Apostol's Mathematical Analysis 1.6

cordyceps
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Homework Statement


Prove that every nonempty set of positive integers contains a smallest member. This is called the well-ordering principle.


The Attempt at a Solution


I'm just starting out with analysis, so I'm not too sure about the format of proofs. Here goes:

Proof. First suppose the set S of positive integers is finite. Then assume S contains no smallest member. It follows that for every x in S, there exists an infinite number of members of S less than x. This contradicts our assumption that S is finite. Thus every finite set of integers contains a smallest member.

Any help/critique is appreciated.
 
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What properties of positive integers do you have to work with?
 
Sorry for the late reply. I have the definition of an inductive set and the definition of a positive integer: a real number which belongs to every inductive set.
 
Oh, also some of the field axioms for real numbers:
1. Exactly one of the relations x = y, x < y, x > y holds.
2. If x > y and y > z, then x > z.
 
So, for the non-empty set S of positive integers, one of the relations x < y or x > y holds between any two members. This means that each member in S is greater than or less than each of the other members. Also, if some member x of S is less than some member y, and y is less than some member z, then x is less than z. It follows that there is a member less than every other member in S.

Is this clear enough? Intuitively, I see why the theorem is true, but I don't think I present my thoughts very rigorously.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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