Talk:Inverse for Real Multiplication

Proof?

I don't see why this proof cannot be applied to $0$. --kc_kennylau (talk) 09:45, 1 November 2016 (EDT)

Because $\dfrac 1 {x_n}$ is not defined when $x_n = 0$. --prime mover (talk) 10:55, 1 November 2016 (EDT)

Consider $x_n := 2^{-n}$. --kc_kennylau (talk) 10:57, 1 November 2016 (EDT)

But $2^{-n} \ne 0$. --prime mover (talk) 11:03, 1 November 2016 (EDT)

But $\left[\!\left[{\left \langle{x_n}\right \rangle}\right]\!\right] = 0$ if $x_n := 2^{-n}$. --kc_kennylau (talk) 11:05, 1 November 2016 (EDT)

Please feel free to plug the gap then. --prime mover (talk) 11:07, 1 November 2016 (EDT)

Proof?

I have found another gap. The inverse is cited without its existence proven. --kc_kennylau (talk) 11:12, 1 November 2016 (EDT)

Is this a gap, though?

1: the proof holds for the non-zero elements of $\R$. Nothing is stated about $x = 0$. Your original question was why it does not apply to zero. It does not matter why it does not apply to zero. This proof makes a statement about the non-zero elements. What happens when $x = 0$ can be studied on another page.

2: While not explicitly stated on this page, (it's a little bit lazy, perhaps we need some "Recall that ..." sort of statements), all the instances of $x_n$ are rational numbers, as you see from the definition of real number multiplication (follow the link). The existence of inverses of rational numbers has been established.

3: The rest of the page demonstrates the behaviour of $x \times \dfrac 1 x$ and thereby shows that $\dfrac 1 x$ is the inverse of $x$, thereby demonstrating the existence of the inverse. --prime mover (talk) 12:32, 1 November 2016 (EDT)

The problem is that $x_n$ can be $0$ for some $n$. --kc_kennylau (talk) 12:53, 1 November 2016 (EDT)

The problem is that $\dfrac 1 {[\![\langle x_n \rangle]\!]}$ is never defined. If we follow the links, division is never defined for $\R$, only as a consequence of $\R$ being a totally ordered field. Which proof relies on this result. So in fact we uncovered a circularity, and I guess a subtlety glossed over in the definition of operations on $\R$ in most books. We'll need to find a source to define it properly (I'm inclined to define said division by $\left[\!\!\left[\left\langle \dfrac 1 {x_n}\right\rangle\right]\!\!\right]$ with proviso $\dfrac 1 0 = 0$ but to show it works is, as usual with subtleties, quite tedious). — Lord_Farin (talk) 13:33, 1 November 2016 (EDT)