0^-1

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Whirligig231
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0^-1

Postby Whirligig231 » Thu Oct 06, 2011 9:01 pm UTC

How is 0^-1 for sure undefined? The only proof I've seen for x^-y = 1/(x^y) assumes x^0 = 1.

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Re: 0^-1

Postby gmalivuk » Fri Oct 07, 2011 3:46 am UTC

That's not so much something to prove as it is the definition.
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Re: 0^-1

Postby antonfire » Fri Oct 07, 2011 4:24 am UTC

You seem to be under the impression that 0^0 is not 1.
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Re: 0^-1

Postby Qaanol » Fri Oct 07, 2011 4:27 am UTC

antonfire wrote:You seem to be under the impression that 0^0 is not 1.

The set of functions from the empty set to itself definitely has cardinality 1.
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Re: 0^-1

Postby skullturf » Fri Oct 07, 2011 5:10 am UTC

I'm of course aware of the reasons behind the convention of sometimes leaving 0^0 undefined, but notice that you need 0^0 = 1 in order for the binomial theorem to work for (x+y)^n when one of x,y is zero.

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Re: 0^-1

Postby yurell » Fri Oct 07, 2011 6:13 am UTC

I thought 0-1 was undefined because in our axioms (or at least the set I learned) we state that every number except 0 has a multiplicative inverse. And if you assume it does have a multiplicative inverse you can prove just about anything, including contradictions.
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Re: 0^-1

Postby Afif_D » Fri Oct 07, 2011 5:08 pm UTC

Actually(0^-1) doesnt make sense. Any negative power of zero doesnt make sense. If it did make then 0^n (n is a positive integer) would not equal zero and could be proved to be an indeterminate form.

Like

0^2=0^(3-1)=(0^3)*(0^-1)=0/0

So the step in which i make a negative power of zero itself is wrong. Thus is doesnt make any sense..
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Re: 0^-1

Postby Diemo » Fri Oct 07, 2011 5:37 pm UTC

What I thought of it is 0-1 =1/01=infinity?
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Re: 0^-1

Postby gmalivuk » Fri Oct 07, 2011 6:38 pm UTC

But it isn't infinity. Defining it as *anything* leads to contradictions.
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Re: 0^-1

Postby eSOANEM » Fri Oct 07, 2011 10:26 pm UTC

gmalivuk wrote:But it isn't infinity. Defining it as *anything* leads to contradictions.


On the Riemann sphere it's well defined.
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Re: 0^-1

Postby Yesila » Sat Oct 08, 2011 3:06 am UTC

Diemo wrote:What I thought of it is 0-1 =1/01=infinity?


but x^{-1}= \frac{1}{x} only for non-zero x. So assuming that 0^{-1} "should" equal 1/0 is the "wrong" decision (since we long ago decided that dividing by zero is a poor choice).

In the case of 0^{-1} I think a case can be made that if it was defined (to be a real number) it "should be" 0^{-1}=0. In that way we preserve the "add the exponents" rule for multiplication of like bases after all IF 0^{-1} is defined to be a real number then

0^{-1}=0^{1-1-1}= 0^1 * 0^{-1} * 0^{-1} = 0

Similarly 0^0 "should" be zero as well.

0^0 = 0^{1-1}=0^1 0^{-1} = 0 * 0^{-1} = 0

Of course as was pointed out above we often times would rather use 0^0=1, which then necessitates that 0^{-1} is not a real number in which case the OP can define it to be whatever they want, so long as they realize that if they want to tack that new element onto R and then "do" things with it they better make some new field operations, and shouldn't expect them to follow the rest of the "rules" that they know and use with real numbers.

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Re: 0^-1

Postby thedufer » Sat Oct 08, 2011 5:55 am UTC

Yesila wrote:Of course as was pointed out above we often times would rather use 0^0=1, which then necessitates that 0^{-1} is not a real number in which case the OP can define it to be whatever they want, so long as they realize that if they want to tack that new element onto R and then "do" things with it they better make some new field operations, and shouldn't expect them to follow the rest of the "rules" that they know and use with real numbers.


Specifically, if we add some element "inf" to the field of reals, then inf needs an additive inverse, so now we add "-inf". Then, we probably need to define our operations such that inf*a is 1 if a=0, inf if a>0, -inf if a<0, and opposite for -inf*a. Also, inf+a is 0 if a=-inf, otherwise 0, and similarly for -inf+a.
Now, inf*(inf + (-inf))=inf*0=1, but we can distribute and get inf*inf + inf*(-inf)=inf + -inf=0, and this 0=1. This is absurd, so the reals must not be a field anymore.
I'm pretty sure there's not a more sensible way to define the operations with inf and -inf, so you'd basically have to redefine addition and multiplication, or give up the various field properties that you take for granted on a regular basis.

