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,$(x+2)/(y+2)>1/3$ and $0<(x-3)/(y-3)<1/10$

Substituting the first equality in the second inequality, $0<(x-3)/(x-2)(x+2)<1/10$

How do I proceed after this?

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,$(x+2)/(y+2)>1/3$ and $0<(x-3)/(y-3)<1/10$

Substituting the first equality in the second inequality, $0<(x-3)/(x-2)(x+2)<1/10$

How do I proceed after this?

escampetaq5

Beginner2022-07-02Added 12 answers

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

antennense

Beginner2022-07-03Added 7 answers

You have an answer already. To add some detail to it, for $x\in \mathbb{Z}$

$\frac{x+2}{{x}^{2}+1}>\frac{1}{3}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{x}^{2}-3x-5<0\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{(x-\frac{3}{2})}^{2}<\frac{29}{4}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}x\u2a7d4$

Similarly, $0<\frac{x-3}{{x}^{2}-4}<\frac{1}{10}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}$

Case ${x}^{2}>4$: gives from $0<{\displaystyle \frac{x-3}{{x}^{2}-4}}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}x>3$

Case ${x}^{2}<4$: gives from

$\begin{array}{}\text{(}\times \text{)}& {\displaystyle \frac{x-3}{{x}^{2}-4}}<{\displaystyle \frac{1}{10}}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}(x-5{)}^{2}<-9\end{array}$

So we are left with $x=4$ as the only possibility.

$\frac{x+2}{{x}^{2}+1}>\frac{1}{3}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{x}^{2}-3x-5<0\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{(x-\frac{3}{2})}^{2}<\frac{29}{4}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}x\u2a7d4$

Similarly, $0<\frac{x-3}{{x}^{2}-4}<\frac{1}{10}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}$

Case ${x}^{2}>4$: gives from $0<{\displaystyle \frac{x-3}{{x}^{2}-4}}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}x>3$

Case ${x}^{2}<4$: gives from

$\begin{array}{}\text{(}\times \text{)}& {\displaystyle \frac{x-3}{{x}^{2}-4}}<{\displaystyle \frac{1}{10}}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}(x-5{)}^{2}<-9\end{array}$

So we are left with $x=4$ as the only possibility.

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