Prove that the Well Ordering Axiom does not apply to the rational and real numbers. I am currently self teaching and have nobody to go to for help. This is not a homework problem.
Show that the Well-Ordering Axiom is not always true if $\mathbb{Z}$ is replaced by $\mathbb{Q}$ or $\mathbb{R}$.
It seems the term Well Ordering Axiom, by that name, is not widely understood. Perhaps it is unique to my book. As stated in it (Algebra by Mark Sepanski): If a nonempty set S$\subset$$\mathbb{Z}$ is bounded above, then S contains a unique largest element, i.e., there is a unique m $\in$ S so that m $\geq$ a for all a $\in$ S. 
Here is my attempt:
Let us consider the set
S={x$\in$Q|0$<$x$<$1}. 
The set S is bounded above by 1. Assume for the sake of contradiction that the Well-Ordering Axiom applies to the rational numbers. Then there should be some maximum element c $\in$ S. Since S $\subset$ $\mathbb{Q}$, c $\in$ $\mathbb{Q}$ and we can write c = $\frac ab$, where a,b $\in$ $\mathbb{Z}$. Furthermore, a < b since 0 < c < 1. Now, we shall construct a counterexample, namely, an element of S greater than c. 
a < b
$\implies$ a+ab < b+ab$\implies$a(b+1) < b(a+1)
$\implies$$\frac ab$<$\frac {a+1}{b+1}$
So, we have found a number greater than c and less than 1 (obviously). Nonetheless, since I am a learner:
a < b $\implies$ a+1 < b+1 $\implies$ $\frac {a+1}{b+1}$ < 1
Finally, since a,b > 0,
0 < $\frac {a+1}{b+1}$ < 1. 
This violates our assumption that c is the maximum element of S. Thus, there is no maximal element of S. Since, $\mathbb{Q}$ $\subset$ $\mathbb{R}$, the same applies to $\mathbb{R}$. QCD
The minimum case I will not bother writing here as this is sufficient for my question. 
Does this method require the proof to be infinite? In this proof I showed that a/b is not the maximum element. Isn't (a+1)/(b+1) a candidate for the new maximum requiring a repition of the proof bringing us to (a+2)/(b+2) and so on.
If my proof is valid then why is this a problem in the case for proving the principle of mathematical induction where the same logic is being applied in a sort of loop? 
Of course, a and b are arbitrary integers so perhaps the logic of my initial proof implicitly applies to anything of the form (integer)/(integer). 
The reason I suspected the possibility of an infinite proof is because to prove the PMI we cannot simply do this:
p1 is given and p(n) => p(n+1). 1+1=2. Therefore p2. Applying the same logic, we can get to any arbitrary n $\in$ $\mathbb{N}$. Therefore, the PMI holds. 
And obviously this spans all of $\mathbb{N}$ QED.
 A: For $\Bbb Q$ and $\Bbb R$ alike we can use $(0,1)$ as an example of a bounded above set without a maximum. If $m \in (0,1)$ were a maximum in $\Bbb R$ or $\Bbb Q$ then $\frac{m+1}{2}$ (the halfway point between $m$ and $1$) is a strictly larger point in the same set (rational if $m$ was) that is still in $(0,1)$. So no maximum exists. 
A: 
Does this method require the proof to be infinite? In this proof I showed that $a/b$ is not the maximum element. Isn't $(a+1)/(b+1)$ a candidate for the new maximum requiring a repetition of the proof bringing us to $(a+2)/(b+2)$ and so on.

No, there can only be one “candidate” for the maximum.  You've essentially shown that for all $a/b \in S$, $(a+1)/(b+1) \in S$ and is larger.  So there's no maximum element.
You can think about a proof by contradiction as a game between two debaters.  You claim there is no maximum element of $S$ and dare your opponent to find one.  Suppose they claim to find one.  Now you construct an element of $S$ that is larger, defying their claim that they found the maximum.  But the opponent doesn't get to change their answer—that's against the rules.
