Why adding the two proper fractions give 1?

Question

I came across this example question while studying Binomial Theorem. Here it is :–

If $n$ is any positive integer, show that the integral part of $(3+√7)^n$ is an odd number.

Suppose $I$ to denote the integral and $f$ the fractional part of $(3+√7)^n$.

Then,

$I+f=3^n+C_13^{n-1}√7+C_23^{n-2}7+C_33^{n-3}(√7)^3+........ —(1)$

Now $3-√7$ is positive and less than $1$, therefore $(3-√7)^n$ is a proper fraction; denote it by $f'$;

$f'=3^n-C_13^{n-1}√7+C_23^{n-2}7-C_33^{n-3}(√7)^3+........ —(2)$

Add together (1) and (2); the irrational terms disappear, and we have

$I+f+f'=2(3^n+C_23^{n-2}7+....)=$ an even integer.

But since $f$ and $f'$ are proper fractions their sum must be 1;

I= an odd integer

The only thing I am not able to understand here is-

Why adding any 2 proper fraction gives 1?

Can anyone provide a convincing proof of this?

Answer 1

You have $$ I+f+f'= 2k\ ,\ \text{or}\\ f+f'=2k-I $$ where $\ k\ $ and $\ I\ $ are integers, and $\ f, f'\ $ are proper fractions—that is $\ 0<f,f'< 1\ $. Thus, $\ 0<f+f'<2\ $. Therefore, since $\ f+f'\ $ is an integer, and $1$ is the only integer in the interval $\ (0,2)\ $, it follows that $\ f+f'=1\ $.