Question

prove that root 3 is irrational number​

Answer 1

Answer:

Let us assume on the contrary that 3

​ is a rational number. Then, there exist positive integers a and b such that

3

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

Now,

3

​=ba​

⇒3=b2a2​

⇒3b2=a2

⇒3∣a2[∵3∣3b2]

⇒3∣a...(i)

⇒a=3c for some integer c

⇒a2=9c2

⇒3b2=9c2[∵a2=3b2]

⇒b2=3c2

⇒3∣b2[∵3∣3c2]

⇒3∣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, root 3 is an irrational number.

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Answer 2

Answer:

Step-by-step explanation:

Let us assume that √5 is a rational no.

We know that the rational no's. Are in the form of p/q form where p,q are integers

So, √5 = p/q

P

p = √5q

We know that 'p' is a rational no. So √5 must be rational no. Since it equals to p

But it doesn't occurs with √5 since it's an integer

Therefore p=/= √5q

This contradicts the fact that √5 is an irrational no.

Hence our assumption is wrong and √5 is an irrational no.

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