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Re: 0^-1

Postby Deveno » Sat Oct 08, 2011 4:47 pm UTC

i don't want to put 1 over 0, because then there's nothing there, and 1 will fall down and maybe hurt himself.

in a word, no. just no. don't go there, do not divide by 0, do not pass go, do not collect $200.

you'd have to throw away so many other genuinely useful rules, that there wouldn't be any arithmetic left.

but, hey: if you'd like a picture, just look at the graph of the hyperbola 1/x. be sure to get a "complete" picture, and not one of those cheap hong kong forgeries that only show a finite part.

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Re: 0^-1

Postby arbiteroftruth » Sat Oct 08, 2011 8:21 pm UTC

thedufer wrote:Specifically, if we add some element "inf" to the field of reals, then inf needs an additive inverse, so now we add "-inf".


I disagree. Similar logic would say that if 0 is a real number, then 0 needs a multiplicative inverse. Either way, we're making an unusual exception to what is a nearly universal rule. It's just as valid to make that exception in the realm of 'inf' having no additive inverse as it is to make that exception in the realm of 0 having no multiplicative inverse.

One may perhaps argue that part of our definitions includes the notion that a multiplicative inverse only exists for "non-zero numbers", but this only demonstrates that common convention has already selected one arbitrary exception over another. One can construct an entirely consistent math system where a multiplicative inverse exists for all real numbers but an additive inverse exists only for "non-inf numbers".

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Re: 0^-1

Postby Robert'); DROP TABLE *; » Sat Oct 08, 2011 9:42 pm UTC

arbiteroftruth wrote: One can construct an entirely consistent math system where a multiplicative inverse exists for all real numbers...

Really? I thought that was much of the point of not defining a div 0 in the first place.
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Re: 0^-1

Postby lightvector » Sat Oct 08, 2011 10:13 pm UTC

arbiteroftruth wrote:One can construct an entirely consistent math system where a multiplicative inverse exists for all real numbers but an additive inverse exists only for "non-inf numbers".


Actually, how well can you really do this, without giving up a lot of other important axioms?

If you add an element inf and declare that 0 and inf are multiplicative inverses then we have 1 = 0 * inf = (-1 * 0) * inf = -1 * (0 * inf) = -1 * 1 = -1, which is a contradiction. The only properties we needed to derive this contradiction were the associative property of multiplication, the fact that -1 * 0 = 0, and the fact that -1 * 1 is not equal to 1. Giving up any of these three properties sounds terrible.

Additionally, what is -1 * inf? We have -1 * inf + inf = -1 * inf + 1 * inf = (-1 + 1) * inf = 0 * inf = 1. The properties used here are the fact that 1 is the multiplicative identity, the distributive property, and (-1+1) = 0. We can then use the associative property of addition, and obtain (-1 + -1 * inf) + inf = -1 + (-1 * inf + inf) = -1 + 1 = 0, which means that (-1 + -1 * inf) is an additive inverse of inf.

This means that inf actually does have an additive inverse, but -1 * inf is not it! That really really sucks. But if we want to avoid that, we have to give up 1 being a multiplicative identity, or the distributive property, or (-1 + 1) = 0, or associativity of addition, or we have to say that -1 * inf is undefined and make multiplication no longer total. Any of these is also terrible.

This mess indicates that the choice to make zero lack a multiplicative inverse is not arbitrary. If you want something that even remotely resembles normal algebra and arithmetic with all the usual useful properties, introducing a single exception for zero is much better than the alternatives.

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Re: 0^-1

Postby arbiteroftruth » Sat Oct 08, 2011 11:54 pm UTC

lightvector wrote:If you add an element inf and declare that 0 and inf are multiplicative inverses then we have 1 = 0 * inf = (-1 * 0) * inf = -1 * (0 * inf) = -1 * 1 = -1, which is a contradiction. The only properties we needed to derive this contradiction were the associative property of multiplication, the fact that -1 * 0 = 0, and the fact that -1 * 1 is not equal to 1.


You have also implicitly used the principle that inf can be meaningfully multiplied by coefficients. Like 0, with which inf is closely related, inf would have the property that multiplication by any finite number (neither infinitely large nor infinitesimally small) results in no change in value. (-1 * 0) * inf = (-1 * inf) * 0 = inf * 0 = 1. When people construct proofs of contradiction that involve hidden cancellations of 0/0, the explanation is typically that 0/0 is not allowed to occur even when hidden by rearrangement of the notation. This would be a similar situation. You have introduced a coefficient of inf and hidden it with the right arrangement of notation.

