This has to do with the proof given in Thurston's The Number System, on page 61. My notation is different, and explained below.

I'm asking about a style of presentation frequently appearing in text books. I often see derivations, etc., written out in ways that seem unnecessarily pedantic, lengthy and/or arcane. Sometimes, after looking at it long enough, I can see a logical or instructive reason for writing things as they were.

This is just an example. I'm hoping someone might explain why the derivation given would be preferred over the alternative I'm suggesting. Details follow.

My question is: What, if any advantage or insight is given by writing:

$$ x\odot \bar{p}\odot a=x\odot p\odot \bar{a}$$ $$ =x\odot \bar{a}\odot p$$ $$ =\bar{x}\odot a\odot p$$ $$ =\bar{x}\odot p\odot a$$ $$ \implies x\odot \bar{p}=\bar{x}\odot p,$$

rather than:

$$ x\odot \bar{a}\odot \bar{p}\odot a=\bar{x}\odot a\odot p\odot \bar{a}$$ $$= x\odot \bar{p}\odot \bar{a}\odot a=\bar{x}\odot p\odot \bar{a}\odot a$$ $$ \implies x\odot \bar{p}=\bar{x}\odot p?$$

I find the first version more difficult to understand and follow. It doesn't reveal to me any significant strategy or structure. It's just more complicated and confusing.

The operation $\odot$ is commutative, associative and cancellative.

The notation $\alpha=\left[\![a,\bar{a}\right]\!]$ is called a dyad, and represents an equivalence class of ordered pairs, where equality between pairs $\mathfrak{p}=\left\langle p,\bar{p} \right\rangle $ and $\mathfrak{x}=\left\langle x,\bar{x} \right\rangle $ means

$$\mathfrak{p}\circeq\mathfrak{x} \iff p\odot\bar{x}=\bar{p}\odot x.$$

To make this easy, in what follows $x\odot y$ may be replaced by $x+y,$ and $\alpha=\left[\![a,\bar{a}\right]\!]$ may be replaced by $a-\bar{a}.$

So $\mathfrak{p}\in\left[\![a,\bar{a}\right]\!]\iff \mathfrak{p}\in\alpha$ simply means $a-\bar{a}=p-\bar{p}\iff a+\bar{p}=\bar{a}+p$.

Our objective is to show that each $\left\langle p,\bar{p} \right\rangle$ is in exactly one dyad.

The rest of this is mostly screen-scrapes from my Mathematica notebook.

We already have a Theorem 2 giving us:

enter image description here

IMO, that more than justifies my suggested derivation. This is just a short outline of the structure of the proof:

enter image description here

I followed Thurston's derivation in the full proof. My suggestion for showing the first inclusion $\left[\![a,\bar{a}\right]\!]\subseteq\left[\![p,\bar{p}\right]\!],$ is take the red equation in (i); transpose it; right-$\odot$ it to the blue equation in (ii); commute $p\odot \bar{a}$ with $a$; and cancel $\bar{a}\odot a$. The 2 in the right column simply refers to the theorem I already mentioned.

enter image description here

I'm certain Thurston was smart enough to see that possibility. I'm wondering why he would chose to do it they way he did.

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    $\begingroup$ Ask him if you want a real answer. We can only speculate. $\endgroup$ – Somos May 31 at 14:31
  • $\begingroup$ He's kind of dead: socalfolkdance.org/master_teachers/thurston_h.htm $\endgroup$ – Steven Thomas Hatton May 31 at 14:43
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    $\begingroup$ Maybe he learned it that way, and figured the guy he learned it from was really smart, so he must have had a reason for doing it that way. And the real motivation behind the original lengthy proof was that the book contract required a certain minimum number of pages. $\endgroup$ – Steven Thomas Hatton May 31 at 15:42
  • $\begingroup$ @StevenThomasHatton. "Humans are creatures of habit." We often do something a certain way because we are used to it, which makes us feel relaxed and comfortable. $\endgroup$ – DanielWainfleet Jun 2 at 0:00
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    $\begingroup$ @DanielWainfleet I had't listed that as a reason in my proposed answer. I guess I wouldn't call that a didactic reason. One might even call that laziness. But it still begs the question of how it originally became a habit. Your suggestion is not all that different from my suggestion "that's the way he learned it." $\endgroup$ – Steven Thomas Hatton Jun 2 at 4:38

After some reflection, I can give two arguments in support of Thruston's choice to not use the most direct and obvious proof.

The most significant is that it leads his student to spend more time on the proof than might otherwise be invested. We might liken this to the difference in experience between driving a car through a park and walking along the same road. When we walk, we don't get to our destination as quickly, but we observe far more of our surroundings.

A second valid motivation is to encourage us to examine different possible approaches to the same problem. If he had presented the simplest, and most obvious proof, we would likely not explore other possibilities. By presenting us with the less obvious and direct proof, we are left to find the obvious and direct on our own.

There are, of course other possible motives. For one, he may have expected the student to discover something significant which has escaped me. Another is simply, that's how he learned it, so that's how he thinks it should be taught.

The most significant drawback to not presenting the direct and obvious proof is that doing so might give the impression that the most direct solution omits something logically essential or important. That is, we might wrongly infer that the most efficient proof is flawed.

I will add that on page 7 Thurston (rightly in my opinion) asserts that what makes the proofs of the laws of basic arithmetic difficult is not that they are complicated, but that they are simple. That is, we aren't accustomed to reasoning about them, we are used to reasoning with them. His goal in presenting a less obvious and efficient proof may have been to encourage his student to reason about the basic laws, and not simply to apply them as expediently as possible.


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