Solving an inequality involving floor.

Question

We have this inequality :

$$\lfloor x+1 \rfloor < \sqrt x +1. $$

I want to find algebraic way to solve this inequality.(not geometric solution)

Note : It is obvious when $x \in \Bbb Z$ we have $0<x<1$ but I can't solve it when $x \notin \Bbb Z$

Answer 1

Note that $\lfloor x+1 \rfloor =\lfloor x \rfloor +1$. Let $n=\lfloor x \rfloor\in\mathbb{N}$ (otherwise the square root is not defined) and $r=x-n\in[0,1)$. Then the inequality $\lfloor x+1 \rfloor < \sqrt x +1$ becomes $$n<\sqrt{n+r}\quad \Leftrightarrow\quad n^2<n+r \Leftrightarrow\quad n(n-1)<r.$$ Now the left-hand side is $0$ for $n=0$ and $n=1$, and it is greater than $1$ for $n\geq 2$.

On the other hand, the right-hand side is $0\leq r<1$.

Hence the inequality is satisfied by $n=0$, $n=1$ and any $r\in(0,1)$, that is for $$x=n+r\in (0,1)\cup (1,2).$$

Answer 2

Hint

Let $k\geq 2$ and solve in $(k,k+1)$.

the equation becomes

$k+1<\sqrt{x}+1$ or $x>k^2>k+1$

$\implies x\notin (k,k+1)$.

Answer 3

To show another possible way , we have that $$ \begin{array}{l} \left\lfloor {x + 1} \right\rfloor < \sqrt x + 1 \\ \left\lfloor x \right\rfloor < \sqrt x \\ x - \left\{ x \right\} < \sqrt x \quad \left| {\;0} \right. \le \left\{ x \right\} < 1 \\ x - 1 < x - \left\{ x \right\} < \sqrt x \\ \end{array} $$ where $ \left\{ x \right\}$ represents the fractional part of $x$.
For the last chain of inequalities to sussist, and since $x$ shall be non-negative, we must have: $$ \left\{ \begin{array}{l} x - 1 < \sqrt x \\ 0 \le x \\ \end{array} \right.\quad \Rightarrow \quad 0 \le x < 1\;\; \vee \;\;\left\{ \begin{array}{l} x^{\,2} - 3x + 1 < 0 \\ 1 < x \\ \end{array} \right. $$ because for $0 \leqslant x<1$ the original inequality is satisfied, for $x=1$ it is not, and for $1<x$ we can square both sides giving the quadratic inequality.
Since the solution of the quadratic equation are $$ \frac{{3 \pm \sqrt 5 }}{2} \approx 0.38,\;2.62 $$ and the lowest is less than $1$, we take only the higher one as the upper bound for $x$. Because of the "squezing" $ x - 1 < x - \left\{ x \right\} < \sqrt x $, that assures that the original $$ x - \left\{ x \right\} < \sqrt x $$ does not have solutions outside of the range $0 < x < \frac{{3 \pm \sqrt 5 }}{2} $, which means $ 0 \leqslant \left\lfloor x \right\rfloor \leqslant 2$. And because $$ \left[ {\begin{array}{*{20}c} {1 < x < 2} & {\left\lfloor x \right\rfloor = 1 < \sqrt x } \\ {2 \le x < \frac{{3 + \sqrt 5 }}{2}} & {\left\lfloor x \right\rfloor = 2 > \sqrt x } \\ \end{array}} \right. $$ we finally get that $$ \left\lfloor {x + 1} \right\rfloor < \sqrt x + 1\quad \Rightarrow \quad 0 < x < 1\;\; \vee \;\;1 < x < 2 $$