问题1569: Take It Further The numbers

1. Let the two numbers be $x$ and $y$.

   • According to the problem, we have two equations based on the given property: $x - y=\frac{x}{y}$ and $y\neq0$.

   • First, rewrite the equation $x - y=\frac{x}{y}$ as an algebraic equation.

     • Multiply through by $y$ (since $y\neq0$) to clear the fraction: $y(x - y)=x$.
     • Expand the left - hand side: $xy-y^{2}=x$.
     • Rearrange the terms to get all terms involving $x$ on one side: $xy - x=y^{2}$.
     • Factor out $x$ on the left - hand side: $x(y - 1)=y^{2}$.
     • Then solve for $x$: $x=\frac{y^{2}}{y - 1}$, where $y\neq1$.

2. Find pairs of numbers by substituting values of $y$:

   • Case 1: Let $y = 4$

     • Substitute $y = 4$ into the formula $x=\frac{y^{2}}{y - 1}$. Then $x=\frac{4^{2}}{4 - 1}=\frac{16}{3}$.
     • Check the difference: $x - y=\frac{16}{3}-4=\frac{16}{3}-\frac{12}{3}=\frac{4}{3}$.
     • Check the quotient: $\frac{x}{y}=\frac{\frac{16}{3}}{4}=\frac{16}{3}\times\frac{1}{4}=\frac{4}{3}$.

   • Case 2: Let $y = 5$

     • Substitute $y = 5$ into the formula $x=\frac{y^{2}}{y - 1}$. Then $x=\frac{5^{2}}{5 - 1}=\frac{25}{4}$.
     • Check the difference: $x - y=\frac{25}{4}-5=\frac{25}{4}-\frac{20}{4}=\frac{5}{4}$.
     • Check the quotient: $\frac{x}{y}=\frac{\frac{25}{4}}{5}=\frac{25}{4}\times\frac{1}{5}=\frac{5}{4}$.

   • Case 3: Let $y=\frac{3}{2}$

     • Substitute $y = \frac{3}{2}$ into the formula $x=\frac{y^{2}}{y - 1}$. Then $x=\frac{(\frac{3}{2})^{2}}{\frac{3}{2}-1}=\frac{\frac{9}{4}}{\frac{1}{2}}=\frac{9}{4}\times2=\frac{9}{2}$, which is the pair we started with.

3. Describe the pattern:

   • If we have a number $y$, the corresponding number $x$ that satisfies $x - y=\frac{x}{y}$ is given by $x=\frac{y^{2}}{y - 1}$, where $y\neq1$.
   • We can rewrite $x$ as follows: $x=\frac{y^{2}-y + y}{y - 1}=\frac{y(y - 1)+y}{y - 1}=y+\frac{y}{y - 1}$.
   • The pairs of numbers $(x,y)$ where $x=\frac{y^{2}}{y - 1}$ ($y\neq1$) all satisfy the property that their difference $x - y$ is equal to their quotient $\frac{x}{y}$. For example, when $y$ is an integer greater than 1, $x$ is a non - integer fraction. When $y$ is a fraction, $x$ can be either an integer or a fraction depending on the value of $y$.

So, some pairs of numbers with the given property are $(\frac{16}{3},4)$, $(\frac{25}{4},5)$, $(\frac{9}{2},3)$, and in general, for $y\neq1$, the pair $(\frac{y^{2}}{y - 1},y)$ satisfies the property that $\frac{y^{2}}{y - 1}-y=\frac{\frac{y^{2}}{y - 1}}{y}$.

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