Author Topic: .9 repeating equals 1?  (Read 21318 times)

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Offline Scipi

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Re: .9 repeating equals 1?
« Reply #30 on: December 12, 2013, 06:44:57 pm »

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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...

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Offline Xeda112358

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Re: .9 repeating equals 1?
« 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 :D


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Re: .9 repeating equals 1?
« 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 :P

EDIT: For the record, I don't think calculators count as valid proof :P
« Last Edit: December 17, 2013, 09:35:15 pm by Deep Thought »




Offline bsl

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Re: .9 repeating equals 1?
« 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?
« Reply #34 on: December 16, 2013, 12:52:31 am »
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Offline Converge

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Re: .9 repeating equals 1?
« 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.