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Given a sorted ascending sequence of fractions, if I add a constant $c$ to each fraction's numerator and denominator, how is the sequence affected?

For example, if I have a sequence in ascending order:

$$\frac12, \frac38, \frac57, \frac ab,\ldots$$

if I add a constant $c=1$

$$\frac{1+c}{2+c}, \frac{3+c}{8+c}, \frac{5+c}{7+c}, \frac{a+c}{b+c},\ldots$$

will my sequence still be sorted in ascending order?

Thank you!

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You need to check that $\frac{a}{b} < \frac{x}{y} \implies \frac{a+c}{b+c} < \frac{x+c}{y+c}$. The latter statement is equivalent to $(a+c)(y+c) < (x+c)(b+c)$ and this to $ay + ac + yc < xb + xc + bc$ as $ay < xb$ is given, you only need to see that $a-b < x-y$. This need not be true.

Consider $\frac{a}{b} = \frac{1}{2}$, $\frac{x}{y} = \frac{3}{5}$, $c=2$. Then $\frac{1+2}{2+2} = \frac{3}{4}$ and $\frac{3+2}{5+2} = \frac{5}{7}$. Finally $3 \cdot 7 > 5 \cdot 4$.

Note that I assumed everything positive. With negative values it is even more obvious. Consider $\frac{1}{2}$ with $\frac{2}{101}$ and $c = -1$.

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Or: Assuming all numbers positve, the fraction $\frac{a+c}{b+c}$ is between $\frac ab$ and $\frac cc=1$, but if $a,b\gg c$, then $\frac ab$ "pulls stronger". Thus $\frac 12<\frac {501}{1000}$, but $\frac 23>\frac{502}{1001}$. On an unrelated sidenote: If $\frac ab$ is in shortest terms, so is $\frac{a+c}{b+c}$. – Hagen von Eitzen Oct 21 '12 at 8:26
What is shortest terms? Do you mean $\gcd(a, b) = 1$ ?. If yes, consider $\frac{3+1}{5+1} = \frac{4}{6} = \frac{2}{3}$. – Karolis Juodelė Oct 21 '12 at 8:59
@Hagen: English doesn’t shorten fractions; it reduces them, to lowest terms. – Brian M. Scott Oct 21 '12 at 10:35

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