If-R-7-4-3-2n-I-f-where-I-N-and-0-lt-f-lt-1-then-R-1-f-equals-

Question Number 108453 by gospelkenny last updated on 17/Aug/20

If R = {(7 + 4 \sqrt{3})}^{2 n} = I + f, where I \in N

and 0 < f < 1, then R (1 - f) equals

Answered by 1549442205PVT last updated on 17/Aug/20

Set Q = {(7 - 4 \sqrt{3})}^{2 n} \Rightarrow \sqrt{RQ} = [{(7 + 4 \sqrt{3})}^{n} {(7 - 4 \sqrt{3})}^{n} = 1 (*)

We prove that S = \sqrt{R} + \sqrt{Q} \in N^{*} . Indeed,

i) For n = 1 we get S = (7 + 4 \sqrt{3}) + (7 - 4 \sqrt{3})

= 14 \Rightarrow State is true for n = 1

ii) Suppose State is true \forall n = \overset{―}{1 \dots k}

\Rightarrow S_{k} = {(7 + 4 \sqrt{3})}^{k} + {(7 - 4 \sqrt{3})}^{k} = m_{k} \in N

iii) Consider S_{k + 1} = {(7 + 4 \sqrt{3})}^{k + 1} + {(7 - 4 \sqrt{3})}^{k + 1}

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

- [(7 + 4 \sqrt{3}) \times (7 - 4 \sqrt{3})] [{(7 + 4 \sqrt{3})}^{k - 1} + {(7 - 4 \sqrt{3})}^{k - 1}]

= {14 S}_{k} - S_{k - 1} = {14 m}_{k} - m_{k - 1} \in N^{*}

(by the introduction hypothesis m_{k}, m_{k - 1} \in N^{*})

This shows that the state is also true for n = k + 1

Hence by the imtroduction principle

the state is true for \forall n \in N^{*} which

means (\sqrt{R} + \sqrt{Q}) \in N^{*} \Rightarrow {(\sqrt{R} + \sqrt{Q})}^{2} \in N^{*}

\Leftrightarrow R + Q = {(\sqrt{R} + \sqrt{Q})}^{2} - 2 \in N^{*}

\Rightarrow {(7 + 4 \sqrt{3})}^{2 n} + {(7 - 4 \sqrt{3})}^{2 n} = q \in N^{*} (* *)

On the other hands, by above we have

[{(7 + 4 \sqrt{3})}^{2 n} \times {(7 - 4 \sqrt{3})}^{2 n}] = 1

\Rightarrow 0 < {(7 - 4 \sqrt{3})}^{2 n} = \frac{1}{{(7 + 4 \sqrt{3})}^{2 n}} < 1

Therefore, if R = {(7 + 4 \sqrt{3})}^{2 n} = I + f and

I \in N and 0 < f < 1 then by (* *)

I = q and f = {(7 - 4 \sqrt{3})}^{2 n} . It follows that

R (1 - f) = R - Rf = {(7 + 4 \sqrt{3})}^{2 n} - [{(7 + 4 \sqrt{3})}^{2 n} \times {(7 - 4 \sqrt{3})}^{2 n}]

= {(7 + 4 \sqrt{3})}^{2 n} - 1

Thus, R (1 - f) = {(7 + 4 \sqrt{3})}^{2 n} - 1