Difference between revisions of "Mediant theorem"

Latest revision as of 22:37, 25 November 2023

Not to be confused with median.

The mediant of two fractions $\frac{a}{c}$ and $\frac{b}{d}$ is defined as $\frac{a+b}{c+d}$ for integers $a,b,c,d (b\ne 0, d\ne 0, b+d\ne 0).$

The Mediant theorem states that for integers $a_1, a_2, ... ,a_n, b_1, b_2, ... ,b_n (b_i\ne 0)$ ,

if $\frac{a_1}{b_1} = \frac{a_2}{b_2} = ... = \frac{a_n}{b_n},$ then

$\frac{a_1}{b_1} = \frac{a_2}{b_2} = ... = \frac{a_n}{b_n} = \frac{a_1+a_2+...+a_n}{b_1+b_2+...+b_n}.$

Let $\frac{a_1}{b_1} = \frac{a_2}{b_2} = ... = \frac{a_n}{b_n} = k.$

Since $a_i = kb_i$ ,

$\frac{a_1+a_2+...+a_n}{b_1+b_2+...+b_n}=\frac{k(b_1+b_2+...+b_n)}{b_1+b_2+...+b_n}=k. \quad \blacksquare$

Furthermore, for nonzero real numbers $x_1,x_2,...,x_n,$

if $\frac{a_1}{b_1} = \frac{a_2}{b_2} = ... = \frac{a_n}{b_n},$ then

$\frac{a_1}{b_1} = \frac{a_2}{b_2} = ... = \frac{a_n}{b_n} = \frac{a_1x_1+a_2x_2+...+a_nx_n}{b_1x_1+b_2x_2+...+b_nx_n}.$

This can also be proved easily by letting $\frac{a_1}{b_1} = \frac{a_2}{b_2} = ... = \frac{a_n}{b_n} = k.$