Question

Show that if x is a real number and n is an integer, then

$\text{a)}x<n\text{if and only if}\lceil x\rceil <n$

$\text{b)}n<x\text{if and only if}\mathrm{n}<\lceil x\rceil$

Short answer

The function is

$\text{a)}x<n\text{if and only if}\lceil x\rceil <n$

$\text{b)}n<x\text{if and only if}\mathrm{n}<\lceil x\rceil$

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: (a)

X is real number .

To prove:$\text{a)}x<n\text{if and only if}\left|x\right|<n$

PROOF

Let x be a integer. If x is an integer, then$\lfloor x\rfloor =x<n$

Thus let us assume that x is not an integer.

Any real number x is less between two consecutive integers.

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

The floor function of x is then the smaller integer of the inequality.

$\lfloor x\rfloor =m$

Thus we have derived:

$\lfloor x\rfloor =m<x<n$

Thus if$x<n\text{, then}\left[x\right]<n\mid$

Step 3:

SECOND PART

PROOF

Let x be a integer. If x is an integer, then$\lfloor x\rfloor =x<n$

Thus let us assume that x is not an integer.

Any real number x is less between two consecutive integers.

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

The floor function of x is then the smaller integer of the inequality.

Since$\lfloor x\rfloor <n$

$\lfloor x\rfloor =mm<n$

Since different integers differ by at least 1.

$m+1\le n$

Using$m<x<m+1$

$x<m+1\le n$

Thus if$\lfloor x\rfloor <n\text{, then}x<n$

Step 4:Given: (b)

X is real number .

To prove:$\text{b)}n<x\text{if and only if}\mathrm{n}<\lceil x\rceil$

PROOF

Let x be a integer. If x is an integer, then$\lfloor x\rfloor =x<n$

Thus let us assume that x is not an integer.

Any real number x is less between two consecutive integers.

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

The ceiling function of x is then the larger integer of the inequality.

$\lfloor x\rfloor =m+1$

Thus we have derived:

$n<x<m+1=\lfloor x\rfloor$

Thus if$n<x\text{, then}n<\lfloor x\rfloor$

Step 5:

SECOND PART

PROOF

Let x be a integer. If x is an integer, then$\lfloor x\rfloor =x<n$

Thus let us assume that x is not an integer.

Any real number x is less between two consecutive integers.

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

The ceiling function of x is then the largeer integer of the inequality.

$\lceil x\rceil =m+1$

Since$n<\lceil x\rceil$

$n<m+1$

Since different integers differ by at least 1.

$n\le m$

Thus if$n<\lfloor x\rfloor \text{, then}n<x$

Hence, the solution is

$\text{a)}x<n\text{if and only if}\lceil x\rceil <n$

$\text{b)}n<x\text{if and only if}\mathrm{n}<\lceil x\rceil$