1
$\begingroup$

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

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.

$\endgroup$
  • $\begingroup$ 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
3
$\begingroup$

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 $\lceil 8.5\rceil=9$.

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.