[Math] Prove 0.999...=1 (Amaze your friends!)

[mgX]

the 1 thing is you using your calculator wrong. 1/3 != 0.3333333... 1/3 being an iregular number, and cannot be described with decimals. Most calculators will assign 1/3 the decimal value 0.33333... since it is merely trying to give an estimate.

2nd 1 is a FAT40. divide (a-b). Since a=b, you would be dividing by zero. Geez, dont you know programming :/


What about, x/0 = infinite?

consider, lim(y->0) x/y approaches infinity?


EDIT: recalled seeing this somewhere, and i found the source: http://en.wikipedia.org/wiki/Invalid_proof

[ReAcTiOnZ]

0.000...01 is not possible... As far as I know.

0.000... means (or atleast in this thread omg ) that the zeroes keep on coming infinitely.
That would mean a number like Infine + 1 which does not exist.

Wrong. 0.0(reoccurring)1

And before you say that isn't possible it is:

0.9r*
+
0.0r*1
=1

r* = Reoccurring

I like this thread, Geekyness is win =]

[kingofrock]

You can't have a recurring 0.0 with a 1 at the end O.o

[ReAcTiOnZ]

You can't have a recurring 0.0 with a 1 at the end O.o

Can you not in the case of a reoccurring number calculation? Anyway 0.9r is not 1 otherwise it would be 1. This thread can never be proved unless you delve into the theory of Maths itself just to provide the ultimate verdict. If you did that you would be the worlds biggest no-life ever.

[mgX]

Can you not in the case of a reoccurring number calculation? Anyway 0.9r is not 1 otherwise it would be 1. This thread can never be proved unless you delve into the theory of Maths itself just to provide the ultimate verdict. If you did that you would be the worlds biggest no-life ever.

geez ****, people are discussing this basic stuff and you dont dismiss it as heracy from the start? the bullcrap the starter of this thread made regarding the 0.9999 hypothesis, is well, bullcrap. Its because he converts things that cannot be converted in to decimals, thus ending up with an aproximation. Obviously if you reverse it, it doesnt make sense, because, ITS AN APROXIMATION DAMNIT.

[Troys]

i dont think i got 1 post in this whole thread :|

[omfgwtflolmfao]

i dont think i got 1 post in this whole thread :|

^ .

[TimmeH32]

And THIS is why I dropped out of high school. >.<'

[d0om]

1/4 = 0.25 FACT (both numbers are EXACTLY EQUAL)

0.999999 recurring = 1 FACT (both numbers are EXACTLY EQUAL)

1/3 = 0.333 recurring FACT (both numbers are EXACTLY EQUAL)

They are just different ways of displaying the same number.

Try to solve 1-0.999, you get 0.000000 recurring with an infinite number of zeros, try doing it by hand, on paper and it becomes quite clear that there will never be a 1 at the end.

the 1=2 proof is invalid as you are dividing by 0 on one line, which means the answer is undefined.

Tl;DR The OP is correct, and you are all a bunch of innumerates who failed basic maths.

[~sInX]

d0om - i understand what you're saying - 0.333r = 1/3. 0.666r = 2/3 But 0.999r is NOT 1 straight off. It's mroe of an intangible thing, you can relate the two numbers but they're regarded as completely differant. Yes, you could say there's nothing in between them, and you would be corect - however only to an extent. The term Infinite is between 0.999r and 1, they are NOT the same number. 1- 0.999r = infinite. 1 - 0.999 = 0.0001, therefore in effect it would equal to 0.0r1, aka infinite, due to it being intangible in itself and not having a literal definition.

As to your 2 = 1 theory:

Proof that 2 = 1

By the common intuitive meaning of multiplication we can see that

4 \times 3 = 3 + 3 + 3 + 3 \,

It can also be seen that for a non-zero x

x = 1_1 + 1_2 + \cdots + 1_x

Now we multiply through by x

x^2 = x_1 + x_2 + \cdots + x_x

Then we take the derivative with respect to x

2x = 1_1 + 1_2 + \cdots + 1_x

Now we see that the right hand side is x which gives us

2x = x \,

Finally, dividing by our non-zero x we have

2 = 1 \,

Q.E.D.

Read and weep

This proof is false because the differentiation is ignoring the 'x' in the subscript (off x_1 + \cdots + x_x). As you are differentiating with respect to x, it clearly cannot be held constant. Once this x is accounted for in the differentiation, using the Leibniz rule, the correct answer is obtained:

x^2 = x_1 + x_2 + \cdots + x_x

We take the derivative with respect to x

2x = (1_1 + 1_2 + \cdots + 1_x) + (x_1 + \cdots + x_1)
2x = x + x

As expected.

It is often claimed that the original proof is false because both sides of the equation in line 3 represent an integer, and so after differentiating you should get 0 = 0 (as the derivative of a constant function is 0). This is fundamentally incorrect on several levels. Firstly, a function that is only defined on the integers is *not* necessarily a constant function; secondly, the equation works perfectly well for non-integer values (for example, \textstyle\pi = \{1 + \cdots + 1\} \, \pi \, \text{times} = 1 + 1 + 1 + 0.141...) as evidenced by the fact that, when the differentiation is done correctly, the paradox is eliminated.

[Xel]

Ok... (recurring) xD

[ReAcTiOnZ]

Today I found out that 0.9r is actually 1. yes, I know iv fought against it but after some long, hard research I have proof.
x = 0.9r
x * 10 = 9.9r
10x - x = 9 = 9y
9y / 9 = 1

[McGoogle]

Chooba yr an idiot...thats exactly the same thing that Xel posted. If you didnt even notice that i dont think you should be replying to this post at all...

[ReAcTiOnZ]

Ok I missed it, Sorry. People make mistakes ok?

[burton992]

it is true 0.999.. does equal one

proof
x = 0.999
10x = 9.999
10x - x = 9.999... -0.999...
9x = 9
x = 1
0.999.. = 1