Is the greatest integer function periodic? Can we call the greatest integer function as a periodic function with no fundamental period or is it just non-periodic.
Please explain your answer.

To my understanding, if we consider $f(x) = [x]$ now, $f(3) = [3] = 3$ and $f(3+0.5) = [3.5] = 3$ So, can't we say that it is periodic (A constant function is periodic with no fundamental period) ? But the problem is to derive the fundamental period.
EDIT: After checking out some aswer I am quite inquisitive to know is it really necessary to have a fundamental period to call a function periodic? However,If you go by my book it is not.
 A: Though the floor function is not periodic, it is quasiperiodic; that is, it is a function that satisfies the functional equation
$$f(x+\alpha)=f(x)+ux+v$$
For the specific case of the floor function, you have
$$\lfloor x+1\rfloor=\lfloor x\rfloor+1$$
so 1 can be thought of as the floor function's quasiperiod.
It's a rather trivial example of a quasiperiodic function, though.
A: A periodic functions repeats it's output values after regular intervals. That is there exists $T$ such that $f(x+T)=f(x)$ for all $x$. The greatest integer function does not satisfy this equation for any $T$. So it is not periodic.
http://en.wikipedia.org/wiki/Periodic_function
The fundamental period is just the smallest (positive) such $T$ (if one exists). 
The sawtooth function which measures the deviation from the greatest integer function (also called floor function) is periodic with period 1.
http://en.wikipedia.org/wiki/Sawtooth_function
A: The answers so far do not catch the fact that the floor function has indeed some built in periodicity: The function $f(x):=\lfloor x\rfloor - x$ is in fact periodic and so can be developped into a Fourier series. "For all practical purposes" one has
$$\lfloor x\rfloor =x-{1\over2} +{1\over\pi}\sum_{k=1}^\infty{\sin(2\pi k t)\over k},$$
but here the right hand side for $t\in\mathbb Z$ does not produce the correct value.
A: A function $f(x)$ is "periodic with period $c$", with $c\gt 0$, if $f(x+c) = f(x)$ for all values of $x$. In particular, $f(x+mc) = f(x)$ for all integers $m$, by using induction.
A function is periodic if it is periodic with period $c$ for some $c\neq 0$.
Every periodic function has lots of periods: if $c$ is a period, then so is $mc$ for any positive integer $m$, at the very least; it may have others. So we may be interested in knowing what is the "quickest" that the function starts repeating. For instance, $\tan x$ certainly repeats every $2\pi$, since $\tan(x+2\pi) = \tan(x)$ for all $x$ (either both are defined and equal, or both are undefined); and by the comment above, it will repeat every $2m\pi$ for any integer $m$. But it actually repeats more often than every $2\pi$: we know that $\tan(x+\pi) = \tan(x)$ for all $x$. So altough it is true that $\tan(x)$ is periodic with period $2\pi$, it is also true that $\tan(x)$ is periodic with period $\pi$.
So we would like to have something more than just a "period"; we would like to be able to talk about the smallest possible period. That's what the "fundamental period" is trying to capture:
We say that a periodic function $f(x)$ has "fundamental period" $k$ if and only if $k$ is the smallest positive number such that $f(x)$ is periodic with period $k$. That means that we require:


*

*$f$ to be periodic with period $k$: so $f(x) = f(x+k)$ for all $x$;

*$k$ to be positive; and

*if $c$ is a positive constant such that $f(x+c) = f(x)$ for all $x$, then $k\leq c$.


So, in general, we would take the set of all positive periods of the function $f$, and look for its minimum (smallest element). The one possible problem is that not every set of positive real numbers has a smallest element, so it may be possible, at least in principle, that a function is periodic but has no fundamental period, because it does have periods, but it doesn't have one that is the smallest period.
The constant functions, $f(x) = a$ for all $x$, are an example of this pathology: they are periodic, because they satisfy the defintion of being periodic. In order to show that they satisfy the definition, we need to show that there is at least one number $c\gt 0$ such that $f(x)=f(x+c)$ for all $x$. Well, we can take $c=1$, because $f(x) = a = f(x+1)$ for all $x$. So the constant function is certainly periodic. But in fact, for every positive number $c$, $f(x+c) = a =f(x)$. That means that if you take any $c\gt 0$, then you can rightly say that "$f(x)$ is periodic with period $c$". So if we look at the set of all periods,
$$P=\{ c \in \mathbb{R}\mid c\gt 0\text{ and $f(x)$ is periodic of period $c$}\}$$
then $P=(0,\infty)$. Since the "fundamental period" is supposed to be the smallest element of $P$, but $P$ does not have a smallest element, then the constant function does not have a fundamental period. But it is periodic (we just saw it satisfies the definition).  
In the case of the greatest integer function, $f$ is not periodic at all. Again, "$f$ is periodic" means "there exists $c\gt 0$ such that $f(x+c)=f(x)$ for all $x$". So the negation of "$f$ is periodic" is "for every $c\gt 0$, there exists at least one $x$ such that $f(x)\neq f(x+c)$."
In order to show that the greatest integer function $f(x)=\lfloor x\rfloor$ is not periodic, we need to show that given any $c\gt 0$ (think of it as a candidate for a period), there is at least one point $x$ for which $f(x)\neq f(x+c)$.
So, let $c\gt 0$ be given. I'm going to come up with some point $x$ (which will depend on $c$) with the property that $\lfloor x\rfloor\neq \lfloor x+c\rfloor$. The idea is simply to take a value of $x$ that is before some integer, but once you add $c$ to it you go above that integer. Then the greatest integer function will give two different values. 
The simplest example to take is $x=-\frac{c}{2}$; this number is negative, so $f(-\frac{c}{2}) = \lfloor -\frac{c}{2}\rfloor\lt 0$; on the other hand, $x+c = -\frac{c}{2}+c = \frac{c}{2}$ is positive, so $f(x+c) = \lfloor \frac{c}{2}\rfloor \geq 0$. Since $f(x)$ is strictly less than $0$ and $f(x+c)$ is at least $0$, $f(x)\neq f(x+c)$. So $f(x)$ is not periodic with period $c$.
But $c$ was an arbitrary positive number. That means that we cannot say "$f(x)$ is periodic with period $c$" for any positive number, and that means, by definition, that $f(x) = \lfloor x\rfloor$ is not periodic.
A: I am inclined to say that the greatest integer function (floor function) is not periodic.
Mathworld [1] tells us that, 
A function f(x) is said to be periodic (or, when emphasizing the presence of a single period instead of multiple periods, singly periodic) with period p if $f(x)=f(x+np)$.
With the floor function, we can see that $\lfloor x+n \rfloor = \lfloor x \rfloor + n$ for $ x \not \in \mathbb{Z}$ and $n \in \mathbb{Z}$. This would give us reason to believe that we cannot find a period $p$ such that $f(x)=f(x+np)$ is satisfied for the greatest integer function.
[1] Weisstein, Eric W. "Periodic Function." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/PeriodicFunction.html
EDIT!
I think I know what you are saying about the constant function now though. I suppose that if you restrict the domain of the function, then you could say that the greatest integer function is periodic. Suppose that we just want to look at the greatest integer function on the interval [0,1). Then we can say that the definition of periodicity is satisfied. However on the entire domain of the greatest integer function, we do not have this property.
A: Even if you choose your period $T$ arbitrarily small, you can't satisfy $f(x+T) = f(x)$ for the arguments that are closer to any integer than T, so yo can't say it's a function with no fundamental period, it is non-periodic.
