Ok so the operation [x] is defined to be equal to the integer such that it is $\leq x$
From this definition it holds that : $$ [x] \leq x $$
I need to prove that $$ [x+n] = [x] + n $$
My proof goes like this:
Since $$ [x+n] \leq x+n $$
It follows that $ [x] + n \leq x + n $ from which we can conclude that $$ [x] \leq x $$ which is true and thus property is true.
I have a feeling that I am missing out on something huge,and that I am making a terrible mistake.Please guide me.
Thanks in advance
Your method of proof not valid - you've tried to deduce that $p$ is true from "$p \Rightarrow q$" holds and $q$ is true. This doesn't work in general: for example, $$(-1)^2 = -1 \\ \Rightarrow \lvert (-1)^2 \rvert = 1$$and it is true that $\lvert (-1)^2 \rvert = 1$, but $(-1)^2 \not = -1$.
To get a (or at least an easy) valid proof, you need to either start with something you know to be true and deduce what you want to prove, or start by assuming something is true and deduce something you know to be false - in which case you know that what you assumed is in fact false (proof by contradiction).
For this;
The definition gives that $[x + n] \leq x + n$. Then $[x + n] - n \leq x$.
Since the LHS is an integer (I'm assuming $n$ is an integer here), it must be the case that $[x + n] - n \leq [x]$. Hence $[x + n] \leq [x] + n$.
Also, we know that $[x] \leq x$, and hence $[x] + n \leq x + n$. Then since the LHS is an integer, have that $[x] + n \leq [x + n]$.
Hence we must have $[x + n] = [x] + n$.
What is $n$? If $n$ is an integer, then $[x+n]$ is the integer that is closest to $x+n$. Since $n$ itself is an integer, this is equal to $n + [x]$.
For example if $x=2.5$ and $n=6$, then $[x+n]=[2.5+6]=6+[2.5]=6+2=8$.
This is easy using the universal property of the floor function, i.e. if $\ x\in \Bbb R\ $ then $$\rm k\,\le\, \lfloor x \rfloor \color{#c00}\,\color{#c00}\iff\, k\,\le\, x,\ \ \ for\ \ all\ \ k\in \mathbb Z\qquad\quad\ $$
$${{\rm\begin{eqnarray} \text{so for }\rm\:n\in \mathbb Z,\,\ x\in \mathbb R, &\rm \quad k &\le&\ \rm\ \color{#0a0}{\lfloor x+n\rfloor} \\ \color{#c00}\iff& \quad\rm k &\le&\ \ \ \ \rm x+n \\ \iff&\quad \rm k-n\,\ &\le&\ \ \rm\ \ x \\ \color{#c00}\iff&\quad \rm k-n\,\ &\le&\ \rm\ \lfloor x\rfloor \\ \color{}\iff&\quad \rm k &\le&\ \rm\ \lfloor x\rfloor+n \\ \color{#c00}\iff&\quad \rm k &\le&\ \rm\ \color{#90f}{\lfloor \lfloor x\rfloor+n\rfloor} \\ \end{eqnarray}}}\qquad\qquad\qquad\ $$
$$\ \ \ \text{thus } \ \ \ \color{#0a0}{\rm \lfloor x+n\rfloor} = \rm\ \color{#90f}{\lfloor\lfloor x\rfloor+n\rfloor} $$
Remark $\ $ See this answer for another example and further discussion. And here is another example from this deleted question: Prove that if $m$ and $n$ are positive integers, and $x$ is a real number, then: $$\left\lfloor\frac{\lfloor x\rfloor+n}m\right\rfloor = \left\lfloor\frac{x+n}m\right\rfloor$$
This is straightforward using the universal property of the floor function, viz. $$\rm k\le \lfloor r \rfloor \color{#c00}\iff k\le r,\ \ \ for\ \ \ k\in \mathbb Z,\ r\in \mathbb R$$ Thus for $\rm\ k,m,n\in \mathbb Z,\,\ m>0,\,\ x\in \mathbb R,\ $
$$\rm\,\begin{array}{rrrl} &\rm k &\le&\!\rm\ \color{#0a0}{\lfloor (\lfloor x\rfloor + n)/m\rfloor} \\ \color{#c00}\iff& \rm k &\le&\ \ \rm (\lfloor x\rfloor + n)/m \\ \iff& \rm mk &\le&\ \ \ \rm \lfloor x\rfloor + n \\ \iff& \rm mk-n &\le&\ \ \rm\ \lfloor x\rfloor \\ \color{#c00}\iff& \rm mk-n &\le&\ \ \rm\ \ \, x \\ \iff& \rm mk &\le&\quad\,\rm x + n \\ \iff& \rm k &\le&\ \ \ \rm (x + n)/m \\ \color{#c00}\iff& \rm k &\le&\ \ \rm\color{blue}{\lfloor (x + n)/m\rfloor} \\ \end{array}$$
So, by the above, we conclude that $\, \rm \color{#0a0}{\lfloor (\lfloor x\rfloor + n)/m\rfloor}\ =\rm\ \ \color{blue}{\lfloor (x + n)/m\rfloor} $
