META: I wrote the explanation for this problem assuming a monospace font... it might be easier to read if you copy and paste it into a text file and view it separately. Or, if you know how, feel free to edit it to have a monospace font with automatic line breaks because I don't know how.
Let 4 variables $a,b,c,d$ be rationals in $[0,1]$ which, when multiplied by $255$, become integers. (That is, $a,b,c,d\in \{\frac{x}{255}\mid 0\leq x\leq 255,\ x\in\mathbb{Z}\}$. Examples of valid values are $1/255$, $2/255$, $3/255$, etc.
The variables are related in one equation. I want to prove that there are no solutions to this equation, by which I mean there are no valid values for the 4 variables that will satisfy the equation. $\frac{ac + (1-a)bd}{a+(1-a)b} = \frac{1}{2}$
Now I'm going to redefine $a,b,c,d$ to be non-negative integers in the domain $[0,255]$. The equation will still hold if I add the denominator $255$ to the variables. $\begin{align*} \frac{\frac{a}{255}\;\frac{c}{255} + \left(1-\frac{a}{255}\right)\frac{b}{255}\;\frac{d}{255}}{\frac{a}{255} + \left(1 - \frac{a}{255}\right)\frac{b}{255}} &= \frac{1}{2}\\ \frac{\frac{ac}{255^2} + \frac{(255-a)bd}{255^3}}{\frac{a}{255}+\frac{(255-a)b}{255^2}} &= \frac{1}{2}\\ \frac{\quad\frac{255ac + (255-a)bd}{255^3}\quad}{\frac{255a + (255-a)b}{255^2}}&=\frac{1}{2}\\ \frac{255 ac + (255-a)bd}{255^3}\;\frac{255^2}{255a+(255-a)b} &= \frac{1}{2}\\ \frac{255ac + (255-a)bd}{255(255a + (255-a)b)}&=\frac{1}{2}\\ \frac{255 ac + (255-a)bd}{255^2a + 255(255-a)b}&=\frac{1}{2}. \end{align*}$
$a,b,c,d$ are non-negative integers in the domain $[0,255]$. Is it possible to prove that there are no solutions to this equation?
One way to determine this is to test all ($255^4=4228250625$) possible combinations, however I'm looking for a more compelling proof.
Both the numerator and denominator will each evaluate to a non-negative integer value. That being said, a part of the set of possible evaluated fractions will look like this: $\frac{1}{2}, \frac{2}{4}, \frac{3}{6},\frac{4}{8},\frac{5}{10},\frac{6}{12},\frac{7}{14},\frac{8}{16},\frac{9}{18},\frac{10}{20},\ldots$
The denominator must evaluate to an even number.
Here are some of the rules of parity (even or odd) arithmetic:
Addition/subtraction:
Even Odd __________ Even |Even Odd Odd |Odd Even
Multiplication:
Even Odd __________ Even |Even Even Odd |Even Odd
The denominator has only two variables $a$ and $b$ that I need to worry about. Let's consider the possible cases of parity and see which combinations result in an even number.
$255^2a + 255(255-a)b$ $(\mathrm{Odd})a + (\mathrm{Odd})((\mathrm{Odd})-a)b$
$a$: Even; $b$: Even (Odd)(Even) + (Odd)((Odd)-(Even))(Even) (Even) + (Odd)(Odd)(Even) (Even) + (Odd)(Even) (Even) + (Even) (Even) a: Odd; b: Even (Odd)(Odd) + (Odd)((Odd)-(Odd))(Even) (Odd) + (Odd)(Even)(Even) (Odd) + (Even)(Even) (Odd) + (Even) (Odd) a: Even; b: Odd (Odd)(Even) + (Odd)((Odd)-(Even))(Odd) (Even) + (Odd)(Odd)(Odd) (Even) + (Odd)(Odd) (Even) + (Odd) (Odd) a: Odd; b: Odd (Odd)(Odd) + (Odd)((Odd)-(Odd))(Odd) (Odd) + (Odd)(Even)(Odd) (Odd) + (Even)(Odd) (Odd) + (Even) (Odd)
Therefore, the denominator is only even when both $a$ and $b$ are even. Let's see the parity of the numerator with $a$ and $b$ both being even.
$255ac + (255-a)bd$
(Odd)(Even)c + ((Odd)-(Even))(Even)d (Even)c + (Odd)(Even)d (Even)c + (Even)d (Even) + (Even) (Even)
Therefore, the numerator must be an even number as well, reducing the set of possible evaluated fractions to those with even numerators: $\frac{2}{4},\frac{4}{8},\frac{6}{12},\frac{8}{16},\frac{10}{20},\ldots$
.. this is the furthest I could go with my insular analysis. Are there any other rules I could use to reduce the set of possible evaluated fractions down to 0?