11 ms·

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

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 mathematics is the most powerful tool for determining truth that humanity has either discovered or invented (hat tip to anyone familiar with that debate). One of the cold, austere, beauties of mathematics* is that if you do not accept a definition, you can reject it as false and reason with the result. To give the traditional example, if you accept as true that two parallel lines never intersect, then along with the other 4 of Euclid's Axioms you can prove all of Euclidian Geometry (what you learn in high school). If you do not accept it as true (an explicitly accept it as false), then you can prove all of Hyperbolic Geometry (one form of non-Euclidian Geometry). In the case here, you're free to reject the convention that defines 0^0 as 1 and reason with the result; you will not break any mathematics. But you should know that mathematicians have never run into any issues with this convention--or can handle it trivially when they arise--so you're only adding a lot of work for yourself. When you really think about it, it fills you with awe. It's awesome. *“Mathematics, rightly viewed, possesses not only truth, but supreme beauty—a beauty cold and austere, like that of sculpture, without appeal to any part of our weaker nature, without the gorgeous trappings of painting or music, yet sublimely pure, and capable of a stern perfection such as only the greatest art can show.” -Bertrand Russell

philbarr 13y ago
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 see the "summation" symbol.
- coldpie 13y ago
  I found abstract algebra to be really interesting. It kind of removes the concept of "numbers" and transitions it to operations on objects like sets or functions. You redefine operations like "addition" and "multiplication" to apply to these new objects, and see what happens. A whole bunch of interesting things fall out, many of them important to information encoding and error correction in practice. For an example, see the concept of a Ring: http://en.wikipedia.org/wiki/Ring_%28mathematics%29 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 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. :)
- krick 13y ago
  > book on mathematics that demonstrates the "beauty" of it I'd say that you're asking wrong question. Yes, there're quite a few books intended to demonstrate the very thing you're asking for: beauty of mathematics. You are even likely to find some showcases interesting, might be somewhat beautiful. But essentially it won't be the same "beauty" that mathematicians see. It will be more like a thousand more-or-less scientific examples of how number φ appears in different places in our life, which isn't the essence of it obviously. So if you're scared by "summation" symbol you're better to reconsider or forget about understanding "beauty of mathematics" forever. Because "beauty of math" is math itself. About FFT: are you sure that you mean FFT? Because it's merely an algorithm to compute Discrete Fourier Transform. If you want to understand the Fourier Transformation itself, here's basic explanation: http://betterexplained.com/articles/an-interactive-guide-to-the-fourier-transform/ http://betterexplained.com/articles/an-interactive-guide-to-... Anyway, here're some easy-to-read books about math considered good by various people: Hardy, "Mathematician's Apology" Hofstadter, "Gödel, Escher, Bach" Klein, "Mathematics and the Search for Knowledge" (he wrote many sci-pop books which I expect to be good, but I read only few of them, so I'll recommend only this one) Сонин, "Постижение совершенства" — a good one, but I don't know if it was ever translated to english, sorry
rntz 13y ago
In a certain sense, "1+1 is 2" is also merely a definition, in the same sense that "0^0 is 1" is a definition. Addition can be formally defined in mathematics; we habitually omit this definition because it is tedious, and because addition is such an intuitive operation that we do not require a definition in order to reason about it. Much as the question "what if the parallel axiom didn't hold?" leads to alternative geometries, the question "what if addition didn't work in the same way?" (or "how can we generalize addition?") leads to some basic notions in abstract algebra.
- rickhanlonii 13y ago
  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.
  - Retric 13y ago
    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.
    - gordaco 13y ago
      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.
      - Retric 13y ago
        You missed the point 1 + 1 = 2 depends on the definition of '1', '+', '=', and '2'. The first case where this maters is probably binary where '2' does not exist, but that's hardly the only time 1+1 != 2.
  - mbrutsch 13y ago
Cacti 13y ago
Invented, obviously. ;)
mathgrad 13y ago
This is not true. A very natural way to define a^b is as the number of functions for a set with "b" elements to a set with "a" elements. In this case there is exactly one function from the empty set to the empty set, and we have proven 0^0 = 1. This is no different than defining addition and then proving 1+1 = 2.
- alecbenzer 13y ago
  I think this definition is somewhat non-standard for laypeople. Most people think of x^y to be defined as "x multiplied by itself y times" or something similar. By this definition, 0^0 is a convention/shorthand, not a truth. I think the "functions from b to a" definition is useful to note too, though
- j2kun 13y ago
  You're just pushing the arbitrariness of defining things one step further to the definition of a number. It makes it no less arbitrary that you've defined it and force us to accept the definition to get to your conclusion.
  - mathgrad 13y ago
    There is a very natural definition of numbers as sets. We define 0 to be the empty set and we define the successor function by S(x) = {x} union x. Then the natural numbers are the smallest set containing 0 and closed under the successor operation. This is the standard way to define the natural numbers within ZFC set theory. This is admittedly very formal and not how the lay person thinks of natural numbers. However, it gets to the point of that although we may lie and say 0^0 = 1 is just a convention it is in fact a theorem within the system we work.
    - gohrt 13y ago
      > within the system we work. within a system we work. Which is the point. The system you are working in determines which value you use. You are more an algebraist than analyst.
    - 13y ago
chacham15 13y ago
> If you do not accept it as true (an explicitly accept it as false), then you can prove all of Hyperbolic Geometry What is the equivalent in this analogy if you do not accept that 0^0=1 (i.e. accept that 0^0=0)?
- deleted 13y ago
  [deleted]
- einhverfr 13y ago
  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 as a limit of 0^x as x -> 0. This may be application dependent. Edit2: Ouch. 0^x can't be defined for negative numbers, so I am back to siding with 1 for reasons that any other answer breaks algebra.
  - ubercow13 13y ago
    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?
    - einhverfr 13y ago
      Because if you can't simplify, you can't treat them as the same, which is a fundamental premise of 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).
einhverfr 13y ago
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 form like this can never be unique to a single point in an otherwise continuous function, if the limit approached from both sides is the same, unless that point as a specific, defined value. This is because a point is arbitrarily small. The limit of 1/x as x->0 is discontinuous because from the negative side and the positive side, the limits are different, as anyone can see when you plot it on a graph. However here we have x/x, and the limit stays at 1 from both sides. It may be formally undefined at the exact point of 0/0, but the limit from both sides is the same, and it remains constant as you get arbitrarily close. Therefore it is simpler, system-wise, to treat it as 1, than to make an exception for it. > In the case here, you're free to reject the convention that defines 0^0 as 1 and reason with the result; But you add complexity which is neither helpful nor desirable.
- j2kun 13y ago
  But what if you're in a context where you're not reasoning about continuous functions at all? Why would you have to be subject to reasoning that doesn't apply to your situation?
  - einhverfr 13y ago
    That's actually an interesting point. And actually it occurs to me that even in continuous equations you have a problem: Does it matter if we are talking 0^x or x^0? I would think that in the context of 0^x, you'd have a constant function of 0, but x^0 you'd have a constant function of 1. These have different limits as x -> 0. I think you have just convinced me that 0^0 is undefined. Edit: Ouch. 0^x can't be defined for a negative x, so that doesn't work. I am back to siding with 1.
- ndeine 13y ago
  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 > But what if you're in a context where you're not reasoning about continuous functions at all? Why would you have to be subject to reasoning that doesn't apply to your situation? In this case you could either do the above, if you have to concern yourself with e.g. a domain of the set of real numbers arbitrarily close to the undefined location. Alternatively you could just define our original function f only for real numbers greater than 0, in which case you escape the necessity of redefining functions to be easier to work with. [a]: http://www.wolframalpha.com/input/?i=domain+of+f%28x%29+%3D+x%2Fx http://www.wolframalpha.com/input/?i=domain+of+f%28x%29+%3D+...
shurcooL 13y ago
Agree, and you've basically described the concluding message in the last minute of this awesome video: https://www.youtube.com/watch?v=TINfzxSnnIE https://www.youtube.com/watch?v=TINfzxSnnIE

