<s>[INST] Question regarding Proof by Contradiction This is taken out of Apostol's Calculus I. It's part of his retelling of Archimedes method for finding the integral of $x^2$. 
At one point, the conclusion is reached
Therefore any number $A$ must also satisfy
$$\frac{b^3}{3} - \frac{b^3}{n} < A < \frac{b^3}{3} + \frac{b^3}{n}$$ 
for every integer $n \geq 1$. Now there are only thee possibilities:
$$A > \frac{b^3}{3}, A < \frac{b^3}{3}, A = \frac{b^3}{3}$$
Afterwards Apostle goes on to prove that the first two possibilities lead to a contradiction and that the third option has to be correct. But this is where I'm confused. Why must it hold for all $n$ in the third case if it doesn't hold for all $n$ in the first two? It's not exactly the same as when you have 3 excluding cases. We're assuming that because it only holds for some $n$ in the first two cases that it therefore must hold for all $n$ in the third. Where am I going wrong here? 
 [/INST] Note that "for all integers $n \geq 1$'' goes with the first sentence of the quote, not the second.  The alternatives in the second sentence do not depend on $n$, but rather compare two constants.
In his version of Archimedes' proof, Apostol assumes $A > \frac{b^3}{3}$ and then uses the first statement of the quote -- that for all integers $n \geq 1$, $A$ is between $\frac{b^3}{3}-\frac{b^3}{n}$ and $\frac{b^3}{3}+\frac{b^3}{n}$ -- to show that all positive integers $n$ must be less than a finite constant $\frac{b^3}{A-\frac{b^3}{3}}$.  There are infinitely many $n$ greater than that constant, so we have infinitely many contradictions and the original assumption must be false.  The original assumption $A > \frac{b^3}{3}$ doesn't depend on $n$.  The second alternative $A < \frac{b^3}{3}$ can be disposed of with a similar argument.  Thus, since $A$ isn't greater than $\frac{b^3}{3}$ and $A$ isn't less than $\frac{b^3}{3}$, $A$ must be equal to $\frac{b^3}{3}$.
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