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I have no idea how to do this.

Suppose $x,y,z,n$ are positive integers. Given that $\frac{x}{y} < \frac{z}{n}$, prove that $$\frac{x}{y} < \frac{x+z}{y+n} $$

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$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ With $\ds{n,x,y,z >0}$:

\begin{align} &\color{#66f}{\large{x \over y} < {z \over n}}\ \imp\ xn<yz\ \imp\ xn + yx<yz + yx\ \imp\ x\pars{n + y}<y\pars{z + x}\ \\[3mm]&\imp\ \color{#66f}{\large{x \over y}<{x + z \over y + n}} \end{align}

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Since we have $$\frac xy\lt \frac zn\iff zy-nx\gt 0,$$ we have $$\frac{z}{n}-\frac{x+z}{y+n}=\frac{z(y+n)-n(x+z)}{n(y+n)}=\frac{zy-nx}{n(y+n)}\gt 0$$and $$\frac{x+z}{y+n}-\frac xy=\frac{y(x+z)-x(y+n)}{y(y+n)}=\frac{zy-nx}{y(y+n)}\gt 0.$$

In the first equation, we show $\frac{x+z}{y+n} < \frac{z}{n}$ by showing that their difference is positive.
In the second equation, we show $\frac{x+z}{y+n} > \frac{x}{y}$ again by showing that their difference is positive.

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  • $\begingroup$ sorry, bt how do you prove that $(x+z)/(y+n) < z/n$ and $> x/y$ ? $\endgroup$
    – Aaron
    Sep 10 '14 at 18:40
  • $\begingroup$ I proved $(z/n)-\{(x+z)/(y+n)\}\gt 0$, which is equivalent to $z/n\gt (x+z)/(y+n).$ Also, I proved $\{(x+z)/(y+n)\}-(x/y)\gt 0$, which is equivalent to $(x+z)/(y+n)\gt x/y$. Hence, I proved that $x/y\lt (x+z)/(y+n)\lt z/n$. $\endgroup$
    – mathlove
    Sep 10 '14 at 18:45
  • $\begingroup$ @Aaron: Note that $A\gt B\iff A-B\gt 0$. So, if we prove $A-B\gt 0$, this means that we prove $A\gt B$. $\endgroup$
    – mathlove
    Sep 10 '14 at 18:51

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