⇒3 divides a...(i) ⇒a=3c for some integer c⇒a2=9c2 ⇒3b2=9c2[∵a2=3b2] 

⇒3 divides a...(i) ⇒a=3c for some integer c⇒a2=9c2 ⇒3b2=9c2[∵a2=3b2] ⇒b2=3c2 

⇒3 divides a...(i) ⇒a=3c for some integer c⇒a2=9c2 ⇒3b2=9c2[∵a2=3b2] ⇒b2=3c2 ⇒3 divides b2[∵3 divides 3c2] 

⇒3 divides a...(i) ⇒a=3c for some integer c⇒a2=9c2 ⇒3b2=9c2[∵a2=3b2] ⇒b2=3c2 ⇒3 divides b2[∵3 divides 3c2] ⇒3 divides b...(ii) 

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