4 ms·

Euler lived in the 1700s. Not sure how that would have affected your training unless you’re much older than average. Cauchy and Taylor both formalized many con

Euler lived in the 1700s. Not sure how that would have affected your training unless you’re much older than average. Cauchy and Taylor both formalized many concepts, again in the 18th and 19th century. What are you thinking is ad box?

soVeryTired 7y ago
In high school mathematics, you're not really given the definition of a limit. Consider the definition of the derivative limit as h -> 0 of (f(x+h) - f(x)) / h. That's well-defined on (0, inf) but not on [0, inf). So you can't just evaluate at h=0 and be done with it. The intuition is 'as h gets smaller and smaller, the ratio gets closer and closer to a new function of x'. But many high-school students aren't given a clear definition of what it means for one function to be 'close' to another, or what it means for x to 'get smaller and smaller'. To see the confusion more clearly, try having the debate about whether 0.999... = 1 with someone who doesn't understand what a limit is.
- dxdy 7y ago
  Let n = 0.999... Then multiply both sides of the equality by 10 so that we have 10n = 9.999...Then subtract n from both sides of the resulting equality to get 9n = 9.000...Finally, divide both sides by 9 and voila we have n = 1 which is what we wanted to show.
  - soVeryTired 7y ago
    My gut feeling is that that proof isn't quite correct, since you haven't used the notion of a limit anywhere. There's a fundamental fact about convergence of geometric series that you need to use. I think your proof goes wrong since you haven't justified how arithemtic operations work with infinite decimals. AFAIK the only way to add non-terminating decimals is to convert them to fractions (or sequences of fractions as with pi, e, etc), add the fractions, and convert them back. So if you convert 0.999... and 9.999... to fractions, you've assumed the conclusion. To play devil's advocate, I can try to rephrase your proof without infinite decimal arithmetic as follows. Assume n = 0.999... = 1 - epsilon, where epsilon is 'infinitesimal' (an ill-defined version of not-quite-zero). We'd like to show that epsilon is zero. 10n = 9.999 = 10 - 10epsilon 9n = 9.999 - (1 - epsilon) = 9 - 9epsilon 9n = 8.999 + epsilon = 9 - 9epsilon The only way to get the epsilons to cancel is to assume epsilon = 0, which is to assume the conclusion.
- darthazad 7y ago
  I had trouble with the whole 0.999... = 1 thing as a young lad until I noticed something about fractions. Any digit over 9 equals a decimal of zero point itself repeating. I.E. 1/9 = 0.111... 2/9 = 0.222... 3/9 = 1/3 = 0.333... ... 1 = 9/9 = 0.999...
  - soVeryTired 7y ago
    This one is better than the other child comment, but you still need to show that the limit of a sum is the sum of the limits. Not as easy as you might think, and not usually done in high school!
- deleted 7y ago
  [deleted]

soVeryTired 7y ago

In high school mathematics, you're not really given the definition of a limit. Consider the definition of the derivative limit as h -> 0 of (f(x+h) - f(x)) / h. That's well-defined on (0, inf) but not on [0, inf). So you can't just evaluate at h=0 and be done with it. The intuition is 'as h gets smaller and smaller, the ratio gets closer and closer to a new function of x'. But many high-school students aren't given a clear definition of what it means for one function to be 'close' to another, or what it means for x to 'get smaller and smaller'. To see the confusion more clearly, try having the debate about whether 0.999... = 1 with someone who doesn't understand what a limit is.

dxdy 7y ago
Let n = 0.999... Then multiply both sides of the equality by 10 so that we have 10n = 9.999...Then subtract n from both sides of the resulting equality to get 9n = 9.000...Finally, divide both sides by 9 and voila we have n = 1 which is what we wanted to show.
- soVeryTired 7y ago
  My gut feeling is that that proof isn't quite correct, since you haven't used the notion of a limit anywhere. There's a fundamental fact about convergence of geometric series that you need to use. I think your proof goes wrong since you haven't justified how arithemtic operations work with infinite decimals. AFAIK the only way to add non-terminating decimals is to convert them to fractions (or sequences of fractions as with pi, e, etc), add the fractions, and convert them back. So if you convert 0.999... and 9.999... to fractions, you've assumed the conclusion. To play devil's advocate, I can try to rephrase your proof without infinite decimal arithmetic as follows. Assume n = 0.999... = 1 - epsilon, where epsilon is 'infinitesimal' (an ill-defined version of not-quite-zero). We'd like to show that epsilon is zero. 10n = 9.999 = 10 - 10epsilon 9n = 9.999 - (1 - epsilon) = 9 - 9epsilon 9n = 8.999 + epsilon = 9 - 9epsilon The only way to get the epsilons to cancel is to assume epsilon = 0, which is to assume the conclusion.
darthazad 7y ago
I had trouble with the whole 0.999... = 1 thing as a young lad until I noticed something about fractions. Any digit over 9 equals a decimal of zero point itself repeating. I.E. 1/9 = 0.111... 2/9 = 0.222... 3/9 = 1/3 = 0.333... ... 1 = 9/9 = 0.999...
- soVeryTired 7y ago
  This one is better than the other child comment, but you still need to show that the limit of a sum is the sum of the limits. Not as easy as you might think, and not usually done in high school!
deleted 7y ago
[deleted]