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Well, here you reduced 1=.9999... to 1/3=0.333... What if I don’t believe that second equation.

by ant6n 7y ago

ginko 7y ago
As in the 0.333... will stop at some point? That would still mean that 3 time 0.333... with a LOT of 3s end up being being 1. I also figure it's a bit more intuitive for pupils to just try out calculating the decimal representation of 1/3 and seeing that it'll just keep going forever.
- deleted 7y ago
  [deleted]
- tomtomtom1 7y ago
  more like 1/3 != 0.3333 .... as in 1/3 does not have a decimal representation. you can only approximate it but never reach it.
bhl 7y ago
What if we recursively defined 1/3? This will allow us to ignore the infinite 0.333... for a second. As an example, let 1/3 = 0.33 + 1/100(1/3). A definition of a third that most people agree with is that if we multiplied that value by 3, we should get 1. Let's check the right hand side: 3 * (0.33 + 1/100(1/3)) = 0.99 + 1/100 * 1 = 0.99 + 0.01 = 1. Great! What other expressions for a 1/3 can we come up with? If you agreed with the previous statement, then you must surely also agree that 1/3 = 0.333 + 1/1000(1/3). Inductively, we should be able to come up with a general formula that 1/3 = bar(3, n) + 1/pow(10, n)(1/3), where bar(3, n) = sum i = 1 to n 3/pow(10, i). We can check that bar(3, 2) = 0.3 + 0.03 = 0.33, and that our first example fits this formula. Intuitively, this formula is giving us a way to represent 1/3 in terms of n decimal places of accuracy and a recursive term. The question is now, what happens when we run that formula with n to infinity? An infinite level of accuracy! That expression is equal to 0.333... as we have defined. The right term, 1/pow(10, n)(1/3), goes to 0, so we can discard that. The left hand side, is a geometric series with 1/10 as the power, and a scalar multiple of 3. Using a closed sum formula for that [1], we can see that the left hand side goes towards 1/3. (Apply the formula from Wikipedia, but remember our index starts off at 1, not 0.) In the end, we have found that 0.333... = n->infty bar(3, n) + 1/pow(10, n)(1/3) = 1/3 [1] https://en.wikipedia.org/wiki/Geometric_series#Sum https://en.wikipedia.org/wiki/Geometric_series#Sum

Well, here you reduced 1=.9999... to 1/3=0.333... What if I don’t believe that second equation.

by ant6n 7y ago

Well, here you reduced 1=.9999... to 1/3=0.333... What if I don’t believe that second equation.

As in the 0.333... will stop at some point? That would still mean that 3 time 0.333... with a LOT of 3s end up being being 1. I also figure it's a bit more intuitive for pupils to just try out calculating the decimal representation of 1/3 and seeing that it'll just keep going forever.

deleted 7y ago
[deleted]
tomtomtom1 7y ago
more like 1/3 != 0.3333 .... as in 1/3 does not have a decimal representation. you can only approximate it but never reach it.

bhl 7y ago

What if we recursively defined 1/3? This will allow us to ignore the infinite 0.333... for a second. As an example, let 1/3 = 0.33 + 1/100(1/3). A definition of a third that most people agree with is that if we multiplied that value by 3, we should get 1. Let's check the right hand side: 3 * (0.33 + 1/100(1/3)) = 0.99 + 1/100 * 1 = 0.99 + 0.01 = 1. Great! What other expressions for a 1/3 can we come up with? If you agreed with the previous statement, then you must surely also agree that 1/3 = 0.333 + 1/1000(1/3). Inductively, we should be able to come up with a general formula that 1/3 = bar(3, n) + 1/pow(10, n)(1/3), where bar(3, n) = sum i = 1 to n 3/pow(10, i). We can check that bar(3, 2) = 0.3 + 0.03 = 0.33, and that our first example fits this formula. Intuitively, this formula is giving us a way to represent 1/3 in terms of n decimal places of accuracy and a recursive term. The question is now, what happens when we run that formula with n to infinity? An infinite level of accuracy! That expression is equal to 0.333... as we have defined. The right term, 1/pow(10, n)(1/3), goes to 0, so we can discard that. The left hand side, is a geometric series with 1/10 as the power, and a scalar multiple of 3. Using a closed sum formula for that [1], we can see that the left hand side goes towards 1/3. (Apply the formula from Wikipedia, but remember our index starts off at 1, not 0.) In the end, we have found that 0.333... = n->infty bar(3, n) + 1/pow(10, n)(1/3) = 1/3 [1] https://en.wikipedia.org/wiki/Geometric_series#Sum https://en.wikipedia.org/wiki/Geometric_series#Sum