Consider a rectangular parity code. Bob is interested in code rates like $2/3$, $3/4$ or in general $x/(x+1)$ for some integer $x>1$. Is it always possible to pick the parameters of the code (i.e. the block size and the number of rows and columns over which to construct the parity bits) so that any such code rate of the form $x/(x+1)$ is achievable. Explain your answer...
I am convinced the answer is no as the rate of a rectangular parity code is $rc/(rc + r +c)$ Thus, for that rate to equal $x/(x+1)$, $r+c$ would have to $= 1$, which is not possible. However, on the solution of this problem posted on MIT Open courseware, the professor works it out this way. Please let me know if I'm missing something or if the solution is simply wrong.
set\begin{align*}\frac{rc}{rc +r +c} = \frac x{x+1} &\Longrightarrow rcx + rc = rcx+ rx +cx\\ &\Longrightarrow rc - rx - cx = 0\\ &\Longrightarrow rc - rx - cx + x^2 = x^2\\ &\Longrightarrow(r-x)(c-x) = x^2\end{align*} if we set $r = 2x$ and $c = 2x$, the equation is satisfied, which means that a rectangular parity code with a rate $x/(x+1)$ exists. This is an interesting observation because although the number of parity bits in a rectangular code grows at least as fast as the square-root of the number of message bits, $2x$, it is still possible to achieve "high" code rates of the form $x/(x+1)$. (if I am correct, then this last line is particularly funny as the professor seems bewildered at his finding !)
Here is the link to the problem set problem 3(b) https://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-02-introduction-to-eecs-ii-digital-communication-systems-fall-2012/readings/MIT6_02F12_chap05.pdf
and here is a link to the solutions: http://web.mit.edu/6.02/www/s2012/handouts/ecc-soln.pdf
You are missing something. The equality $$\frac{p}{p + q}=\frac{x}{x+1}$$
(for integer variables) does not imply $q=1$. Rather, we should divide the numerator and denominator on the left by $q$ to get the correct implication:
$$\frac{\frac{p}{q}}{\frac{p}{q}+1}=\frac{x}{x+1} \implies \frac{p}{q}=x$$
In our case: $$x=\frac{rc}{r+c} $$ Of course, it's not inmmediately obvious that for any integer $x$ there exists such integers $r,c$. But setting $r=c$ you get the solution posted. More in general, we could write
$$ (r-x)(c-x)=x^2-x(r+c)+rc=x^2$$
If $x$ can be factored as $x=ab$ we can set $r=a^2+x$ and $c=b^2+x$ and we get other solution. In particular $r=1+x$ and $c=x(1+x)$ works. If $x$ is not prime, then there are more solutions.