Trouble finding the floor function of a given expression.

Question

I am currently learning a bit about number theory - currently studying continuous fractions and came up with the following task:

Task. Show that the floor function of $$ \frac{n + \sqrt{n^2+4}}{2},$$ where $n$ is a positive integer, i.e., $n \in \Bbb N$ is $n$.

Obviously, the objective of this is to say that $ \frac{n + \sqrt{n^2+4}}{2} = [n; (...)]$, but I can't make this first step. How would one prove what's wanted?

I have tried doing it by induction and also tried squaring the expression, but couldn't find something that helped me.

Thanks for any help in advance.

Answer 1

The major observation is that the addition of $4$ does not increase the value of $n^2$ by very much—after taking square roots and dividing by $2$, this $+4$ contributes little to the term. In more detail:

Observe that $$ n^2 + 4 = (n+2)^2 - 4n, $$ where $4n > 0$ (since $n > 0$). Then, as all of the relevant terms are positive, \begin{align} n^2 < n^2 + 4 < (n+2)^2 &\iff n < \sqrt{n^2 + 4} < n+2 && \text{(take square roots)}\\ &\iff 2n < n + \sqrt{n^2+2} < 2n + 2 && \text{(add $n$)} \\ &\iff n < \frac{n+\sqrt{n^2+4}}{2} < n+1. && \text{(multiply by $\tfrac{1}{2}$)} \end{align} This is precisely what was desired.

Answer 2

First we prove the following two properties of the square-root function:

(1) For any positive real numbers $a$ and $b$, if $a < b$, then $\sqrt{ a} < \sqrt{b}$.

Proof:

We note that, if $b > a > 0$, then $b-a > 0$ and $\sqrt{b} + \sqrt{a} > 0$, which implies that $$ \sqrt{b} - \sqrt{a} = \frac{ \left( \sqrt{b} \right)^2 - \left( \sqrt{a} \right)^2 }{ \sqrt{b} + \sqrt{a} } = \frac{ b-a }{ \sqrt{b} + \sqrt{a} } > 0, $$ and hence $$ \sqrt{b} > \sqrt{a}, $$ as required.

Alternatively, we note that derivative of the function $f(x) = \sqrt{x}$ for $x > 0$ is given by $$ f^\prime (x) = \frac{ 1 }{ 2 \sqrt{x} } > 0, $$ which implies that the square root function is strictly increasing.

Our second property is as follows:

(2) If $a > 0$ and $b > 0$, then we have $\sqrt{ a+b} < \sqrt{a} + \sqrt{b}$.

Proof:

As $a > 0$ and $b > 0$, so we have $a + b > 0$, and thus $$ \sqrt{ a} + \sqrt{ b} + \sqrt{a + b} > 0 $$ also. Therefore we have $$ \left( \sqrt{ a } + \sqrt{ b } \right) - \sqrt{ a+b } = \frac{ \left( \sqrt{ a } + \sqrt{ b } \right)^2 - \left( \sqrt{ a+b } \right)^2 }{ \sqrt{ a} + \sqrt{ b} + \sqrt{a + b} } = \frac{ a + b + 2 \sqrt{a}\sqrt{b} - a - b}{ \sqrt{ a} + \sqrt{ b} + \sqrt{a + b} } = \frac{2 \sqrt{a}\sqrt{b} }{ \sqrt{a} + \sqrt{b} + \sqrt{ a + b} } > 0, $$ which implies that $$ \sqrt{ a } + \sqrt{ b } > \sqrt{ a+b}. $$

Now we show that, for each $n = 1, 2, 3, \ldots$, we have $$ n < \frac{ n + \sqrt{ n^2 + 4 } }{2} < n+1. $$ We note that \begin{align} n &= \frac{n + n}{2} \\ &= \frac{n + \sqrt{n^2} }{2} \\ &< \frac{ n + \sqrt{ n^2 + 4 } }{2} \qquad [\mbox{ because of our property (1) }] \\ &< \frac{ n + \sqrt{ n^2 } + \sqrt{ 4 } }{2} \qquad [\mbox{ because of our property (2) }] \\ &= \frac{ n + n + 2 }{2} \\ &= n+1. \end{align} Therefore we must have $$ \left\lfloor \frac{ n + \sqrt{ n^2 + 4 } }{2} \right\rfloor = n $$ for each $n = 1, 2, 3, \ldots$.

Answer 3

Easy way: $\ \color{#c00}{x>1},\ \overbrace{x^2 = n\:\!x +1}^{\text{by quad. formula}}\,\underset{\!\!\div\,x}\Longrightarrow\, \overbrace{x = n + \frac{1}x}^{\large \color{#c00}{n\, <\, x \,<\, n+1\!\!\!\!\!\!}}\,\Rightarrow\, \color{#c00}{\lfloor x\rfloor = n}$