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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 eq

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+...
- einhverfr 13y ago
  > 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.
  - lotharbot 13y ago
    > "x * f(x) = x cannot be equivalent to f(x) = x/x" Sure it's equivalent, over a domain not including x=0. This does not break algebra any more than, say, restricting the domain of the square root (when working in the reals) to non-negative numbers. We work in restricted domains in mathematics all the time. > " f'(x) = 2x/x" f'(x) = lim (h->0) [2(x+h)-2x]/h. Since h is approaching (and therefore not equal to) zero, there is no problem. Any appearance of 0/0 in the problem is a result of an attempted (but unsuccessful, that is, indeterminate) evaluation -- it's not actually 0/0, it's 2h/h where h is close to but not equal to 0. We don't need to define 0/0=1 in order to have either algebra or calculus work. We choose to define 0/0=1 in certain circumstances which make certain calculations go more smoothly, and we choose not to define 0/0 in other circumstances where it's either unnecessary or potentially misleading.
zodiac 13y ago
I see that you are defining 0/0 to be a value such that f(x) = x/x has no discontinuity at x=0. But your chosen value means that the function g(x) = x^2 / x is discontinuous at x=0, and the function h(x) = -x/x also has a discontinuity, etc.
- einhverfr 13y ago
  I am defining continuity such that single point that is undefined does not break continuity if and only if it has the same limit when approached from both sides. This is because, as I say, a point is arbitrarily small and therefore if the limit is the same on both sides, it's simpler to treat it as still a continuous function. Otherwise I think you get some problems which break algebra, because multiplying of dividing by variables can break the domains of functions. If we have f(x) = 1, then this should be the same even if we multiply both sides by x. x * f(x) = x, which results in f(x) = x/x. You break algebra by allowing such straight-forward manipulations (which really are the core of algebra) to tamper with the original domain of the function. Just as you can't just treat a square root as a core algebraic process (because it is not a function), you would no longer be able to divide by a variable, since that would be undefined wherever the variable is 0. Edit Also it occurs to me that treating x/x as undefined at 0 would also man that the first derivative of f(x) = x would also be undefined at 0, which can't work either. So therefore treating x/x as discontinuous at 0 breaks both algebra and calculus.

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+...

einhverfr 13y ago
> 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.
- lotharbot 13y ago
  > "x * f(x) = x cannot be equivalent to f(x) = x/x" Sure it's equivalent, over a domain not including x=0. This does not break algebra any more than, say, restricting the domain of the square root (when working in the reals) to non-negative numbers. We work in restricted domains in mathematics all the time. > " f'(x) = 2x/x" f'(x) = lim (h->0) [2(x+h)-2x]/h. Since h is approaching (and therefore not equal to) zero, there is no problem. Any appearance of 0/0 in the problem is a result of an attempted (but unsuccessful, that is, indeterminate) evaluation -- it's not actually 0/0, it's 2h/h where h is close to but not equal to 0. We don't need to define 0/0=1 in order to have either algebra or calculus work. We choose to define 0/0=1 in certain circumstances which make certain calculations go more smoothly, and we choose not to define 0/0 in other circumstances where it's either unnecessary or potentially misleading.

zodiac 13y ago

I see that you are defining 0/0 to be a value such that f(x) = x/x has no discontinuity at x=0. But your chosen value means that the function g(x) = x^2 / x is discontinuous at x=0, and the function h(x) = -x/x also has a discontinuity, etc.

einhverfr 13y ago
I am defining continuity such that single point that is undefined does not break continuity if and only if it has the same limit when approached from both sides. This is because, as I say, a point is arbitrarily small and therefore if the limit is the same on both sides, it's simpler to treat it as still a continuous function. Otherwise I think you get some problems which break algebra, because multiplying of dividing by variables can break the domains of functions. If we have f(x) = 1, then this should be the same even if we multiply both sides by x. x * f(x) = x, which results in f(x) = x/x. You break algebra by allowing such straight-forward manipulations (which really are the core of algebra) to tamper with the original domain of the function. Just as you can't just treat a square root as a core algebraic process (because it is not a function), you would no longer be able to divide by a variable, since that would be undefined wherever the variable is 0. Edit Also it occurs to me that treating x/x as undefined at 0 would also man that the first derivative of f(x) = x would also be undefined at 0, which can't work either. So therefore treating x/x as discontinuous at 0 breaks both algebra and calculus.