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Old 2021-04-22, 16:52   #1
drkirkby
 
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Default 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
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Old 2021-04-22, 17:04   #2
charybdis
 
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(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.
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Old 2021-04-22, 18:20   #3
Nick
 
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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).
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Old 2021-04-23, 05:35   #4
Happy5214
 
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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?
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Old 2021-04-23, 09:35   #5
drkirkby
 
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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/Elementary-Nu...dp/0321500318/

although my edition is quite old. However, used copies of the latest edition are not too expensive on eBay.
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Old 2021-04-23, 10:09   #6
drkirkby
 
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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.
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Old 2021-04-25, 06:55   #7
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Quote:
Originally Posted by Nick View Post
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 View Post
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.
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?
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Old 2021-04-25, 07:01   #8
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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.
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Old 2021-04-25, 08:30   #9
retina
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Quote:
Originally Posted by charybdis View Post
(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
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Old 2021-04-25, 09:39   #10
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Quote:
Originally Posted by retina View Post
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 ≠ 0 & b ≠ 0 => a*b ≠ 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) ≠ sedenion(0,0,0,0,0,0,0,1,0,0,0,0,1,0,0,0), according to that article).
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Old 2021-04-25, 13:39   #11
charybdis
 
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Quote:
Originally Posted by retina View Post
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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