fractional part

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High School Math Grades 9-12, Ages 13-18

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High School Math Grades 9-12, Ages 13-18

Grades 9-12, Ages 13-18

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#1 h Apr 13, 2025, 2:51 PM

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{x^2}+{x}=0.64

How many positive real values of x satisfy this equation?

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#4 h Apr 14, 2025, 11:59 AM

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Infinite,all +ve integers of the form n+0.4433981132056604 where n belong to N

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#5 h Apr 14, 2025, 2:03 PM

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How did you get it

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#6 h Apr 15, 2025, 9:26 PM

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Actually, $\{x^2\}\neq\{x\}^2$ in general.

Let $f(x)=\{x^2\}+\{x\}$ . For $n\geq 0$ and $0\leq k\leq 2n$ , define the sets $S_{nk}=\left[\sqrt{n^2+k},\sqrt{n^2+k+1}\right)$ and the image sets $T_{nk}=f(S_{nk})$ . Note that $f:S_{nk}\to T_{nk}$ is bijective.

Then $\{x\}=x-n$ and $\{x^2\}=x^2-n^2-k$ .

Letting $x=n+y$ gives $x^2=n^2+2ny+y^2$ and so $\{x\}=y$ and $\{x^2\}=2ny+y^2-k$ and $T_{nk}=\left[\sqrt{n^2+k}-n,1+\sqrt{n^2+k+1}-n\right)$

The upper bound of $T_n$ is $1+\sqrt{n^2+k+1}-n>1$ .

Lower bound of $T_{nk}$ :
$\sqrt{n^2+k}-n\leq\frac{16}{25}\implies 25\sqrt{n^2+k}\leq 25n+16\implies 625n^2+625k\leq 625n^2+800n+256\implies k\leq\frac{800n+256}{625}$ .

$(2n+1)y+y^2-k=\frac{16}{25}\implies 25y^2+25(2n+1)y-25k-16=0\implies y=\frac{-10n-5+\sqrt{100n^2+100n+100k+89}}{10}$ .

So the overall solution set is $\left\{\left.\frac{-5+\sqrt{100n^2+100n+100k+89}}{10}\right|n\geq 0,0\leq k\leq\left\lfloor\frac{800n+256}{625}\right\rfloor\right\}$

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