Question

prove that √3 is an irrational number it's very urgent don't spam please​

Answer 1

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 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.

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.

Answer 2

Answer:

√3 cannot be express in p/q forms except with 1

Step-by-step explanation:

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