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Omnimaga
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.9 repeating equals 1?
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Topic: .9 repeating equals 1? (Read 19972 times)
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Scipi
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Re: .9 repeating equals 1?
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Reply #30 on:
December 12, 2013, 06:44:57 pm »
Quote from: Jim Bauwens on December 12, 2013, 04:01:46 am
She also posted this video
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Late last night, Quebec was invaded by a group calling themselves, "Omnimaga". Not much is known about these mysterious people except that they all carried calculators of some kind and they all seemed to converge on one house in particular. Experts estimate that the combined power of their fabled calculators is greater than all the worlds super computers put together. The group seems to be holding out in the home of a certain DJ_O, who the Omnimagians claim to be their founder. Such power has put the world at a standstill with everyone waiting to see what the Omnimagians will do...
Wait... This just in, the Omnimagians have sent the UN a list of demands that must be met or else the world will be "submitted to the wrath of Netham45's Lobster Army". Such demands include >9001 crates of peanuts, sacrificial blue lobsters, and a wide assortment of cherry flavored items. With such computing power stored in the hands of such people, we can only hope these demands are met.
In the wake of these events, we can only ask, Why? Why do these people make these demands, what caused them to gather, and what are their future plans...
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Re: .9 repeating equals 1?
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Reply #31 on:
December 12, 2013, 06:53:56 pm »
It is most definitely 1, and that is a cool feature of how the Reals work. Think of it this way:
Try to find a number between .9999... and 1. You can't and that is why they are equal!
I had to take a proof-writing course with a pretty neat professor and he once asked me something like this for homework:
Let f(x) be the function that returns the value of the first digit after the decimal point in base 10. Is this a function? (Meaning that f(x)=f(x))
On initial inspection, I thought it was indeed a function and I wrote the proof for it. That was until I realised the morning it was due that .999... =1 and f(.99...) returns 9 and f(1) returns 0, yet the inputs are identical! Crazy shtuff, that
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Re: .9 repeating equals 1?
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Reply #32 on:
December 12, 2013, 10:41:31 pm »
I've had three math teachers prove it to me in three different ways (algebraically, with calculus, and then with real analysis), but I've personally always accepted it as fact without any proof. It might have something to do with using calculators a lot—type
.99999999999
on your graphing calculator and see what it comes up with. Calculators never lie
EDIT:
For the record, I don't think calculators count as valid proof
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Last Edit: December 17, 2013, 09:35:15 pm by Deep Thought
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bsl
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Re: .9 repeating equals 1?
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Reply #33 on:
December 15, 2013, 04:05:46 pm »
Standard analysis they are equal
Non-standard analysis [the hyper-reals], they are not equal
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Re: .9 repeating equals 1?
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Reply #34 on:
December 16, 2013, 12:52:31 am »
Relevant xkcd
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Re: .9 repeating equals 1?
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Reply #35 on:
January 10, 2014, 02:45:02 pm »
The following is an argument that makes sense to me: two real numbers are different if and only if their difference (the result of subtracting them) is not 0. If we look at the result of subtracting .9repeating from 1, it is certainly nonnegative, but it is also less than .1, .01, .001, and so on. So whatever this number is, if it is positive, we can find a (negative) power of 10 that is smaller than it and that it must be smaller than. Since two numbers x and y cannot have both x>y and x<y, we have arrived at a contradiction. Therefore, this number cannot be positive, and must be 0.
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.9 repeating equals 1?