Question

Prove that V3 is an irrational number​

Answer 1

Let us assume on the contrary that root3 is a rational number. 

Then, there exist positive integers a and bsuch that

root3=ba where, a and b, are co-prime i.e. their HCF is 1

Now,

root3=ba

⇒3=b2a2 

⇒3b2=a2 

⇒3 divides a2[∵3 divides 3b2] 

⇒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) 

From (i) and (ii), we observe that a and bhave 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, root 3 is an irrational number

thanks........

Answer 2

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.

Step-by-step explanation:

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