If a b= 0, how do I prove a and/or b =0?
I have another proof question, which is an engineer
I consider blindingly obvious, but I can’t prove. If a and b are integers, and a times b = 0, then obviously a and/or b = 0. But how can I prove that, starting with the basic axioms? Dave 
(Assuming you mean the axioms from the book you gave in your other thread)
Hint: use the cancellation law. [spoiler]Suppose ab = 0 and b =/= 0. The book gives a proof that anything multiplied by 0 is 0, so ab = 0b. Now the cancellation law gives a = 0.[/spoiler] 
This is an early example of an important general principle, namely
that the nunber system a problem is posed in may not be the best one to solve it in. In this case, it is better to work with the rational numbers instead of the integers (I don't have your book, so I don't know if you have axioms for those yet). 
charybdis's proof applies to all integral domains, of which the integers are the algebraic prototype. (In fact, the nonordering integer axioms you posted in the [URL="https://www.mersenneforum.org/showthread.php?t=26714"]other thread[/URL] form the definition of an integral domain.) It is much easier in a field like the rationals, though, with access to multiplicative inverses. Are these exercises you have to do requiring use of the integer axioms?

charybdis's proof seems pretty understandable to me. Thank you for that.
The book, states this proof, plus some others (see attachment), should start from the axioms from integers, although I am not going to attempt 1d. 😢😢 These exercise are only on the 7th page, at which point no axioms for irrational numbers are given. In fact, skimming the book, I don’t see any axioms for irrational numbers. I don’t know how good/bad this book is. There are PDFs of the 1st and 6th editions on [url]https://www.pdfdrive.com/[/url] The 6th edition seems to get fairly reasonable reviews on Amazon, but is blinking expensive [url]https://www.amazon.com/ElementaryNumberTheoryItsApplication/dp/0321500318/[/url] although my edition is quite old. However, used copies of the latest edition are not too expensive on eBay. 
I can't seem to get a copy of the page to be attached  either the original is too big, or a compressed version is considered invalid. I've tried a compress tool on my iphone, as well as Gimp on Windoze 10.

[QUOTE=Nick;576525]In this case, it is better to work with the [B]rational[/B] numbers instead of
the integers (I don't have your book, so I don't know if you have axioms for those yet).[/QUOTE] [QUOTE=drkirkby;576598]The book, states this proof, plus some others (see attachment), should start from the axioms from integers, although I am not going to attempt 1d. 😢😢 These exercise are only on the 7th page, at which point no axioms for [B]irrational[/B] numbers are given. In fact, skimming the book, I don’t see any axioms for irrational numbers. [/QUOTE] Note the difference. The axioms for the rational numbers (an ordered field) are the same as the real numbers (essentially the only Dedekindcomplete ordered field) minus the least upper bound property/Dedekind completeness, so if your book has axioms for the real numbers, you can derive the axioms for the rational numbers from those. 
Mods: Please fix the quoted emojis above for everyone's sanity. My browser is not letting me edit my post anymore because of them, so I can't fix them.