philbarr 13y ago

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 see the "summation" symbol.

coldpie 13y ago
I found abstract algebra to be really interesting. It kind of removes the concept of "numbers" and transitions it to operations on objects like sets or functions. You redefine operations like "addition" and "multiplication" to apply to these new objects, and see what happens. A whole bunch of interesting things fall out, many of them important to information encoding and error correction in practice. For an example, see the concept of a Ring: http://en.wikipedia.org/wiki/Ring_%28mathematics%29 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 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. :)
krick 13y ago
> book on mathematics that demonstrates the "beauty" of it I'd say that you're asking wrong question. Yes, there're quite a few books intended to demonstrate the very thing you're asking for: beauty of mathematics. You are even likely to find some showcases interesting, might be somewhat beautiful. But essentially it won't be the same "beauty" that mathematicians see. It will be more like a thousand more-or-less scientific examples of how number φ appears in different places in our life, which isn't the essence of it obviously. So if you're scared by "summation" symbol you're better to reconsider or forget about understanding "beauty of mathematics" forever. Because "beauty of math" is math itself. About FFT: are you sure that you mean FFT? Because it's merely an algorithm to compute Discrete Fourier Transform. If you want to understand the Fourier Transformation itself, here's basic explanation: http://betterexplained.com/articles/an-interactive-guide-to-the-fourier-transform/ http://betterexplained.com/articles/an-interactive-guide-to-... Anyway, here're some easy-to-read books about math considered good by various people: Hardy, "Mathematician's Apology" Hofstadter, "Gödel, Escher, Bach" Klein, "Mathematics and the Search for Knowledge" (he wrote many sci-pop books which I expect to be good, but I read only few of them, so I'll recommend only this one) Сонин, "Постижение совершенства" — a good one, but I don't know if it was ever translated to english, sorry

rntz 13y ago

In a certain sense, "1+1 is 2" is also merely a definition, in the same sense that "0^0 is 1" is a definition. Addition can be formally defined in mathematics; we habitually omit this definition because it is tedious, and because addition is such an intuitive operation that we do not require a definition in order to reason about it. Much as the question "what if the parallel axiom didn't hold?" leads to alternative geometries, the question "what if addition didn't work in the same way?" (or "how can we generalize addition?") leads to some basic notions in abstract algebra.

