221–230 of 256 posts
Note, I also forgot to check when P(x) = -1, assuming that Q(x) is even there ;P
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This is simply wrong. 1+1=2 is the same kind of statement as 2^1=2 - it follows from the definition of the operation and from a general rule (it would be absurd here to say that 2^1 is the definition of 2, or that 2^1=2 is part of the definition of ^). 0^0 is just a meaningless symbol unless given explicit definition. There is no popular set of axioms where 1+1=2 is taken as an axiom - it is always proven. The symbol…
What do you think the definition of ^ is then? The ^ operator is defined as: 0^0 = 1 x^y = exp(ylog(x)) if x != 0 0^y = 0 if y != 0 Or similarly set theoretically.
m^0 = 1
m^n = m^(n-1).
After that, I consider the question of what x^y means for non naturals as motivated largely by considerations of algebraic and analytic extension. We get negative exponents simply by extending the operation with the requirement that
m^0 = 1
m^(n+p) = m^n*m^p.
(m^n)^p = m^(n*p).
The tension then is that the real and complex extensions don't "confirm" the original reasonable setting of 0^0=1, but I personally find it more objectionable poking a hole in the function, when it was so clear in the case of natural numbers that 0^0 should be 1 (and see the set theoretic and category theoretic accounts for more reasons why this is clear).
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Well meaningful as soon as someone finds a donkapple.
Not at all. Complex numbers are used to solve real world engineering problems, despite the fact that you cannot "find" any complex numbers in nature.
(I realise that this is kindof a horrible question to try and answer on an internet forum, please try your best physicists ;)
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But here's your problem: infinity * n = infinity, right? 0 * n = 0, right? 0 * infinity = ? Ok, this is relevant here particularly because: Lim 1/x as x -> 0 from the positive side is infinity, right? So 0 * that is..... lim 1/x as x -> from the negative side is negative infinity, right? So 0 * that is..... That's why 0/0 doesn't work as such. You don't know how 0 is derived or what it means. If we have x^2/x, and ta…
You are conflating the limiting behavior of a function with the value of a function. In (standard) analysis, there is no actual value called infinity - it's just used to describe how a function behaves arbitrarily close to a given value. When you have a 0^0 limiting form (or 0/0, or 0 * infinity), the function's behavior is indeterminate. The "0" and "infinity" you're looking at aren't precisely 0 or infinity, but on…
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I think you break algebra if you start treating y = x^2/x as distinct from y = x, and the both the former and x/x have the same problem with x=0. Edit: As has occurred to me, you might actually have a point, in that x/x may be a continuous function, but so is 0^x, which gives you a different limit. So it may actually determine whether in context it makes sense to treat 0^0 as x/x or whether it makes sense to treat it…
y=x^2/x and y=x aren't the same thing though. The first is undefined at x=0, the second is not. How does this break algebra?
I.e. the promise/premise of algebra is that if you take an equation, it remains equally valid when you add, subtract, multiply, or divide both sides by the same expression. If you treat these as distinct, then every division operation where you divide by a variable expression reduces the domain of possible answers.
This could have some very nasty corner cases where a division operation removes from the domain the answer you are looking for, and therefore renders an equation which could otherwise be solved undefined.
Hence my simple observation that in both algebra and calculus, we treat simplify equations before determining if they are continuous or not.
Here's a trivial example:
Solve for f(x) in the following equation (x + 2) * f(x) = x^2 + 4x + 4, where x = -2.
The problem you run into is that your first step is to simplify, so you divide both sides by x + 2, and therefore you end up with a division by 0 unless you are allowed to simplify before evaluating. Otherwise, you eliminate -2 from your domain and the answer is undefined. (The answer should be 0).
It's very important to note here that 0^0=1 is a shorthand and not a truth . Mathematicians are absolutely not stating that they have proven, or that it is true, that 0^0=1. It is a definition, not a claim of equality. They're not saying "0^0 is 1" in the sense that they say "1+1 is 2" or "0.999... is 1". They're saying "we define 0^0 to be 1". The difference is more than just pedantry, it strikes at the core of why…
Can anyone recommend a really good book on mathematics that demonstrates the "beauty" of it? I'd love to really learn and understand about Fast Fourier Transforms and the like, but any book I have is from my old college days and is just so bloody tedious. I used to really love maths when I was younger but it got beaten out of me by endless repetition and "now do Questions 1 - 50" tedium. Now I get scared whenever I s…
For an example, see the concept of a Ring:
http://en.wikipedia.org/wiki/Ring_%28mathematics%29
Less directly related to math, I spent a joyous couple of hours back in college working through the Lambda Calculus wiki page, trying to every piece it. Once it "clicked" it became really fun to work through the examples.
http://en.wikipedia.org/wiki/Lambda_calculus
> Now I get scared whenever I see the "summation" symbol.
I'm not sure why, but I guess for me, removing the idea of "numbers" makes math suddenly fascinating. It's like dropping down a level to redefine what you thought was a well-defined concept (addition of numbers) as a generic operation for which numbers are simply one special case. Weird. :)
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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.
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This is simply wrong. 1+1=2 is the same kind of statement as 2^1=2 - it follows from the definition of the operation and from a general rule (it would be absurd here to say that 2^1 is the definition of 2, or that 2^1=2 is part of the definition of ^). 0^0 is just a meaningless symbol unless given explicit definition. There is no popular set of axioms where 1+1=2 is taken as an axiom - it is always proven. The symbol…
>There is no popular set of axioms where 1+1=2 is taken as an axiom - it is always proven. I would love to see a proof of 1+1=2. For example in the case of elliptic curves, as far as I know addition is simply an axiom. E.g. A+A'=0= inf , where inf is the point at infinity and A' is the reflection of A. Math is not related to nature. Try to reason your way to matrix multiplication with apples: A * B != B * A
data Nat = Zero | Suc Nat
plus :: (Nat, Nat) -> Nat
plus (Zero, y) = y -- axiom 1
plus (Suc x, y) = Suc (plus (x, y)) -- axiom 2
one = Suc Zero
two = Suc one
plus (Suc Zero, Suc Zero)
= Suc (plus (Zero, Suc Zero)) [by axiom 2]
= Suc (Suc Zero) [by axiom 1]Earlier quoted context omitted.
Maybe your parent meant that "1 + 1 = 2" is the definition of "2", in the sense that we write the symbol "2" to denote the Peano successor of "1", the Peano successor of "0", which is not a successor.
Maybe. Formally this is inaccurate as the successor function is not the addition function, and one has to prove that S(S(0))=S(0)+S(0). Edit: it should also be noted that this equality shouldn't seem "obvious": the name "successor" here is misleading, as it automatically links to our everyday understanding of a successor as being "the next number", in which case it is indeed obvious. But formally it's just a meaningl…
I disagree with your assertion that 1+1=2 and 0^0=1 are of completely diferent kinds. That depends on what intuitive angle you approach them from; my intuition for ^ is that it is the iteration of multiplication, and in this context the only sensible definition is that x^0 = 1 (the unit of multiplication), for any x. The generalization to continuous bases and exponents should preserve this property.
But this is certainly a matter of taste, and certainly you'll find more disagreement here than in the matter of 1+1=2. What this tells us about human nature, I think, is not much: humans will disagree more when things get more complicated (^ is more complicated/abstract than +) and there is more room for disagreement.
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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.
It took Alfred Whitehead and Bertrand Russell 379 pages to prove that it "logically follows", and that was before they even defined addition! http://quod.lib.umich.edu/u/umhistmath/aat3201.0001.001/401?...