Waldronjw

2022-07-01

Find the fraction.
Let us consider a fraction whose denominator is smaller than the square of the numerator by unity. If we add 2 to the numerator and the denominator, the fraction will exceed 1/3; now if we subtract 3 from the numerator and the denominator, the fraction remains positive but smaller than 1/10. Find the fraction.
My work:
Let $x/y$ be the fraction. Thus $y+1={x}^{2}$. Also,$\left(x+2\right)/\left(y+2\right)>1/3$ and $0<\left(x-3\right)/\left(y-3\right)<1/10$
Substituting the first equality in the second inequality, $0<\left(x-3\right)/\left(x-2\right)\left(x+2\right)<1/10$
How do I proceed after this?

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escampetaq5

Expert

HINT From $\frac{x+2}{y+2}>\frac{1}{3}$ you get $\frac{x+2}{{x}^{2}+1}>\frac{1}{3}$ and from here ${x}^{2}-3x-5<0,x\in \mathbb{N}$

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antennense

Expert

You have an answer already. To add some detail to it, for $x\in \mathbb{Z}$
$\frac{x+2}{{x}^{2}+1}>\frac{1}{3}\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}{x}^{2}-3x-5<0\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}{\left(x-\frac{3}{2}\right)}^{2}<\frac{29}{4}\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}x⩽4$
Similarly, $0<\frac{x-3}{{x}^{2}-4}<\frac{1}{10}\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}$
Case ${x}^{2}>4$: gives from $0<\frac{x-3}{{x}^{2}-4}\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}x>3$
Case ${x}^{2}<4$: gives from
$\begin{array}{}\text{(}×\text{)}& \frac{x-3}{{x}^{2}-4}<\frac{1}{10}\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\left(x-5{\right)}^{2}<-9\end{array}$
So we are left with $x=4$ as the only possibility.

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