prove that 6+2 root 3 is an irrantional number
Answers
Answer :-
Required to prove :-
6 + 2√3 is an irrational number
Conditions used :-
Here conditions refers to the properties ,
- p, q are integers
- q ≠ 0
- p,q are co-primes .
Similarly,
An irrational number is not equal to a rational number
Solution :-
6 + 2√3
Here,
Let's assume on the contradictory that 6 + 2√3 is a rational number .
So,
Equal the number 6 + 2√3 with p by q ( where p and q are integers , q ≠ 0 , p and q are co - primes )
Transpose +6 to the right side
By taking LCM we get,
Now, transpose the 2 to the right side
So,
we know that √3 is an irrational number but since it is not given in the question so we have too prove that √3 is an irrational number .
Hence,
Let's assume that √3 is a rational number
So, equal the value of √3 to a by b ( where a,b are integers , b ≠ 0 and a,b are co-primes )
Squaring on both sides
Recall the fundamental theorem of
According to which,
If a divides q²
Then, a divide q also.
So,
3 divides a²
3 divides a ( also )
Similarly,
Let's take the value of a as 3k
So,
squaring on both sides
So,
3 divides b²
3 divides b also .
Hence,
we can conclude that ,
a,b have common factor as 3 .
But we know that,
a, b should have common factor as 1 because a,b are co - primes .
So,
This contradicton is due to the wrong assumption that ,
√3 is a rational number.
So, our assumption is wrong .
Hence,
√3 is an irrational numbers
So, it is proved that √3 is an irrational number .
Hence
This is because,
An irrational number is not equal to a rational number .
So,
6 + 2√3 is an irrational number .
For similar question refer to,
https://brainly.in/question/17856936
Let us assume that , 6 + 2√3 is an rational number
So ,
Here , √3 is an irrational number but (p - 6q)/(2q) is rational number
Since , irrational ≠ rational
Thus , our assumptions is wrong