mersenneforum.org If a b= 0, how do I prove a and/or b =0?
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 2021-04-22, 16:52 #1 drkirkby   "David Kirkby" Jan 2021 Althorne, Essex, UK 2×229 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 canโ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
 2021-04-22, 17:04 #2 charybdis     Apr 2020 3FE16 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.
 2021-04-22, 18:20 #3 Nick     Dec 2012 The Netherlands 2·7·131 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).
 2021-04-23, 05:35 #4 Happy5214     "Alexander" Nov 2008 The Alamo City 22·5·72 Posts charybdis's proof applies to all integral domains, of which the integers are the algebraic prototype. (In fact, the non-ordering 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?
 2021-04-23, 09:35 #5 drkirkby   "David Kirkby" Jan 2021 Althorne, Essex, UK 2×229 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. ๐ข๐ข These exercise are only on the 7th page, at which point no axioms for irrational numbers are given. In fact, skimming the book, I donโt see any axioms for irrational numbers. I donโ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/Elementary-Nu...dp/0321500318/ although my edition is quite old. However, used copies of the latest edition are not too expensive on eBay.
 2021-04-23, 10:09 #6 drkirkby   "David Kirkby" Jan 2021 Althorne, Essex, UK 2×229 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.
2021-04-25, 06:55   #7
Happy5214

"Alexander"
Nov 2008
The Alamo City

98010 Posts

Quote:
 Originally Posted by Nick 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).
Quote:
 Originally Posted by drkirkby 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. ๐ข๐ข These exercise are only on the 7th page, at which point no axioms for irrational numbers are given. In fact, skimming the book, I donโt see any axioms for irrational numbers.
Note the difference. The axioms for the rational numbers (an ordered field) are the same as the real numbers (essentially the only Dedekind-complete 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.

Last fiddled with by retina on 2021-04-25 at 08:24 Reason: Getting those emojis correct is very important, right?

 2021-04-25, 07:01 #8 Happy5214     "Alexander" Nov 2008 The Alamo City 22×5×72 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.
2021-04-25, 08:30   #9
retina
Undefined

"The unspeakable one"
Jun 2006
My evil lair

677010 Posts

Quote:
 Originally Posted by charybdis (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.
If we use the sedenion algebra, just using the cancellation law is not enough.

There can be two non-zero 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 non-zero 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 2021-04-25 at 08:39 Reason: Added example to show this

2021-04-25, 09:39   #10
Happy5214

"Alexander"
Nov 2008
The Alamo City

22×5×72 Posts

Quote:
 Originally Posted by retina If we use the sedenion algebra, just using the cancellation law is not enough. There can be two non-zero 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 non-zero 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
I'm sure you're confusing him, as I'm confused. First of all, the proof that there can't be two non-zero values that multiply to 0 in the integers is the OP's question itself (or alternatively its contrapositive, a โ  0 & b โ  0 => a*b โ  0). The sedenions aren't a domain, so the cancellation property doesn't even apply to it (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, but sedenion(0,0,0,0,0,1,0,0,0,0,0,0,0,0,1,0) โ  sedenion(0,0,0,0,0,0,0,1,0,0,0,0,1,0,0,0), according to that article).

2021-04-25, 13:39   #11
charybdis

Apr 2020

102210 Posts

Quote:
 Originally Posted by retina If we use the sedenion algebra, just using the cancellation law is not enough. There can be two non-zero 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?
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 (a-b)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).

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