Answer: since, √3 is a irrational number Step-by-step explanation: Let us assume √3 is rational. If it is rational,then there must exist two integers r and s (s≠0) such that √3=r/s. If r and s have a common factor other than 1. Then, we divide by the common factor to get √3=a/b,where a and b are co-prime. So, b√3=a. On squaring both sides and rearranging, we get 3b²=a². Therefore, 3 divides a². Now,by Theorem 1.6, it follows that since 3 is dividing a², It is also divides a. So, we can write a=3c for some integer c. Substituting for a, we get 3b²=9c², that is, b²=3c². This means that 3 divides b², and so 3 divides b. Therefore, both a and b have 3 as a common factor. But this contradicts the fact that a and b are co-prime. This contradiction has arisen because of our assumption that √3 is rational. Thus our assumption is false. So, we conclude that √3 is irrational. Reply
Answer: Therefore there exists no rational number r such that r2=3. Hence the root of 3 is an irrational . Reply
Answer:
since, √3 is a irrational number
Step-by-step explanation:
Let us assume √3 is rational.
If it is rational,then there must exist two integers r and s (s≠0) such that √3=r/s.
If r and s have a common factor other than 1. Then, we divide by the common factor to get √3=a/b,where a and b are co-prime. So, b√3=a.
On squaring both sides and rearranging, we get 3b²=a². Therefore, 3 divides a².
Now,by Theorem 1.6, it follows that since 3 is dividing a², It is also divides a.
So, we can write a=3c for some integer c.
Substituting for a, we get 3b²=9c², that is, b²=3c².
This means that 3 divides b², and so 3 divides b.
Therefore, both a and b have 3 as a common factor.
But this contradicts the fact that a and b are
co-prime.
This contradiction has arisen because of our assumption that √3 is rational. Thus our assumption is false. So, we conclude that √3 is irrational.
Answer:
Therefore there exists no rational number r such that r2=3. Hence the root of 3 is an irrational .