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?

  • $\begingroup$ It will probably take more context of the proof to answer your question. Can you put up a picture of the page or copy the whole paragraph? $\endgroup$ Commented Jun 7, 2017 at 17:36
  • $\begingroup$ The inequality $n \lt b^3/(A - b^3/3)$ would have to be true for every n. But the RHS is some (possibly large) value that does not depend on n. So we could find an $n$ that is bigger than that, which contradicts the initial inequality: the contradiction proves that the assumption $A > b^3/3$ is false. $\endgroup$
    – NickD
    Commented Jun 7, 2017 at 17:53
  • $\begingroup$ So is it the case that if we get some constant being greater than $n$, then it provides a contradiction since we can always find an $n$ greater than the constant for which $\{n\quad|\quad n\in \mathbb Z \qquad and \ \qquad n\gt0\}$? If so, what can we meaningfully say for $n \gt$ some constant? From my understanding, it doesn't look like a contradiction in this case. I ask only because I want to see if I actually understand the meaning behind $n$ as we've defined it, since that seems to be the core of where my problems lie. $\endgroup$
    – svarq
    Commented Jun 7, 2017 at 17:59
  • $\begingroup$ The method of exhaustion rests upon the fact that if $x$ is a number and $y$ is a number then either $x<y$, $x>y$ or $x=y$. So if you can rule out $x<y$ and $x>y$ then it must be the case that $x=y$ since the alternatives have been exhausted. In proving that the area $A$ under $y=x^2$ on the interval $[0,b]$ is $\dfrac{b^3}{3}$ one first assumes that $A<\dfrac{b^3}{3}$ equaling $\dfrac{b^3}{3}-\epsilon$ for example. Then one shows that if $n$ is large enough, there is lower sum $s_n$ with area larger than $A=\dfrac{b^3}{3}-\epsilon$. But a lower sum should have area less than $A$. $\endgroup$ Commented Jun 8, 2017 at 5:12
  • $\begingroup$ So assuming $A<\dfrac{b^3}{3}$ leads to a contradiction. Then one assumes that $A>\dfrac{b^3}{3}$, say $A=\dfrac{b^3}{3}+\epsilon$. Then one finds an $n$ large enough that the $n$th upper sum $S_n$ is smaller than $A=\dfrac{b^3}{3}+\epsilon$. But an upper sum must be larger than $A$. So this is also a contradiction. So if it is false that $A<\dfrac{b^3}{3}$ and false that $A>\dfrac{b^3}{3}$ then it must be the case that $A=\dfrac{b^3}{3}$ because the alternatives have been exhausted. $\endgroup$ Commented Jun 8, 2017 at 5:19

1 Answer 1


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 $A = \frac{b^3}{3}$.

  • 3
    $\begingroup$ Use LaTeX please! $\endgroup$ Commented Nov 2, 2018 at 14:56
  • $\begingroup$ Credits to Seigun Ojo for noticing that x should be n. $\endgroup$
    – hteica
    Commented Feb 19, 2019 at 23:25

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