Nature of Integral Part of Expression N = (a + sqrt b)^n

Solved Examples

Example 1: If $P=(7+4 \sqrt{3})^9$ and $P=\mathrm{I}+\mathrm{f}$, where I is the integer just less than P, then the value of $P(1-f)$ is equal to

1) $1 $

2) $2^n$

3) $2 $

4) $2^{2 n}$

Solution:

Now,

$P=(7+4 \sqrt{3})^9 $

Let $P^{\prime}=(7-4 \sqrt{3})^9 \quad \& \quad 0<7-4 \sqrt{3}<1$

Using Binomial addition for $P+P'$

$ P+P^{\prime}=2 \cdot\left[{ }^9 C_0 7^9+{ }^9 C_2 7^7 \cdot(4 \sqrt{3})^2+{ }^9 C_4 7^5 \cdot(4 \sqrt{3})^4 \cdots \cdots\right] $

and it is given that $P+I+f$, where $0<f<1$

Hence,

$ I+f+P^{\prime}=2 \cdot\left[{ }^9 C_0 7^9+{ }^9 C_2 7^7 \cdot(4 \sqrt{3})^2+{ }^9 C_4 7^5 \cdot(4 \sqrt{3})^4 \ldots \ldots\right] $

So, $I+f+P^{\prime}=$ even integer

Now it is clear that $I$ is an integer so $f+P^{\prime}$ should be an integer

$f$ lies between $(0,1)$ and $P^{\prime}$ also lies between $(0,1)$

hence, $f+P^{\prime} \in(0,2)$

But their sum is an integer and 1 is the only integer that lies between this interval

hence, $f+P^{\prime}=1$

$ P^{\prime}=1-f $

we need to calculate $P(1-f)$

$ \text { or } P(1-f)=P P^{\prime}=\left[(7+4 \sqrt{3})^9(7-4 \sqrt{3})^9\right] $

$P(1-f) =\left[(7+4 \sqrt{3})^9(7-4 \sqrt{3})^9\right] $

$ =[(7+4 \sqrt{3})(7-4 \sqrt{3})]^9 $

$ =[49-48]^9 $

$ =1^9=1 $

Hence, the answer is option 1.

Example 2: If $P=(10+3 \sqrt{11})^n$ and $P^{\prime}=(10-3 \sqrt{11})^n$, then which one of the following is not true?

1) $P P^{\prime}=1$

2) $P+P^{\prime}$ is an even integer

3) $P^{\prime}=1-f$ where, $f$ is the fractional part of $P$

4) $f+P^{\prime} \in(0,1)$

Solution:

Now

i) $P P^{\prime}=(10+3 \sqrt{11})^n(10-3 \sqrt{11})^n=\left(10^2-(3 \sqrt{11})^2\right)^n=1^n=1$

ii) $P+P^{\prime}=2\left[{ }^n C_0(10)^n+{ }^n C_2(10)^{n-2}(3 \sqrt{3})^2+{ }^n C_2(10)^{n-4}(3 \sqrt{3})^4 \cdots \cdots\right]$

hence, it is an even integer

iii) Let $P=[P]+f$, where $[\mathrm{P}]$ is the integer just less than P

Now, from (ii)

$ P+P^{\prime}=\text { even integer } $

$ P+P^{\prime}=[P]+f+P^{\prime}=\text { Integer }$

Example 3: If $x=(7+4 \sqrt{3})^{2 n}=[x]+f$, then $x(1-f)$ is equal to $\qquad$ .

1) $1 $

2) $2 $

3) $3 $

4) $0 $

Solution:

We have ${ }^{7-4 \sqrt{3}}=\frac{1}{7+4 \sqrt{3}}$

$\therefore 0<7-4 \sqrt{3}<1 \Rightarrow 0<(7-4 \sqrt{3} 3)^{2 n}<1 $

Let $F=(7-4 \sqrt{3})^{2 n}$, then

$ x+F=(7+4 \sqrt{3})^{2 n}+(7-4 \sqrt{3})^{2 n} $

$ =2\left[{ }^{2 n} C_0 7^{2 n}+{ }^{2 n} C_2 7^{2 n-2}(4 \sqrt{3})^2+{ }^{2 n} C_4 7^{2 n-4}(4 \sqrt{3})^4+\ldots+{ }^{2 n} C_{2 n}(4 \sqrt{3})^{2 n}\right] $

$=2 m$, where $m$ is some positive integer

$\Rightarrow[x]+f+\mathrm{F}=2 m $

$ \Rightarrow f+F=2 m-[x] $

Since, $0 \leq f<1$ and $0<F<1$, we get $0<f+\mathrm{F}<2$. Also, since $f+F$ is an integer, we must have $f+\mathrm{F}=1$

Thus, $x(1-f)=x F=(7+4 \sqrt{3})^{2 n}(7-4 \sqrt{ } 3)^{2 n}=(49-48)^{2 n}=1^{2 n}=1$

Hence, the answer is option (1).

Example 5: Let $R=(5 \sqrt{5}+11)^{2 n+1}$ and $f=R-[R] {\text { where [.] denotes the greatest }}$ integer function. The value of $R . f$ is

1) $4^{2 n+1}$

2) $4^{2 n}$

3) $4^{2 n-1}$

4) $4^{-2 n}$

Solution:

Since $f=R-[R] \quad R=f+[R][5 \sqrt{5}+11]^{2 n+1}=f+[R] {\text { where }[R] \text { is integer }}$

Now let $f^{\prime}=[5 \sqrt{5}-11]^{2 n+1}, 0<f^{\prime}<1$

$ f+[R]-f^{\prime}=[5 \sqrt{5}+11]^{2 n+1}-[5 \sqrt{5}-11]^{2 n+1}= $

$2\left[{ }^{2 n+1} C_1(5 \sqrt{5})^{2 n}(11)^1+{ }^{2 n+1} C_3(5 \sqrt{5})^{2 n-2}(11)^3+\ldots \ldots .\right] $

$ =2 \cdot(\text { Integer })=2 K(K \in N)=\text { Even integer } $

Hence, $f-f^{\prime}=$ even integer $-[R]$, but $-1<f-f^{\prime}<1$.
Therefore, $f-f^{\prime}=0 \quad \therefore f=f^{\prime}$

Hence R. $f=R \cdot f^{\prime}=(5 \sqrt{5}+11)^{2 n+1}(5 \sqrt{5}-11)^{2 n+1}=4^{2 n+1}$.

Hence, the answer is option (1).