20210422, 16:52  #1 
"David Kirkby"
Jan 2021
Althorne, Essex, UK
2^{6}·7 Posts 
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 
20210422, 17:04  #2 
Apr 2020
7^{2}·11 Posts 
(Assuming you mean the axioms from the book you gave in your other thread)
Hint: use the cancellation law. 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. 
20210422, 18:20  #3 
Dec 2012
The Netherlands
2·3·293 Posts 
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). 
20210423, 05:35  #4 
"Alexander"
Nov 2008
The Alamo City
1100000101_{2} Posts 
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 other thread 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?

20210423, 09:35  #5 
"David Kirkby"
Jan 2021
Althorne, Essex, UK
2^{6}·7 Posts 
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 https://www.pdfdrive.com/ The 6th edition seems to get fairly reasonable reviews on Amazon, but is blinking expensive https://www.amazon.com/ElementaryNu...dp/0321500318/ although my edition is quite old. However, used copies of the latest edition are not too expensive on eBay. 
20210423, 10:09  #6 
"David Kirkby"
Jan 2021
Althorne, Essex, UK
2^{6}×7 Posts 
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.

20210425, 06:55  #7  
"Alexander"
Nov 2008
The Alamo City
773_{10} Posts 
Quote:
Quote:
Last fiddled with by retina on 20210425 at 08:24 Reason: Getting those emojis correct is very important, right? 

20210425, 07:01  #8 
"Alexander"
Nov 2008
The Alamo City
773 Posts 
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.

20210425, 08:30  #9  
Undefined
"The unspeakable one"
Jun 2006
My evil lair
2^{2}×3^{2}×5^{2}×7 Posts 
Quote:
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: 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 Last fiddled with by retina on 20210425 at 08:39 Reason: Added example to show this 

20210425, 09:39  #10  
"Alexander"
Nov 2008
The Alamo City
773 Posts 
Quote:


20210425, 13:39  #11  
Apr 2020
539_{10} Posts 
Quote:
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). 

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