rickhanlonii 13y ago
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.
- Retric 13y ago
  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.
  - gordaco 13y ago
    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.
    - Retric 13y ago
      You missed the point 1 + 1 = 2 depends on the definition of '1', '+', '=', and '2'. The first case where this maters is probably binary where '2' does not exist, but that's hardly the only time 1+1 != 2.
- mbrutsch 13y ago

Cacti 13y ago

Invented, obviously. ;)

mathgrad 13y ago

This is not true. A very natural way to define a^b is as the number of functions for a set with "b" elements to a set with "a" elements. In this case there is exactly one function from the empty set to the empty set, and we have proven 0^0 = 1. This is no different than defining addition and then proving 1+1 = 2.

alecbenzer 13y ago
I think this definition is somewhat non-standard for laypeople. Most people think of x^y to be defined as "x multiplied by itself y times" or something similar. By this definition, 0^0 is a convention/shorthand, not a truth. I think the "functions from b to a" definition is useful to note too, though
j2kun 13y ago
You're just pushing the arbitrariness of defining things one step further to the definition of a number. It makes it no less arbitrary that you've defined it and force us to accept the definition to get to your conclusion.
- mathgrad 13y ago
  There is a very natural definition of numbers as sets. We define 0 to be the empty set and we define the successor function by S(x) = {x} union x. Then the natural numbers are the smallest set containing 0 and closed under the successor operation. This is the standard way to define the natural numbers within ZFC set theory. This is admittedly very formal and not how the lay person thinks of natural numbers. However, it gets to the point of that although we may lie and say 0^0 = 1 is just a convention it is in fact a theorem within the system we work.
  - gohrt 13y ago
    > within the system we work. within a system we work. Which is the point. The system you are working in determines which value you use. You are more an algebraist than analyst.
  - 13y ago

chacham15 13y ago

> If you do not accept it as true (an explicitly accept it as false), then you can prove all of Hyperbolic Geometry What is the equivalent in this analogy if you do not accept that 0^0=1 (i.e. accept that 0^0=0)?

deleted 13y ago
[deleted]
einhverfr 13y ago
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 as a limit of 0^x as x -> 0. This may be application dependent. Edit2: Ouch. 0^x can't be defined for negative numbers, so I am back to siding with 1 for reasons that any other answer breaks algebra.
- ubercow13 13y ago
  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?
  - einhverfr 13y ago
    Because if you can't simplify, you can't treat them as the same, which is a fundamental premise of 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).

einhverfr 13y ago

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 form like this can never be unique to a single point in an otherwise continuous function, if the limit approached from both sides is the same, unless that point as a specific, defined value. This is because a point is arbitrarily small. The limit of 1/x as x->0 is discontinuous because from the negative side and the positive side, the limits are different, as anyone can see when you plot it on a graph. However here we have x/x, and the limit stays at 1 from both sides. It may be formally undefined at the exact point of 0/0, but the limit from both sides is the same, and it remains constant as you get arbitrarily close. Therefore it is simpler, system-wise, to treat it as 1, than to make an exception for it. > In the case here, you're free to reject the convention that defines 0^0 as 1 and reason with the result; But you add complexity which is neither helpful nor desirable.

j2kun 13y ago
But what if you're in a context where you're not reasoning about continuous functions at all? Why would you have to be subject to reasoning that doesn't apply to your situation?
- einhverfr 13y ago
  That's actually an interesting point. And actually it occurs to me that even in continuous equations you have a problem: Does it matter if we are talking 0^x or x^0? I would think that in the context of 0^x, you'd have a constant function of 0, but x^0 you'd have a constant function of 1. These have different limits as x -> 0. I think you have just convinced me that 0^0 is undefined. Edit: Ouch. 0^x can't be defined for a negative x, so that doesn't work. I am back to siding with 1.
ndeine 13y ago
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 > But what if you're in a context where you're not reasoning about continuous functions at all? Why would you have to be subject to reasoning that doesn't apply to your situation? In this case you could either do the above, if you have to concern yourself with e.g. a domain of the set of real numbers arbitrarily close to the undefined location. Alternatively you could just define our original function f only for real numbers greater than 0, in which case you escape the necessity of redefining functions to be easier to work with. [a]: http://www.wolframalpha.com/input/?i=domain+of+f%28x%29+%3D+x%2Fx http://www.wolframalpha.com/input/?i=domain+of+f%28x%29+%3D+...

shurcooL 13y ago

Agree, and you've basically described the concluding message in the last minute of this awesome video: https://www.youtube.com/watch?v=TINfzxSnnIE https://www.youtube.com/watch?v=TINfzxSnnIE