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#19
Jul3110, 06:18 PM

P: 73

Im aware that high school depictions can be mislead. I really just think that this whole thing is more or less a matter of opinion.
To say a division problem, has to result in a single value, is more or less a rule someone has said. Show me the mathematical proof for this and ill back down from that statement. I think that the idea of 0/0= any value should be played with to see if it has any application, because this is just a matter of opinion. 


#20
Jul3110, 08:19 PM

PF Gold
P: 1,622

It's not a matter of opinion. An operation is defined in terms of functions, making it impossible for one input to produce multiple outputs. If you would like to define something differently, you can go ahead and do so, but you're probably not going to get very far.



#21
Jul3110, 11:32 PM

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P: 3,682

The obvious definition caleb is herading toward is the preimage of [real] multiplication (here I'll use the symbol "/" for this, to remind you that it's divisioninspired while not actually division), which is a "/" b = {a/b} for nonzero b, a "/" 0 = {} for nonzero a, and 0 "/" 0 = R, where R is the set of real numbers.



#22
Aug310, 11:37 AM

PF Gold
P: 1,622

CRGreathouse, I'm not sure that that's the definition caleb is heading towards. Saying that 0"/"0 = R is considerably different than saying 0/0 = any element of R.



#23
Aug310, 12:46 PM

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P: 21,408

Arithmetic is so basic and has such a long history in humankind that there aren't a lot of books that present arithmetic at an axiomatic and theoretic level. The only one I'm aware of is Principia Mathematica, by Bertrand Russell and Alfred North Whitehead. I haven't read this, but I understand that they don't get around to proving that 1 + 1 = 2 until a ways into the second volume. All four arithmetic operations that are defined produce a single value. The only arithmetic operation that is not defined is division by zero. Why would you want to make an exception for this particular operation and say that the result could be any number? 


#24
Aug310, 01:59 PM

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http://us.metamath.org/mpegif/2p2e4.html The arithmetic above takes place in the complex numbers, and so the proof involves Dedekind cuts or the like underneath; not too simple. (I think complex numbers are pairs of reals, which are themselves Dedekind cuts of rationals, which are themselves ratios of integers, which are themselves pairs of natural numbers.) I seem to remember that there are simpler versions if you just want to add naturals. 


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