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The inequality in question is below:

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

Essentially, I must prove the above for every real number $x$. To begin this proof, I broke it into two cases: $x$ is an integer or $x$ is a real number.

Case 1 ($x$ is an integer):

I break this part up into two components to prove the left and the right side of the inequality.

For the left component,

For $x - 1 < \lfloor x\rfloor \le x$, we know that $\lfloor x\rfloor = x$ since $ x \in \mathbb Z$ therefore $ x - 1 < x \le x$ holds.

For the right component,

For $x \le \lceil x \rceil \le x + 1$, we know that $\lceil x \rceil = x$ since $ x \in \mathbb Z$ therefore $x \le \lceil x \rceil \le x + 1$ holds.

Case 2 ($x$ is a real number and non-integer):

Here is the part where I have trouble proving 'formally'.

For the left component,

For $x - 1 < \lfloor x\rfloor \le x$, we know that if $x = 4.5$ we know that $\lfloor x\rfloor = 4$, which is greater than $ x - 1 = 3.5 $. So, we know that $x - 1 < \lfloor x\rfloor \le x$ holds. This works for all $ x \in \mathbb {R} $.

For the right component,

For $x \le \lceil x \rceil \le x + 1$, we know that if $x = 4.5$ we know that $\lfloor x\rfloor = 5$, which is greater than $ x $. We know that $x + 1 > 5$, so $x \le \lceil x \rceil \le x + 1$ holds. This works for all $ x \in \mathbb {R} $.

Since both the left and the right components of the inequality hold, the whole inequality holds.

I am new to proofs, so I'm not sure if I'm doing this proof by induction or deduction (or something). All I know is that I was able to construct cases in which I know that this inequality works. However, I do not feel that my proof is 'all there'. I feel that Case 1 is sound enough to be considered a proof, but I feel my Case 2 is lacking. Is there any way I can prove this more formally if the above work is unacceptable as a proof?

EDIT:

Adding proof to show that the inequality holds:

The floor is defined as $\lfloor x \rfloor = n \le x < n + 1$.

If $x$ is a real number and $n$ is an integer, then $\lfloor x \rfloor$ is defined as the smallest integer less than or equal to $x$. (Credit to kccu) Since the smallest integer would be equivalent to $x$, we know that $x - 1 < \lfloor x \rfloor$ is less than $x$. Therefore, the left hand side of the inequality $x - 1 < \lfloor x\rfloor \le x$ holds.

Since $\lceil x \rceil$ is defined to be the smallest integer that is greater than or equal to $x$, $n + 1$ is an integer greater than or equal to $x$. Since $n$ is not $/re x$, we know that $n + 1$ is the smallest integer greater than or equal to $x$ (Again, credit to kccu). Since we know that $n + 1$ is the smallest integer greater than or equal to $x$, the right hand side of the equation holds since $\lceil x \rceil$ is equivalent to x since it is the smallest integer greater than or equal to $x$. Thus, the right hand side holds as well.

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  • $\begingroup$ If $x \notin \mathbb{Z}$, then there exists some integer $n \in \mathbb{Z}$ such that $n<x<n+1$. Now write $\lfloor x\rfloor$ and $\lceil x \rceil$ in terms of $n$. $\endgroup$ – kccu Jan 10 '16 at 4:37
  • $\begingroup$ Would saying $x \notin \mathbb {Z}$ be proof by contradiction? $\endgroup$ – TheLongDark Jan 10 '16 at 4:40
  • $\begingroup$ No, I'm just starting at Case 2 because that is the case you said you had a problem with. Case 1 is just fine the way it is. $\endgroup$ – kccu Jan 10 '16 at 4:41
  • $\begingroup$ Ah, I see. For the left component of the inequality involving the floor, $\lfloor x \rfloor = n \le x < n + 1 $. The right component of the inequality is defined in terms of $n$ as $\lceil x \rceil = n < x \le n + 1 $. $\endgroup$ – TheLongDark Jan 10 '16 at 4:47
  • $\begingroup$ Almost... the ceiling should be $\lceil x \rceil = n+1$. $\endgroup$ – kccu Jan 10 '16 at 4:50
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This all depends on your definition of the Floor and Ceiling functions... once you choose a definition you can show the different forms are synonymous, but you must choose a definition. Perhaps the easiest definition I have seen for you purposes comes from Wikipedia

Since there is exactly one integer in a half-open interval of length one, for any real x there are unique integers m and n satisfying $x-1 < m \leq x \leq n < x+1$

From this we can define $\lfloor x \rfloor = m$ and $\lceil x \rceil = n$ and your proof becomes trivial... a self evidence by a substitution

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  • $\begingroup$ So does this mean the proof is satisfied because m is some integer less than or equal to x but greater than x - 1 (for the left side of the inequality, same logic on the right)? Or is there more to it than that? $\endgroup$ – TheLongDark Jan 10 '16 at 15:32
  • $\begingroup$ @TheLongDark this proof works because this is how we define the floor function.... There is always a unique m and n for any given x satisfying the property above.... Think about it?, either your interval of 1 goes from say, 2.5-3.5 and only crosses 3 or it goes from 3-4, but is only either 3 or 4 since once side of the interval is open (the choice of the side you leave open is irrelevant), and we define m as the floor and n as the ceiling. Then, just plug in $m=\lfloor x \rfloor$ and $n = \lceil x \rceil$ $\endgroup$ – Brevan Ellefsen Jan 10 '16 at 15:34
  • $\begingroup$ @TheLongDark remember that a half open set is something like $(0,1]$ where one endpoint is included but the other is not.... A fundamental property of real numbers says only integer lies in any half open interval of length 1. We define the closed side of one interval to be $x$ in our proof and the open side to be x-1, giving an interval of length one.... Thus some integer m must exist.. Do the same for n $\endgroup$ – Brevan Ellefsen Jan 10 '16 at 15:47
  • $\begingroup$ If you define the floor as the greatest integer less than or equal to x and the ceiling the smallest integer greater than or equal than x all you have to do is note that the m and n satisfied by my definition are unique and are the first integers encountered on either side except for maybe x itself.... The equivalance is therefore shown $\endgroup$ – Brevan Ellefsen Jan 10 '16 at 15:53
  • $\begingroup$ you could first express everything in terms of some mathematical logic and show the equivalance, but this isn't really any more formal since we defined the floor and ceiling that way.. All that you depend on is the construction of the real numbers $\endgroup$ – Brevan Ellefsen Jan 10 '16 at 15:55
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We can alter the equation in the following ways:

original:

$$x - 1 < [x] <= x <= (x) < x + 1 $$

multiply by negative one:

$$1-x > -[x] >= -x >= -(x) > -x-1$$

add x to all parts:

$$1 > x - [x] >= 0 >= x - (x) > -1$$

x minus the floor of x is the definition of x modulo 1, or the remainder of x when divided by 1. It's obvious that the remainder is always less than 1 and never goes below 0. X minus the ceiling is equal to -(x % 1). The inner modulo has the same range as the first portion. Taking the negative flips the direction of the inequality, resulting in the correct equation.

By comparing them to modulo I am able to use the rules of the remainder to confirm the inequality.

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