Closed awindsor closed 1 year ago
I've waffled back and forth over the years on the introduction of 0, 1, $-u$, and $u^{-1}$ before knowing these objects are unique. I guess I settled on going that route at some point. I just looked back at an old version of that chapter and I made a different choice. I won't try hard to defend it as it seems a matter of preference. I think it is okay that there is a number called 0 that has a special property and then later we prove that it is the unique number with this property.
I do currently object to the wording of Theorems 5.2 and 5.3 since the use of the word "the" to start the sentence is problematic. I think something like the following is preferable:
There is a unique additive/multiplicative identity in R.
As for the proof of Theorem 5.3, here is what I would do. Suppose 1 and 1' are both multiplicative identities. Since 1 is a multiplicative identity, we know $1\cdot a = a$ for all real numbers $a$. In particular, $1\cdot 1' = 1'$. Similarly, since $1'$ is a multiplicative identity, $a\cdot 1'=a$ for all real numbers $a$, and in particular, $1\cdot 1'=1$. We have $1'=1\cdot 1'=1$, and so $1=1'$.
It just occurred to me that I very much misunderstood what you meant by "How does one show that the number 1 is unique?" I see what you were getting at there. I may not remove mention of 0, 1, $-u$, and $u^{-1}$ in list of axioms, but I will ponder ways to tidy up the confusion.
I've tweaked things a little in an attempt to clarify existence vs uniqueness.
I think it is far cleaner to reserve notation like $-u$ for the additive inverse or $u^{-1}$ for the multiplicative inverse until after uniqueness is established. I normally have less of a problem with the use of $0$ and $1$ but in this case we have $0$ as being the (not yet unique) additive inverse AND simultaneously a number in $\mathbb{R}$, which is very confusing. How does one show that the number 1 is unique?