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/sci/ - Science & Math

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4898488 No.4898488[DELETED]  [Reply] [Original]

>People still think .999... = 1
>mfw

>> No.4898492

0/10

>> No.4898502

I'm willing to consider your opinion. Please offer a proof.

>> No.4898503

We know that 1 is a rational number, that is, it can be expressed as a/b. Now if 0.999... is equal to 1, it too must be a rational number. This means that there exists c/d such that c/d = 0.999... However, no such c and d exist. So 0.999... is not a rational number. Therefore 0.999... cannot be equal to 1.

>> No.4898504
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4898504

>Trolls are still obvious
>mfw

>> No.4898506

>>4898502
1,0] is a closed set. In this set all real numbers between 1 and 0 are included.
(1,0] is an open set. In this set 1 is not included, but .999.... is.
Error here. Circular reasoning by stating that since .999.... is not 1. If it is 1, then obviously .999... is NOT in the set.
How can two numbers that are equal not be in the same set? It is because they are not.
Here is why:
A set S in Rm is open if for each xεS:
Ǝex>0:Bex(x)cS
All .99… does is make ex ¬ infinitely small, but it still is bigger than 0.
Therefore these two numbers are not equal.

QED

>> No.4898507

>>4898488
Seriously OP? My faith in humanity has been lost.

>> No.4898527

>>4898503
>This means that there exists c/d such that c/d = 0.999... However, no such c and d exist.

this is a decent argument, but you need to actually show that c and d don't exist.

>>4898506
>(1,0] is an open set. In this set 1 is not included, but .999.... is.

1) weird notation. learn latex.
2) you have to show .999... is in [0, 1).

>> No.4898535

http://en.wikipedia.org/wiki/0.999......
stop fooling yourselves

>> No.4898537

If two decimal numbers first differ in the nth decimal place, and the difference between the decimals is ≥ 2, we definitely have two disparate real numbers with a difference d ≥ 10^(-n).

But, if the decimals differ merely by 1, d is smaller than 10^(-n) and we need to check the next decimal in order to get a grip of the difference. There are two cases:

1. The decimal of the larger number is ≥ 1, or the decimal of the smaller number is ≤ 8.
=> d ≥ 10^(-(n+1)).

2. The decimal of the larger number is 0 and the decimal of the smaller number is 9.
=> d < 10^(-(n+1)) and we need to check the next decimal as before.

We therefore eventually obtain two disparate real numbers unless case (2) repeats in infinity! If it does, the numbers differ by less than 10^-k for all integers k, which is to say NOT AT ALL.

/thread

>> No.4898542

>>4898535
0/10

>> No.4898545
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4898545

>>4898527
>1) weird notation. learn latex.

Learn interval notation.
(-inf, 2] U (3, inf)

Does not include -inf, includes 2.
Does not include 3, Does not include 2.

>> No.4898548

1/9=.111, 2/9=.222, ... 9/9=1
How's that?

>> No.4898558

>>4898545
if b > a, who the fuck writes (b, a] as an interval instead of [a, b)?

that's the weird thing.

and learn latex. you should never write x "epsilon" S for "x is an element of S" or more properly "<span class="math">x \in S[/spoiler]".

>> No.4898561

fucking read Rudin's analysis you dickweeds...

My jimmies have been rustled