Question

Show that if x is a real number, then$x-1<\lfloor x\rfloor \le x\le \lfloor x\rfloor <x+1$

Short answer

The function is$x-1<\lfloor x\rfloor \le x\le \lfloor x\rfloor <x+1$

Step-by-step solution

Step 1:

Ceiling function [x] : smallest integer that is greater than or equal to x.

Floor function [x] : largest integer that is less than or equal to x.

Step 2:Given:

X is real number .

To prove:$x-1<\lfloor x\rfloor \le x\le \lfloor x\rfloor <x+1$

PROOF

Let x be a real number. Any real number x is lies between two consecutive integers(inclusive, as the real number can also be an integer)

$\mathrm{\exists }n\in Z:n\le x\le n+1$

FIRST CASE

The ceiling function of x is then the larger integer, while the floor function of x is the smaller integer.

$\begin{array}{r}\lfloor x\rfloor =n+1\\ \lfloor x\rfloor =n\end{array}$

Since $n<x<n+1\text{and}\lfloor x\rfloor =n+1\text{and}\lfloor x\rfloor =n$.

$\lfloor x\rfloor <x<\lfloor x\rfloor$

If x is an integer, then$x=\lfloor x\rfloor =\lfloor x\rfloor$

$\lfloor x\rfloor <x<\lfloor x\rfloor \mid$

Step 3:Part (b)

SECOND CASE

We had found the inequality$n<x<n+1$ .let us subtract 1 from each side of this in equality:

$n-1<x-1<n+1$

simplification

$x-1<n$

Since$\lfloor x\rfloor =n$ ,

$x-1<\lfloor x\rfloor$

Step 4:

THIRD CASE

We had found the inequality .let us add 1 from each side of this in equality:

$n+1<x+1<n+2$

simplification

$n+1<x+1$

Since$\lfloor x\rfloor =n+1,$ ,

$\lfloor x\mid <x+1$

Conclusion since $\lfloor x\rfloor <x<\lfloor x\rfloor \text{and}x-1<\lfloor x\rfloor \text{and}\lfloor x\rfloor <x+1$

Hence, the solution is

$x-1<\lfloor x\rfloor \le x\le \lfloor x\rfloor <x+1$