Prove or disprove $\lceil$$\lfloor$ x $\rfloor$$\rceil$ = $\lfloor$ x $\rfloor$ for all real numbers x.
How do I do this? I know that it is true (I think) because I tried to counterexample it and it didn't work. Any advice would be helpful, such as what type of proof to use etc. Note that x IS NOT an integer, but a real number, which I think makes it a little more difficult.
EXAMPLE PROOF:
Let x be a real number. Show that $\lfloor$ 3x $\rfloor$ = $\lfloor$ x $\rfloor$ + $\lfloor$ x + (1/3) $\rfloor$ + $\lfloor$ x + (2/3) $\rfloor$
There are three cases we should consider.
Case 1): n $\leq$ x $<$ n + (1/3) for some integer n
n + 1/3 $\leq$ x + 1/3 $<$ n + 2/3
n + 2/3 $\leq$ x + 2/3 $<$ n + 1
3n $\leq$ 3x $<$ 3n + 1
So, $\lfloor$ x $\rfloor$ + $\lfloor$ x + (1/3) $\rfloor$ + $\lfloor$ x + (2/3) $\rfloor$ = n + n + n = 3n = $\lfloor$ 3x $\rfloor$
Case 2): n + 1/3 $\leq$ x $<$ n + 2/3 for some integer n
n + 2/3 $\leq$ x + 1/3 $<$ n +1
n + 1 $\leq$ x + 2/3 $<$ n + 1 + 1/3
3n + 1 $\leq$ 3x $<$ 3n + 2
So, $\lfloor$ x $\rfloor$ + $\lfloor$ x + 1/3 $\rfloor$ + $\lfloor$ x + 2/3 $\rfloor$ = n + n + (n + 1) = 3n + 1 = $\lfloor$ 3x $\rfloor$
Case 3): n + 2/3 $\leq$ x $<$ n + 1 for some integer n
n + 1 $\leq$ x + 1/3 $<$ n + 1 + 1/3
n + 1 + 1/3 $\leq$ x + 2/3 $<$ n + 1 + 2/3
3n + 2 $\leq$ 3x $<$ 3n + 3
So, $\lfloor$ x $\rfloor$ + $\lfloor$ x + (1/3) $\rfloor$ + $\lfloor$ x + 2/3 $\rfloor$ = n + (n + 1) + (n + 1) = 3n + 2 = $\lfloor$ 3x $\rfloor$
To conclude, it is shown that for all real numbers x, $\lfloor$ 3x $\rfloor$ = $\lfloor$ x $\rfloor$ + $\lfloor$ x + (1/3) $\rfloor$ + $\lfloor$ x + (2/3) $\rfloor$