Tom Apostol - Calculus Vol. 1: Method of Exhaustion in Introduction I'm new to proof based math, and I really want to get better from pursuing the discipline by myself. I actually had a fear, last year, of proof-based math after my first semester in college. It was pretty brutal - I had never seen or experienced proofs before, and there was no attempt to help the students familiarize themselves. It seemed like either we knew the methods of proof, and how to write a good proof from the beginning, or spend all of your time being utterly confused throughout the material of the class. It was during this time I came to appreciate that I really need math literacy, and to get more experience with the techniques and underlying assumptions that are implicit in reading certain proofs. So, I'm now trying to get over this fear, by tackling the same content with the same textbook - Tom M. Apostol's Calculus Vol. 1.
In the book, in the introduction portion I $1.3$, the method of exhaustion is used to calculate the area under a parabolic segment of function $f(x)=x^2$ and base length $b$. 
Near equation I$.10$, we proceed to prove that $\frac{b^3}{3}$ is the only value between $s_n$ and $S_n$, the areas below and above the segment respectively, and hence must be the value of the area under the parabolic segment from $0$ to $b$.
My confusion of this part of the proof comes in when he assumes that $A>b^3/3$ to find the following contradiction: $$n < \frac{b^3}{A - b^3 / 3}$$
for $\{n\quad|\quad n\in \mathbb Z \qquad and \ \qquad n\gt0\}$, where $A$ is the guessed value for the area under the segment.  The contradiction lies in the fact that obviously $$n\ge \frac{b^3}{A-b^3/3}$$
but I'm not sure why -- it seems like it's reasonable for $\frac{b^3}{A-b^3/3}$ to be larger than 2 if we assume $b$ is some integer greater or equal to 1, and $A-\frac{b^3}{3}$ is at least less than half of 1, and n is allowed to be 1, so how does this contradict what has been said before?
 A: Remember, he states before it that, for all $n\ge1$, $A < \frac{b^3}{3} + \frac{b^3}{n}$. This is actually the same with the statement: 
$$∀n\ge1,\Biggl(A < \frac{b^3}{3} + \frac{b^3}{n}\Biggr)$$
Thus since we assumed that $A > \frac{b^3}{3}$, this yields to the fact that $A - \frac{b^3}{3} > 0$, thus we can change something inside the universal quantifier statement (like multiplying with $\frac{x}{x}$ or any other algebra operation).
So now, $∀n \ge 1, \bigl(A < \frac{b^3}{3} + \frac{b^3}{n}\bigr)$ means $∀n\ge1, \bigl(A - \frac{b^3}{3} < \frac{b^3}{n}\bigl)$, since the LHS is positive, we divide both sides and get $∀n \ge 1, \biggl(1 < \frac{\frac{b^3}{n}}{A - \frac{b^3}{3}}\biggr)$, we proceed to multiply both sides by $n$, which is okay, since $n \ge 1$ we finally arrives at a contradiction, namely: 
$$∀n \ge 1 \Biggl(n < \frac{b^3}{A - \frac{b^3}{3}}\Biggr)$$.
How to see this contradiction more clearly? First, note that the RHS is positive, so $\frac{b^3}{A - \frac{b^3}{3}} > 0$, we add $1$ to both sides and get $\frac{b^3}{A - \frac{b^3}{3}} + 1 > 1$, let $n = \frac{b^3}{A - \frac{b^3}{3}} + 1$, which is larger than $1$ (so this is one of the quantified value of the universal quantifier), but this $n$ is not smaller than $\frac{b^3}{A - \frac{b^3}{3}}$, for if it does, then $1 < 0$ which is not possible, therefore, a contradiction, as required. You can do the same with the other direction, showing that the only possibility is for $A = \frac{b^3}{3}$.