[QUOTE=charybdis;576507](Assuming you mean the axioms from the book you gave in your other thread)
Hint: use the cancellation law. [spoiler]Suppose ab = 0 and b =/= 0. The book gives a proof that anything multiplied by 0 is 0, so ab = 0b. Now the cancellation law gives a = 0.[/spoiler][/QUOTE]If we use the [url=https://en.wikipedia.org/wiki/Sedenion]sedenion[/url] algebra, just using the cancellation law is not enough. There can be two nonzero values a and b that satisfy ab == 0. So simply using cancellation isn't enough to prove it. What other axiom(s) must be invoked to make the proof correct? ETA: Example of two nonzero sedenions multiplied together to make zero: [c]sedenion(0,0,0,0,0,1,0,0,0,0,1,0,0,0,0,0) * sedenion(0,0,0,0,0,0,1,0,0,1,0,0,0,0,0,0) == 0[/c] 
[QUOTE=retina;576816]If we use the [url=https://en.wikipedia.org/wiki/Sedenion]sedenion[/url] algebra, just using the cancellation law is not enough.
There can be two nonzero values a and b that satisfy ab == 0. So simply using cancellation isn't enough to prove it. What other axiom(s) must be invoked to make the proof correct? ETA: Example of two nonzero sedenions multiplied together to make zero: [c]sedenion(0,0,0,0,0,1,0,0,0,0,1,0,0,0,0,0) * sedenion(0,0,0,0,0,0,1,0,0,1,0,0,0,0,0,0) == 0[/c][/QUOTE] I'm sure you're confusing him, as I'm confused. First of all, the proof that there can't be two nonzero values that multiply to 0 in the integers is the OP's question itself (or alternatively its contrapositive, a ≠ 0 & b ≠ 0 => a*b ≠ 0). The sedenions aren't a domain, so the cancellation property doesn't even apply to it ([c]sedenion(0,1,0,0,0,0,0,0,0,0,1,0,0,0,0,0) * sedenion(0,0,0,0,0,1,0,0,0,0,0,0,0,0,1,0) == sedenion(0,1,0,0,0,0,0,0,0,0,1,0,0,0,0,0) * sedenion(0,0,0,0,0,0,0,1,0,0,0,0,1,0,0,0) == 0[/c], but [c]sedenion(0,0,0,0,0,1,0,0,0,0,0,0,0,0,1,0) ≠ sedenion(0,0,0,0,0,0,0,1,0,0,0,0,1,0,0,0)[/c], according to that article). 
[QUOTE=retina;576816]If we use the [url=https://en.wikipedia.org/wiki/Sedenion]sedenion[/url] algebra, just using the cancellation law is not enough.
There can be two nonzero values a and b that satisfy ab == 0. So simply using cancellation isn't enough to prove it. What other axiom(s) must be invoked to make the proof correct?[/QUOTE] The cancellation law is actually equivalent to the property that there are no zero divisors (i.e. ab = 0 implies a = 0 or b = 0). I've already given the proof that cancellation => no zero divisors; conversely, if there are no zero divisors, then ac = bc gives (ab)c = 0 which then implies either c = 0 or a = b. So no zero divisors => cancellation. Notice that we did not use associativity or commutativity of multiplication, only distributivity, so this holds in a very general setting (in particular, for the sedenions). 
[QUOTE=Happy5214;576813]Note the difference. The axioms for the rational numbers (an ordered field) are the same as the real numbers (essentially the only Dedekindcomplete ordered field) minus the least upper bound property/Dedekind completeness, so if your book has axioms for the real numbers, you can derive the axioms for the rational numbers from those.[/QUOTE]
This isn't quite correct. The axioms of the reals minus the least upper bound property gives you an ordered field, but it is not sufficient to characterize the rationals. Specifically, it need not be Archimedean, which the rationals are. E.g. see the surreal numbers, for which these axioms apply, but which are certainly not isomorphic to the rationals. 
[QUOTE=jyb;576952]This isn't quite correct. The axioms of the reals minus the least upper bound property gives you an ordered field, but it is not sufficient to characterize the rationals. Specifically, it need not be Archimedean, which the rationals are. E.g. see the surreal numbers, for which these axioms apply, but which are certainly not isomorphic to the rationals.[/QUOTE]
Correct. The rationals are the [I]smallest[/I] ordered field, in the sense that they are embedded in every other ordered field, so that would be the missing axiom. I couldn't tell you how that implies the Archimedean property (I proved it for the reals in my analysis class, but that proof is based on completeness). 
[QUOTE=Happy5214;576953]I couldn't tell you how that implies the Archimedean property[/QUOTE]
In that case, it's time for our 2nd important general principle of the thread: given any property, always consider whether the set of all elements with that property is just a subset or a substructure (subgroup, subring, subfield or whatever). Here: all you have to do is show that the set of rational numbers that have the Archimedean property form a subfield. 
The Archimedean property is that if x and y are in an ordered field, x > 0 and y > 0, there is a positive integer n such that n*x > y.
Assuming the ordered field is the field of rational numbers with the usual ordering, I note that if x and y are positive rational numbers, there is a positive integer M such that M*x and M*y are both positive integers. Then taking n = M*y + 1, we have n*(M*x) >= n*1 = M*y + 1 > M*y, so that n*x > y. If we ignore the ordering, and use instead a "nonArchimedean valuation" (padic valuation), anything dependent on ordering (like "upper bound" or "least upper bound," and therefore "Dedekind completeness") goes out the window. Luckily, "Cauchy completeness" (every Cauchy sequence in the field has a limit in the field) can still be used to embed the padic rationals (and their extensions) into fields that are (Cauchy) complete WRT a nonArchimedean valuation. 
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