$\lfloor \sqrt{\lceil x \rceil} \rfloor = \lfloor \sqrt{x} \rfloor, \forall x \in \mathbb{R}$

My Attempt:
Let $a = \lfloor \sqrt{\lceil x \rceil} \rfloor$

$$a \leq \sqrt{\lceil x \rceil} < a + 1\\ a^2 \leq \lceil x \rceil < (a+1)^2$$

Since $a^2 = \lceil x \rceil \Rightarrow a^2 \leq \lceil x \rceil < a^2+1$, It follows

$$a^2 \leq x < (a+1)^2\\ a \leq \sqrt{x} < a+1$$

It follows,

$$a = \lfloor \sqrt{x} \rfloor$$

Since equal (=), is also an equivalence $\iff$, I reckon I should also prove the converse, but it is easy to see from here. I'm new to Discrete Math and floor/ceil functions, thus I need someone to confirm if my proof is right.

Note: There's also a known inequality (I've seen from one of the posts here), which is not in my book $k \geq \lceil r \rceil \iff k \geq r$ and $k \leq \lfloor r \rfloor \iff k \leq r$, I rcekon I could also use this one, but since it's not in my book I'm abstaining from using it.

  • $\begingroup$ Doesn't seem to be unconditionally true. Let $x=0.01$. The right side is $0$, the left is $1$. $\endgroup$ – André Nicolas Jul 12 '14 at 4:57
  • $\begingroup$ @AndréNicolas How would you formally approach this not using a counter example? $\endgroup$ – JoeyAndres Jul 12 '14 at 5:03
  • $\begingroup$ One more or less has to produce a counterexample, or show indirectly that one exists. A proof is out of the question, since the result is not correct. One might try to characterize nicely the $x$ for which it is correct. Or try to find a related equality that is correct. $\endgroup$ – André Nicolas Jul 12 '14 at 5:08
  • $\begingroup$ @AndréNicolas Thanks again, that cleared things up more. I was thinking of proof by contradiction, but this assumes that the proposition is already true like you said. Cheers. $\endgroup$ – JoeyAndres Jul 12 '14 at 5:11
  • $\begingroup$ You are welcome. I guess my point of view is to look at the "geometry" before worrying about how to prove. Once one knows what's happening, proof details come fairly naturally. $\endgroup$ – André Nicolas Jul 12 '14 at 5:21

The result, as it is written now, is not true. There are many counterexamples. Let $x=8.5$. Then the right-hand side is $2$, while the left-hand side is $3$, since $\lceil 8.5\rceil=9$.

  • $\begingroup$ I do see your point, but can you pls point out the fallacy in my proof not using a counter example? $\endgroup$ – JoeyAndres Jul 12 '14 at 5:06
  • $\begingroup$ @Joey: You should try plugging in $x$ into your attempt and see which step fails. $\endgroup$ – Rahul Jul 12 '14 at 5:10
  • $\begingroup$ One needs to trace. Use say $x=0.01$ of my earlier comment, or $8.5$. Certainly $a^2\le x$ is not true. $\endgroup$ – André Nicolas Jul 12 '14 at 5:14
  • $\begingroup$ @JoeyAndres:$a^{2}\leq x$ is true. It is $x<(a+1)^{2}$ that is false, should be $\leq$ instead. There is also another errors at $a^{2}=\lceil x\rceil$, but it's completely irrelevant to the proof, as you don't use that fact for anything later. $\endgroup$ – Gina Jul 12 '14 at 5:34
  • $\begingroup$ From my reading of the OP's definition of $a$, using my $x=8.5$, we have $a=3$, so it is not true that $a^2\le x$. $\endgroup$ – André Nicolas Jul 12 '14 at 5:38

After your first two sets of inequalities you assert that $$a^2 =\lceil x\rceil,$$ which is false. If it were true the math display below it would then be false. There is a third logical error in your conclusion.

If you're having trouble with the floor and ceiling functions it might help to pick an explicit value for $x$ and see what your proof's steps look like in a single case. This can help your intuition (but don't fall prey to proof by example either). In this problem a counter example is the quickest solution, and easy enough to come up with.


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