One result of this rule that inf cannot have coefficients is that inf is an exception to the distributive property, as a side-effect of the principle that it has no additive inverse. The lack of an additive inverse makes it unsurprising that another rule involving addition might change. The analogous side-effect of the traditional option of saying 0 has no multiplicative inverse is that 0 is an exception to the property that x^0 = 1, or to the property that x^a * x^b = x^(a+b). If we allow both properties to apply to 0 like it does to all the other real numbers, then 1 = 0^0 = 0^(-1) * 0^1 = 0^(-1) * 0, meaning 0 has a multiplicative inverse.

One thing the above example may bring to mind is that another possible contradiction would be: 0 = 1 * 0 = (inf * 0) * 0 = inf * (0 * 0) = inf * 0 = 1. Meaning that inf must also be an exception to the commutative property of multiplication.

The end result is that inf can be included consistently in one's math system, but the only valid operation one can ever do to it is to multiply it by 0 and get 1. So it's inclusion in the system gains you nothing, but it creates no contradiction.

Calling zero's lack of a multiplicative inverse "arbitrary" was overly strong wording. My point was that it isn't a fundamental mathematical truth, but a convention of our usage.

EDIT: Technically, one can perform operations of exponentiation without contradiction. But the inability to apply multiplicative commutativity means this is still restricted to inf^a * 0^a = 1, with no right to equate 0^a = 0 before first multiplying by inf^a. So it's still useless.

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Re: 0^-1

Postby jestingrabbit » Sun Oct 09, 2011 1:28 am UTC

arbiteroftruth wrote:One thing the above example may bring to mind is that another possible contradiction would be: 0 = 1 * 0 = (inf * 0) * 0 = inf * (0 * 0) = inf * 0 = 1. Meaning that inf must also be an exception to the commutative property of multiplication.


You haven't used commutativity in your calculation here, you've used associativity, which lightvector already pointed out causes problems.

arbiteroftruth wrote:The end result is that inf can be included consistently in one's math system, but the only valid operation one can ever do to it is to multiply it by 0 and get 1. So it's inclusion in the system gains you nothing, but it creates no contradiction.


So you're saying that you're defining inf such that a*inf is undefined unless a is 0? That's considerably more useless than the standard treatment.

http://en.wikipedia.org/wiki/Real_projective_line

arbiteroftruth wrote:Calling zero's lack of a multiplicative inverse "arbitrary" was overly strong wording. My point was that it isn't a fundamental mathematical truth, but a convention of our usage.


It perhaps isn't fundamental, but it isn't far from it. If a is any element of a set where there are binary operations addition and multiplication, then

[math]\begin{align*}0 &= (a \cdot 0) - (a \cdot 0) \qquad &\text{(existence of additive inverse)}\\
&= (a \cdot (0+0)) - (a \cdot 0) \qquad &\text{(0 is the additive identity)}\\
&= ((a \cdot 0) + (a\cdot 0)) - (a \cdot 0) \qquad &\text{(multiplication distributing over additition)}\\
&= (a \cdot 0) + ((a\cdot 0) - (a \cdot 0)) \qquad &\text{(associativity of addition)}\\
&= (a \cdot 0) + 0 \qquad &\text{(additive inverse cancellation)}\\
&= (a \cdot 0) \qquad &\text{(0 is the additive identity)} \end{align*}[/math]

so we only need 0 being the additive identity, additive inverses, associativity of addition and mulitiplication distributing over addition to get that a*0 = 0. Break any of those axioms and you're getting a fairly unstructured arithmetic.
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Re: 0^-1

Postby Deveno » Sun Oct 09, 2011 9:14 am UTC

of all those laws, i would miss the distributive law the most. the others? meh, lots of binary operations in the sea....

pfft! ok, since y'all are gonna be srs abt this:

i think the first thing to ask when considering a possible definition for 0^(-1) is:

"what do i hope to gain from this, and what am i prepared to give up"?

algebraically, the consequences are usually limiting, you lose more than you gain. for example:

imagine addition is of no consequence to us, and suppose we set 0^(-1) = inf, and set:

a*inf = inf for all a except 0. the question arises: how should we define 0*inf?

if we set 0*inf = 1, then:

a = a*1 = a*(0*inf) = (a*0)*inf = 0*inf = 1 ???? huh? oh darn, we just lost associativity of multiplication.

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Re: 0^-1

Postby Whirligig231 » Mon Oct 10, 2011 5:03 pm UTC

The real projective line is useful to work with, but the important thing to keep in mind is that you still have to leave certain things undefined, namely 0/0, ∞/∞, ∞+∞, ∞-∞, ∞*0, and 0*∞. (∞ = infinity) These also (mostly) correspond to the "indeterminate forms" found in calculus.

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Re: 0^-1

Postby thedufer » Mon Oct 10, 2011 9:27 pm UTC

arbiteroftruth wrote:I disagree. Similar logic would say that if 0 is a real number, then 0 needs a multiplicative inverse.


Nope. In a field, every element except the additive identity needs a multiplicative inverse. We can let 0 be special because it is special, being the inverse.


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