Question

If \[{\left( {7{\text{ }} + {\text{ }}4\sqrt 3 } \right)^n}{\text{ }} = {\text{ }}I{\text{ }} + {\text{ }}F\] where \[I\] and $n$ are positive integers and \[O{\text{ }} < {\text{ }}F{\text{ }} < {\text{ }}1\] then show that,
i) \[I\] is an odd integer.
ii) \[\left( {1{\text{ }} + {\text{ }}F} \right)\left( {1{\text{ }}-{\text{ }}F} \right){\text{ }} = {\text{ }}1\].

Answer 1

Hint: The given question is based on integers. To solve that try to prove the result. Binomials are also used. To solve this we consider another function having a negative sign. Then we add up both functions and apply a binomial on that is...
${[^n}{c_o}{x^n}{y^o}{ + ^n}{c_2}{x^{n - 2}}{(y)^2} + - - - - - - - - - - ]$

Complete step by step answer:
Integers: Integers are all positive and negative numbers on a number line.
Positive integers are all the positive numbers on a number line like \[1,2,3,...\] so on.
Negative integers are the negative numbers on a number line like $ - 1,{\text{ }} - 2,{\text{ }} - 3,....$ so on.

i) Consider the given function
Let \[{\left( {7{\text{ }} + {\text{ }}4\sqrt 3 } \right)^n}{\text{ }} = {\text{ }}I{\text{ }} + {\text{ }}F - - - - - - - - - (a)\]
Where \[{\text{ }}I\] and \[F\] are the integral and fractional parts respectively.
It means $0f < 1$ given that.
Now we can say that $7 - 4\;\sqrt 3 $ it also lies between $0$ and $1$. As $7 - 4\;\sqrt 3 $ another function which was considered by us.
So,
$0{\text{ }} < $$7 - 4\;\sqrt 3 $ $ < {\text{ }}1 - - - - - - - - - - - - (b)$
We apply the power $n$ on the above ($b$) equation it becomes
${0^n} < {(7 - 4\;\sqrt 3 )^n} < {1^n}$
As ${0^n}$ it becomes $0$ and ${1^n}$ it becomes $1$ so the equation is like...
$0 < {(7 - 4\;\sqrt 3 )^n} < 1 - - - - - - - - - - - (c)$
Let ${(7 - 4\;\sqrt 3 )^n}$ be ${f^1}$ so we can write is as
 ${(7 - 4\;\sqrt 3 )^n} = {f^1} - - - - - - - - - - (d)$
$0{\text{ }} < {f^1} < {\text{ }}1$ as ${f^1}$ is also lies between $0{\text{ }}$ and $1$.
Now adding the terms $(a)$ and $(d)$
We get
$1 + F + {f^1} = {(7 - 4\;\sqrt 3 )^n} + {(7 - 4\;\sqrt 3 )^n}$
Solving the right-hand side by using identity ${[^n}{c_o}{x^n}{y^o}{ + ^n}{c_2}{x^{n - 2}}{(y)^2} + - - - ]$
So it becomes…
$
\Rightarrow 2{[^n}{c_0}{7^n}{ + ^n}{c_2}{7^{n - 2}}{(4\sqrt 3 )^2} + - - - ] \\
\Rightarrow 1 + F + {f^1} \\
\Rightarrow 02 \\
 \;0 < {\text{ F + }}{f^1}{\text{ < }}2 \\
  I + 1 = 2{[^n}{c_0}{7^n}{ + ^n}{c_2}{7^{n - 2}}{(4\sqrt 3 )^2} + - - - ]$

ii) As we can say $1 + F + {f^1}$ are even integers so, it can lie between $0$ and $2$ as $2$ is an even number
$\;0 < {\text{ F + }}{f^1}{\text{ < }}2$
\[F{\text{ }} + {\text{ }}{f^1}{\text{ }} = {\text{ }}1\]
As $I + 1 = $ Even integer
I is an odd integer as the sum of two odd numbers is an even number.

Note: The above question is solved by using integer theory. Integers are all the numbers that lie on a number in integers, positive and negative numbers also included. The number lies in a particular interval 0 and 1, we are trying to solve it by adding and equating the functions. In that way, we concluded our result.