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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 thu

by ndeine 12y 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 12y 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 12y 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.
  - einhverfr 12y ago
    > 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. Sure it does, because if that is the case, you restrict your domain when you divide by a variable expression. If you divide both sides by x-1, then you effectively rule out 1 from the domain. That's the problem. Now this is not the same as 0/0. The point is that 0/0 is only undefined when it persists after simplification and only because you can't define a relationship between the two zeros. I.e. 0/0 is undefined because 2x/x, 52x/x, and x^2/x give you three different answers as x->0. That doesn't mean that every function which has not been reduced and can transiently evauate to 0/0 is treated as non-continuous. Regarding derivatives, this highlights the problem because to solve the first derivative of a variable to a simple exponent, you multiply by the exponent and divide by the variable (x^2 becomes 2x, 2x becomes 2, and so forth). What this means is that you may be dealing with a limit but the limit defines a function, which is something like 2x^2/x for the derivative of x^2 and 2x/x for the second derivative. Unless you allow simplification before determining whether the function is continuous, these things don't make sense. If you allow reduction first, then x/x^2 is undefined where x = 0, but x/x is not, because you can reduce it to 1 before applying any further logic. Both may appear to evaluate to 0/0 however. There are a huge number of things that seem to break in algebra and calculus if one treats x/x as non-continuous and undefined. The simpler solution is to allow reduction to 1 before determining that it is undefined. (of course 52x/x would reduce to 52 instead, again showing why 0/0 is oversimplifying the problem).
    - tomp 12y ago
      > Sure it does, because if that is the case, you restrict your domain when you divide by a variable expression. If you divide both sides by x-1, then you effectively rule out 1 from the domain. > That's the problem. That's the problem that a mathematician must handle. The solution to the equation x * f(x) = x is very simple: x is either 0 or such that f(x) = 1.
    - lotharbot 12y ago
      > "you restrict your domain when you divide by a variable expression" Why is this a problem? Whenever you perform an operation that has a restricted domain, you restrict your domain. This may result in an actual "not defined at x=1" result, or simply "the value at x=1 is found through an alternative method" result. > "0/0 is only undefined when it persists after simplification" When you're working in the context of limits, it wasn't an actual 0/0 to begin with; it was near-0/near-0, which is perfectly OK to simplify. The limit defines a function that already has a restricted domain -- h->0 means h is not actually zero. The expression naively evaluating to 0/0 simply tells you that you need to do more work to properly evaluate it -- 0/0 is not the actual result. Note that using the limit to find the derivative gives you a function that you'd like to be continuous in x, but the divide-by-zero is in h. Consider f(x)=x^2. The derivative is lim h->0 [(x+h)^2 - x^2 ] /h lim h->0 [ x^2 + 2xh + h^2 - x^2 ] /h lim h->0 [ 2xh + h^2 ] / h lim h->0 [2x + h] * h/h since h does NOT equal zero, we can treat h/h=1, and the limit trivially collapses to 2x. Note that we never had the variable x in the denominator of our fraction; we never placed a restriction on x or suggested anything about a discontinuity relative to x. We only restricted h, which was already restricted by the limit itself.

by ndeine 12y ago

einhverfr 12y 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 12y 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.
- einhverfr 12y ago
  > 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. Sure it does, because if that is the case, you restrict your domain when you divide by a variable expression. If you divide both sides by x-1, then you effectively rule out 1 from the domain. That's the problem. Now this is not the same as 0/0. The point is that 0/0 is only undefined when it persists after simplification and only because you can't define a relationship between the two zeros. I.e. 0/0 is undefined because 2x/x, 52x/x, and x^2/x give you three different answers as x->0. That doesn't mean that every function which has not been reduced and can transiently evauate to 0/0 is treated as non-continuous. Regarding derivatives, this highlights the problem because to solve the first derivative of a variable to a simple exponent, you multiply by the exponent and divide by the variable (x^2 becomes 2x, 2x becomes 2, and so forth). What this means is that you may be dealing with a limit but the limit defines a function, which is something like 2x^2/x for the derivative of x^2 and 2x/x for the second derivative. Unless you allow simplification before determining whether the function is continuous, these things don't make sense. If you allow reduction first, then x/x^2 is undefined where x = 0, but x/x is not, because you can reduce it to 1 before applying any further logic. Both may appear to evaluate to 0/0 however. There are a huge number of things that seem to break in algebra and calculus if one treats x/x as non-continuous and undefined. The simpler solution is to allow reduction to 1 before determining that it is undefined. (of course 52x/x would reduce to 52 instead, again showing why 0/0 is oversimplifying the problem).
  - tomp 12y ago
    > Sure it does, because if that is the case, you restrict your domain when you divide by a variable expression. If you divide both sides by x-1, then you effectively rule out 1 from the domain. > That's the problem. That's the problem that a mathematician must handle. The solution to the equation x * f(x) = x is very simple: x is either 0 or such that f(x) = 1.
  - lotharbot 12y ago
    > "you restrict your domain when you divide by a variable expression" Why is this a problem? Whenever you perform an operation that has a restricted domain, you restrict your domain. This may result in an actual "not defined at x=1" result, or simply "the value at x=1 is found through an alternative method" result. > "0/0 is only undefined when it persists after simplification" When you're working in the context of limits, it wasn't an actual 0/0 to begin with; it was near-0/near-0, which is perfectly OK to simplify. The limit defines a function that already has a restricted domain -- h->0 means h is not actually zero. The expression naively evaluating to 0/0 simply tells you that you need to do more work to properly evaluate it -- 0/0 is not the actual result. Note that using the limit to find the derivative gives you a function that you'd like to be continuous in x, but the divide-by-zero is in h. Consider f(x)=x^2. The derivative is lim h->0 [(x+h)^2 - x^2 ] /h lim h->0 [ x^2 + 2xh + h^2 - x^2 ] /h lim h->0 [ 2xh + h^2 ] / h lim h->0 [2x + h] * h/h since h does NOT equal zero, we can treat h/h=1, and the limit trivially collapses to 2x. Note that we never had the variable x in the denominator of our fraction; we never placed a restriction on x or suggested anything about a discontinuity relative to x. We only restricted h, which was already restricted by the limit itself.