Solution -: We will consider that
I + f where I is an integer and 0, f < 1. We have to show that I is odd and that (I + f) (I - f) = 1
1st we will solve this expression by multiplication we get
(2 + √3)
n (2 - √3)
n = (4 - 3)
n = 1 by using ( a-b)(a+b) = a^2 - b^2
or
(2 + √3)
n(2 - √3)
n =1. It is thus required to prove that (2 - √3)
n = I-f
We know that by theorem
(2 + √3)n + (2 + √3)n = [2n - C1.2n-1.√3 + C22n-2.(√3)2 - ...]
+ [2n - C1.2n-1.√3 + C22n-2.(√3)2 - ...]
So we will get an even term
2[2
n - C
2.2
n-2.3 + C
42
n-4.3
2 - ...]
Now it is true that
0 < (2 - √3) < 1
Apply power n both sides we get
0 < (2 - √3)
n < 1
So w e can say that
If (2 - √3)
n = f' then I + f + f'
Will be even .
Again
0 < f < 1 and 0 < f' < 1 we get
f + f' = integer
So it will be
f + f' = 1 (·.· 0 < f + f' < 2) as above,
It means
I is odd and f' = I - f which will give
(I + f)(I - f) = 1 which is odd and it is true .