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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) $\rfloor$

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  • $\begingroup$ What happens when you take the floor (or the ceiling) of an integer? $\endgroup$ – Dave L. Renfro Nov 1 '16 at 21:02
  • $\begingroup$ Since $\left \lfloor x \right \rfloor$ is an integer, the ceiling function does nothing to it. $\endgroup$ – Kaynex Nov 1 '16 at 21:03
  • $\begingroup$ No, this is exactly what I'm trying to avoid! x is NOT an integer. $\endgroup$ – knowledge_is_power Nov 1 '16 at 21:06
  • $\begingroup$ @Gabby: Nobody said anything about $x$ being an integer. $\endgroup$ – user14972 Nov 1 '16 at 21:07
  • $\begingroup$ I agree, the floor of x should be an integer, and ceiling function does nothing, but how do I PROVE that this equals just the floor of a REAL NUMBER x? Sure, it's going to be an integer too, but that is not enough to prove it. $\endgroup$ – knowledge_is_power Nov 1 '16 at 21:07
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This is trivial.

1) for any $x \in \mathbb R$ there is a unique integer $\lfloor x \rfloor$ so that $\lfloor x \rfloor \le x < \lfloor x \rfloor + 1$.

2) if $n \in \mathbb Z$ then $\lfloor n \rfloor = n$. This should be clear as $n \le n < n+1$

3) for any $x \in \mathbb R$ there is a unique integer $\lceil x \rceil$ so that $\lceil x \rceil- 1 < x \le \lceil x \rceil$

4) if $n \in \mathbb Z$ then $\lceil n \rceil = n$. This should be clear as $n-1 < n \le n$.

So.....

$\lfloor x \rfloor \in \mathbb Z$ by 1).

And so $\lceil \lfloor x \rfloor \rceil = \lfloor x \rfloor$ by 4).

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Let's have $\lfloor$ x $\rfloor$ = n, where n is an integer.

LHS = $\lceil$ n $\rceil$

RHS = n

Since n is an integer, the ceiling of n is simply n.

Therefore, n = n, LHS = RHS, QED.

I have proved that $\lceil$$\lfloor$ x $\rfloor$$\rceil$ = $\lfloor$ x $\rfloor$ for all real numbers x.

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  • $\begingroup$ n is not "some" integer. n is specifically [x] which is a very specific integer. But the point is it IS an integer. But yes, that is an adequate answer. $\endgroup$ – fleablood Nov 1 '16 at 21:29

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