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Re: 0^0

#161

Whenever you try to cross-breed an apple tree with a donkey, you get nonsense. But for convenience, whatever the offspring is, we may call it a donkapple. That's they beauty of math.

The utility of math is that you can then do a bunch of calculations with donkapples and at the end, get a meaningful real-world answer.

Well meaningful as soon as someone finds a donkapple.

Re: 0^0

#162
post #44

Earlier quoted context omitted.

The symbols we use to represent math are arbitrary but that doesn't mean the rules behind them are. Many concepts in math are fundamental truths.

There are an infinite number of fundamental logical truths out there, but we arbitrarily picked useful ones to make a system of math.

Picking the useful ones is the opposite of arbitrary.

Re: 0^0

#163

Earlier quoted context omitted.

I didnt discuss a function, I discussed a mapping. They are different constructs. Imagine the question like this: you have two groups of people, students and teachers. How many possible ways are there to assign students to teachers? To put it another way, how many arrows would it take pointing from the student to the teacher to illustrate every possible assignment? Now, if there are no students and no teachers, you h…

I am a mathematician I don't need examples. What is your formal definition of a mapping. In mathematic it is normally used as a synonym for a morphism in a given category and in the category of sets this would be a function. So in our context a mapping is a function.

You're being obtuse. If you read what he was saying, it would be obvious that he didn't mean a formalized "mapping" and meant a binary relation. Instead of using this as an opportunity to evangelize maths and teach, you criticized him with a pretentious self-aggrandizing glory that Mathematicians already have such a bad reputation for.

Re: 0^0

#164
post #155

Earlier quoted context omitted.

I see it as somewhat complex but in the end I sort of agree that it is a definition. Let's start with a problem: 0/0 The problem with this is not that the equation itself is meaningless. the limit of n/x as x->0 is infinity for any positive real number, and negative infinity for any negative real number. In essence 0/0 ends up reducing to 0 * infinity, which isn't very helpful. I would argue that discontinuity in a f…

The relationship f(x)=x/x is only defined for x /= 0 and thus is not equivalent to 1. We can see this in [a] when we ask for the domain of the function, and thus we can redefine some function f' as a piecewise function which is defined to have f'(0)=1, but in proofs we must thus make sure to first prove that using f' as a substitute for f does not affect our result. In one of the below posts we have the suggestion >…

> The relationship f(x)=x/x is only defined for x /= 0 and thus is not equivalent to 1.

If that's the case then:

x * f(x) = x cannot be equivalent to f(x) = x/x which breaks algebra in pretty fundamental ways (since the former would certainly be defined for 0 but the latter would not).

The second major problem is that it also breaks calculus. Let's start with a straight line: f(x) = 2x.

Now let's take the first derivative of this: f'(x) = 2x/x.

Does the line at the point where x = 0 have a slope or not? If this is discontinuous, then you have also broken calculus.

This gets as to why 0/0 is undefined, namely because when you cannot express it as a limit, and have no idea how both zeros are derived (and hence what they mean) you cannot give a specific number. You can come up with equations which for some value reduce to 0/0 but whose limits range from negative infinity all the way to positive infinity. But that doesnt mean that x/x is undefined where x = 0. x/x reduces to 1. Always. Anything else breaks higher mathematics generally.

Re: 0^0

#165
post #91

Earlier quoted context omitted.

I don't think the fifth axiom is considered particularly mysterious anymore. The traditional fifth axiom clearly isn't a logical result of the first four, since it can be replaced with other parallel postulates to yield non-Euclidean geometries which are themselves perfectly workable and consistent. In fact, that section of the Wikipedia article notes that Beltrami proved the independence of the parallel postulate.

> clearly It's mysterious because it is only clear in retrospect. The fact that you can negate the parallel postulate and get a system that is still self-consistent is incredibly mysterious. This is what sets mathematics apart. We "make it up" like other human accomplishments, and yet we can't actually just make it up. If we arbitrarily defined 0^0 = π, we'd just be speaking nonsense.

It's not mysterious that negating the parallel postulate is consistent; you just have to divorce the axioms from their originally intended meanings, and realize that they apply to other objects besides those meanings.

It's kind of like this. A biologist independently discovers directed graphs, but refers to the nodes as "organisms" and edges as "parenthoods", and then spends five hundred years trying to prove that every organism has only finitely many parenthoods, thinking it must be true since it's obvious biologically! Without realizing that graphs are far more general and apply elsewhere.

Arbitrarily defining 0^0=π would not be speaking nonsense, it would just be speaking a little arbitrarily. Nonsense would be defining 0^0=rainbow

Re: 0^0

#166

The real problem here is that x^y is a single shorthand which refers to a few fundamentally different mathematical concepts (which happen to have significant overlap with each other). First, it refers to a function f:C x N --> C, defined in terms of repeated multiplication. f(x,0) is 1 for all x != 0, and so we adopt the convention that f(0,0) is also 1. But it also refers to a function g:C x C --> C, defined as g(x,…

[deleted]

Re: 0^0

#167
post #157

Earlier quoted context omitted.

I agree with the point you're making, but I want to be a little pedantic: It's true that addition is a definition, but 1+1=2 is not--it logically follows from the definition of addition.

Not quite, just as in calculus you end up rediscovering some old geometry formulas there are branches of mathematics where you end up rediscovering 2 such that 1 + 1 equals it. Other options include 1 + 1 = 0 or 1+1 = 1.

But 1+1=0 or 1+1=1 happens because you are redefining the operator + (sometimes subtly, like using it over Z2), so what rickhanlonii said still holds. 1+1 is 2 if you're using the usual definition of sum, i.e., addition over integers or a superset of integers.

Re: 0^0

#168
post #25

I feel like in discrete mathematics (especially combinatorics), since we don't use continuous functions, it's useful to say 0^0 is 1, along with 0! = 1, and so on. Makes a lot of things around Binomial theorem and the like easier. I'm not so sure if it's safe to use that when doing and calculus proofs or anything along those lines, but there, you have more useful tools for dealing with limits that might approach 0^0.

0! really is 1 in a much more reasonable sense. The empty product is the multiplicative identity. The continuous notion of ! also agrees: http://en.wikipedia.org/wiki/Gamma_function

And it's far easier to prove: 1!=1, (n+1)!=n!·(n+1), thus 0!=1!/1=1.

Re: 0^0

#169
I read upto "High School Teacher"... skimmed through "Calculus Teacher"... and scrolled through "Mathematician"...

Re: 0^0

#170

Are there any examples where the x^0=1 definition turns out to make other definitions more complicated to write down? For example, you'd have a general definition and need a special definition for when you get an exponent that equals 0.

Suppose you want a compact, clever formula for the function: f(x)=1 if x!=0, f(x)=0 if x=0

If 0^0 were defined as 0, you could write the above function as f(x)=x^0.

With 0^0 defined as 1, you're forced to write something like f(x)=1-0^|x| (absolute values to avoid division by zero), a bit more complicated.

This is silly though, and of no importance anywhere.

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