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Old 2020-10-12, 05:33   #14
paulunderwood's Avatar
Sep 2002
Database er0rr

70418 Posts

Originally Posted by Nick View Post
Feedback on version dated 11 October 2020 (wow you've written a lot!)
Thank you very much for taking tome to read the text. I have made some changes to my local copy but will upload after more changes.
p5 The proof that root 2 is irrational relies on the uniqueness of prime factorization.
As this is a special property of the integers (not true in all number systems), I would at least mention it.
I mention it now in the both the paragraph on integers and when arguing about the irrationality of root 2.
p6 Ordinary sets are unordered and may not have repeating elements (this is so that they correspond with properties - if you select all objects satisfying a certain condition, you want what you get to be a set). Orderings and multisets can be constructed from ordinary sets if needed (in fact, so can everything in mathematics).
I have added the word "not".
p6 "subtracting all elements of one set from another" sounds confusing to me, as if you are calculating x-y for each x in the first set and y in the second. I would consider "removing"
I replaced the word "subtract" with "remove",
p7 The group axioms \(e\circ g=g\) needs to be \(g\circ e=g\), or state it and the next one both ways around.
I have subtended e rather than prepending it and in all subsequent instances where identity existence in mentioned including the section on rings,
p7 under multiplication the units of a commutative ring with 1 form an abelian group, but these are not all the elements of the ring (except in the trivial case where the ring has just 1 element).
I found this difficult. I have made mention of the center, although it might be out of context:
\item multiplicative identity : $(\exists 1\in {\cal S})(\forall a\in{\cal S})\hspace{.1in}a\times 1=a$.
The units form an Abelian group apart from a ring with one element, called the {\em center}.
p8 If you want to be able to relax condition 11 then you need to write condition 12 both ways around!
p8 A field is also required to have 1 not equal to 0.
I have added "where $1\ne 0$".
I'm out of time now - I'll look again in the week.
Again, many thanks,

Edit I have uploaded the latest copy -- 12 Oct.

Last fiddled with by paulunderwood on 2020-10-12 at 14:42
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