# Prove (or disprove): If $$\lceil x\lfloor x\rfloor\rceil+\lfloor x\lceil x\rceil\rfloor=A$$ has real solutions of $$x$$ with positive range of $$\alpha\le x\le\beta$$, then $$A=\alpha\lceil\ \sqrt{A/2}\ \rceil+\beta\lfloor\ \sqrt{A/2}\ \rfloor$$.

Like the title said:

Is it true that if $$\lceil x\lfloor x\rfloor\rceil+\lfloor x\lceil x\rceil\rfloor=A$$ has real solutions of $$x$$ with positive range of $$\alpha\le x\le\beta$$, then $$A=\alpha\left\lceil\ \sqrt{\frac A2}\ \right\rceil+\beta\left\lfloor\ \sqrt{\frac A2}\ \right\rfloor$$ must be true?

Inspired by Raghav Vaidyanathan which was inspired by Pi Han Goh.

Note by Pi Han Goh
2 years, 11 months ago

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Hmm , nice inspirations !

- 2 years, 11 months ago

  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 #include #include #include #include #include double fccf(double x) { double d=ceil(x*floor(x))+floor(x*ceil(x)); return d; } int main() { clrscr(); double x,l[2],g[2]={0,0},n; int r=0; cin>>n; l[0]=floor(sqrt(n/2));l[1]=ceil(sqrt(n/2)); for(x=l[0];x<=l[1];x+=0.0001) { double k=fccf(x); if(k==n&&r==0) { g[0]=x; r=1; } else if(k>n&&r==1) { g[1]=x; break; } } cout<

C++ code... Tested for a lot of numbers.. seems that it isn't always true... especially in the vicinity of numbers of form $$2n^2$$

- 2 years, 11 months ago

So you've found a counterexample? Can you show me which values of $$A$$ shows that is not true?

- 2 years, 11 months ago

I think $$A=129$$ doesn't satisfy. And also, numbers like $$A=128$$ have $$\alpha=\beta$$.

- 2 years, 11 months ago

There's no solution for $$x$$ when $$A=129$$ so it doesn't satisfy the condition. Well technically, if $$A$$ is in the form of $$2B^2$$ for some integer $$B$$, then $$\alpha = \beta$$, so by squeeze theorem, $$x = \alpha = \beta$$ which doesn't contradict the statement. In other words, it's trivial to pointless to disprove this statement when $$x$$ is an integer.

- 2 years, 11 months ago

I agree with you on numbers of the form $$2n^2$$ and also numbers which don't yield a solution. Now try $$A=137$$...

- 2 years, 11 months ago

Because $$8 < x < \frac{73}9$$ is the range for $$x$$ when $$x=137$$, by applying the inequality, we can have $$\alpha =8 + \epsilon, \beta = \frac{73}{9} - \epsilon$$ for an extremely small positive value $$\epsilon$$. In other words $$\alpha \gtrsim 8, \beta \lesssim \frac{73}9$$.

So my statement still holds.

If this is a cop out explanation, then you've successfully disproven my statement! (and I can't have that, haha!)

- 2 years, 11 months ago

The values of $$\alpha, \beta$$ are one thing, but you may notice that: $$9\alpha+8\beta \ne 137$$. Counterexample.

- 2 years, 11 months ago

Follow up (or modified) question: Is it true that if $$\lceil x\lfloor x\rfloor\rceil+\lfloor x\lceil x\rceil\rfloor=A$$ has real solutions of $$x$$ with positive range in a closed interval $$[\alpha, \beta ]$$ for $$\alpha \ne \beta$$, then $$A=\alpha\left\lceil\ \sqrt{\frac A2}\ \right\rceil+\beta\left\lfloor\ \sqrt{\frac A2}\ \right\rfloor$$ must be true?

That means to say that we can't take $$A=137$$ as an example because the range of solution of positive $$x$$ is an open interval: $$\left(8, \frac{73}9\right)$$.

- 2 years, 11 months ago

Oh I failed my math. D=

Okay this is disproven! Thank you for your cooperation. I knew this is too good to be true.

- 2 years, 11 months ago