Algebraic proof by contradiction I have the following problem:
For the following predicate from domain of all positive real numbers 
$$ \frac{}{x+1} < \frac{+1}{x+2}$$
a. Write down the predicate logic statement for the theory above
b. Negate your predicate logic statement in (a)
c. Show that negated predicate statement in (b) cannot be correct and
therefore predicate in part (a) must be correct (prove by contradiction)
My solution is as follows:
a)
$ \Bbb R _>0$ = {x=$\Bbb R $ | $ x > 0 $}
P(y) = $ \frac{y}{y+1} < \frac{y+1}{y+2}$
$\forall x \in \Bbb R_>0 (P(x)) $
b)
$\exists x \in \Bbb R_>0 (\lnot P(x)) $
c)
This is where I'm stuck. How should I go about doing a proof by contradiction here?
 A: Assume that $x$ is positive and:
$$
\frac{x}{x+1} \geq \frac{x + 1}{x + 2}
$$
This leads to:
\begin{align}
\frac{x}{x+1} \geq&\ \frac{x + 1}{x + 2}\\
x(x + 2) \geq&\ (x + 1)^2 \\
x^2 + 2x \geq&\ x^2 + 2x + 1 \\
0 \geq 1
\end{align}
This is universally false meaning that the original statement, $\frac{x}{x+1} < \frac{x + 1}{x + 2}$ is universally true...but for all real numbers--not just positives.
p.s. My above reasoning is wrong...
You need to do the following:
\begin{align}
\frac{x}{x+1} \geq&\ \frac{x + 1}{x + 2}\\
\frac{x}{x +1} - \frac{x + 1}{x + 2} \geq&\ 0 \\
\frac{x^2 + 2x - x^2 - 2x - 1}{(x + 1)(x + 2)} \geq&\ 0 \\
\frac{-1}{(x + 1)(x + 2)} \geq&\ 0
\end{align}
You then need to make a sign chart:
$$
\frac{-1}{(x + 1)(x + 2)} \begin{cases}
< 0 & -\infty < x < -2 \\
> 0 & -2 < x < -1 \\
< 0 & -1 < x < \infty
\end{cases}
$$
Meaning that assuming that $\frac{x}{x+1} \geq \frac{x + 1}{x + 2}$ leads to the conclusion that $-2 < x < -1$ which contradicts the original assumption that $x$ is a positive real...although I don't see how this result is amenable to proof by contradiction since, at least the way I did it, required directly proving the desired result.
