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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
1 year, 10 months ago

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Hmm , nice inspirations !

@Raghav Vaidyanathan , @Pi Han Goh Azhaghu Roopesh M · 1 year, 10 months ago

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#include<iostream.h>
#include<conio.h>
#include<stdlib.h>
#include<math.h>
#include<iomanip.h>

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<<g[0]<<" "<<g[1]<<endl<<setprecision(5)<<(g[0]*l[1]+l[0]*g[1]);
    getch();
    return 0;
    }
C++ code... Tested for a lot of numbers.. seems that it isn't always true... especially in the vicinity of numbers of form \(2n^2\) Raghav Vaidyanathan · 1 year, 10 months ago

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@Raghav Vaidyanathan So you've found a counterexample? Can you show me which values of \(A\) shows that is not true? Pi Han Goh · 1 year, 10 months ago

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@Pi Han Goh I think \(A=129\) doesn't satisfy. And also, numbers like \(A=128\) have \(\alpha=\beta\). Raghav Vaidyanathan · 1 year, 10 months ago

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@Raghav Vaidyanathan 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. Pi Han Goh · 1 year, 10 months ago

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@Pi Han Goh I agree with you on numbers of the form \(2n^2\) and also numbers which don't yield a solution. Now try \(A=137\)... Raghav Vaidyanathan · 1 year, 10 months ago

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@Raghav Vaidyanathan 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!) Pi Han Goh · 1 year, 10 months ago

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@Pi Han Goh The values of \(\alpha, \beta\) are one thing, but you may notice that: \(9\alpha+8\beta \ne 137\). Counterexample. Raghav Vaidyanathan · 1 year, 10 months ago

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@Raghav Vaidyanathan 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: \( \left(8, \frac{73}9\right)\). Pi Han Goh · 1 year, 10 months ago

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@Raghav Vaidyanathan Oh I failed my math. D=

Okay this is disproven! Thank you for your cooperation. I knew this is too good to be true. Pi Han Goh · 1 year, 10 months ago

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