# A question regarding the didactic motive for making a proof more complicated than logically necessary.

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:

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

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.

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.

• Ask him if you want a real answer. We can only speculate. – Somos May 31 at 14:31
• He's kind of dead: socalfolkdance.org/master_teachers/thurston_h.htm – Steven Thomas Hatton May 31 at 14:43
• 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. – Steven Thomas Hatton May 31 at 15:42
• @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. – DanielWainfleet Jun 2 at 0:00
• @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." – Steven Thomas Hatton Jun 2 at 4:38
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