Poll: 0.999... = 1

[Sprntr_Zomby]

The term your looking for is significant values and the fact that all tems in an equation need the same amount of significant numbers. The original proof is incorrect as your'e adding on one extra value to 10x so they are not equal.

A series where 2=1
x=1/n from 0 to infinity, I'm only proving it out to n=10, continue on if you don't believe it. Note that multiplying by 2 is equal to multiplying by 2/1 so 1/2*2=2/2 not 2/4.
x=1-(1/2)+(1/3)-(1/4)+(1/5)-(1/6)+(1/7)-(1/8)+(1/9)-(1/10)
2x=2-(2/2)+(2/3)-(2/4)+(2/5)-(2/6)+(2/7)-(2/8)+(2/9)-(2/10)
2x=2-1+(2/3)-(1/2)+(2/5)-(1/3)+(2/7)-(1/4)+(2/9)-(1/5)
rearrainging the numbers makes2-1)-(1/2)+((2/3)-(1/3))-(1/4)+((2/5)-(1/5)+(2/7)+(2/9)
2x=1-(1/2)+(1/3)-(1/4)+(1/5)+(2/7)+(2/9).
so 2=1.
If I had done it out to n=20 the series would be perfect out to 10 and would end with a 2/11 2/13 2/15 2/17 2/19.
as x goes to infinity the error values get smaller and smaller and with an infinite amount of values would equal 0.

 

[emeraldrafael]

Snip

its provide a real world example (since math should always work in the real world).

and still, you're going ot come to you're rounding up, or making a number something it isnt. Saying that 0.9 repeating (to infinity) isnt making it infinity, its just saying you keep adding a nine to a value thats less then one.

Snip

He's still giving the same variable(x) two different values, which you cant do. Again, go find a limit on a graphing calculator at one and tell me when it gets to one. Go ahead, I'll wait. Just be sure you have a picture to back that up.

 

[Rabid Toilet]

This is why I wish more people would support my idea to introduce a base-12 numerical system

That actually wouldn't solve the problem. A classmate and I had a discussion about it. The only thing that would change would be that instead of .99... equaling 1, it would be .(whatever the number before 0 is)... equals 1.

 

[emeraldrafael]

Damn, that should have been a nine instead of an eight when it rounds up.

he said it Implies. he's changing the equation and the value of X. X as variable can not hold the same value. and again, if you want to prove it, look into the concept of graphical limits.

Uh..please don't use my choice of word in the argument. I was just putting in random words to continue with my equation. I didn't really think about the word I was using. >.>

Thats not the point of it anyway. The overall point is that youre giving a variable(x) two different values, which you cant.

 

[zoulza]

I think the problem most people who disagree with 1=.9999... have is that they think infitesimals exist. As in, there is some "smallest possible number," and adding this to .999... yields 1.

Sorry to break it to you all, but no such thing exists. There is no such thing as an infinite number of zeroes followed by a one on the end, because if the string of zeroes is infinite, there is no end!

It all comes down to infinity messing everything up. Don't try to apply your intuition to infinity; it's not going to work.

 

[PureChaos]

wasn't this done before and had something like 1000 responses?

 

[acer840]

If x = 0.999999...
Then 10x = 9.9999...
Therefore, 10x - x = 9
Which implies 9x = 9
Thus, x = 1
x also = 0.99999...

In conclusion, I have just proven 1 = 0.9999...

Well, the algebra equation is incorrect:

If x = 0.99999*
Then 10x = 9.99999*
Therefore 10x - x = 9x
Which implies 9x = 8.99999*
Thus x = 0.99999*

In conclusion, I have just proven 1x = 0.99999*

I honestly have no idea what you did there, but your algebra is most certainly not correct.

If x = 0.99999* (Right so far)
Then 10x = 9.99999* (Still good)
Therefore 10x - x = 9x (What? How does 9.99* - x = 9x?)
Which implies 9x = 8.99999* (And how does 9x = 8.999*?)
Thus x = 0.99999* (Did you divide both sides by 9? If so, how did you get that 8.99*/9 = .99* unless you assumed that .99* = 1?)

In the part:
10x - x = 10, is incorrect. Lets say "x" = "apple". 10 "apple", take away "apple" = 9?
Shouldn't this be either 10 OR 9 "apple"?

 

[blankedboy]

Every math major I have talked to and showed that to has described that as "shady".

I myself have my doubts about it, but I just can't find anything wrong in any step of the proof! Every step is perfectly logical.

There is an even simpler proof.

1/3 = 0.333...

1/3 + 1/3 + 1/3 = 3/3 = 1

But 0.333... + 0.333... + 0.333 = 0.999...

Hence 0.999 = 1

But 1/3 is greater than 1.33333...
So you've failed this round.

 

[Rabid Toilet]

Snip

its provide a real world example (since math should always work in the real world).

and still, you're going ot come to you're rounding up, or making a number something it isnt. Saying that 0.9 repeating (to infinity) isnt making it infinity, its just saying you keep adding a nine to a value thats less then one.

Snip

He's still giving the same variable(x) two different values, which you cant do. Again, go find a limit on a graphing calculator at one and tell me when it gets to one. Go ahead, I'll wait. Just be sure you have a picture to back that up.

First, he's not giving a number two different values, since .99.. = 1.

Second, you can indeed give a variable two different values.

x^2 = x
Solve for x

x = 1 and 0

Third, it gets to one at infinity. You can't technically reach infinity, but then you can't technically have an infinite amount of nines.

However, since we are talking about pure math (which is theoretical and usually has no real life applications), since you have an infinite number of nines, you have reached infinity, and it equals one.

 

[Rabid Toilet]

If x = 0.999999...
Then 10x = 9.9999...
Therefore, 10x - x = 9
Which implies 9x = 9
Thus, x = 1
x also = 0.99999...

In conclusion, I have just proven 1 = 0.9999...

Well, the algebra equation is incorrect:

If x = 0.99999*
Then 10x = 9.99999*
Therefore 10x - x = 9x
Which implies 9x = 8.99999*
Thus x = 0.99999*

In conclusion, I have just proven 1x = 0.99999*

I honestly have no idea what you did there, but your algebra is most certainly not correct.

If x = 0.99999* (Right so far)
Then 10x = 9.99999* (Still good)
Therefore 10x - x = 9x (What? How does 9.99* - x = 9x?)
Which implies 9x = 8.99999* (And how does 9x = 8.999*?)
Thus x = 0.99999* (Did you divide both sides by 9? If so, how did you get that 8.99*/9 = .99* unless you assumed that .99* = 1?)

In the part:
10x - x = 10, is incorrect. Lets say "x" = "apple". 10 "apple", take away "apple" = 9?
Shouldn't this be either 10 OR 9 "apple"?

He never says 10x - x = 10, he says that it equals 9, which it does.

 

[Rabid Toilet]

No,I am pretty damn sure that .999999... is .99999...

You're right, .9999... does in fact equal .9999...

But .9999... equals 1.

 

[Ravek]

God, not this thread again. People who don't understand maths should just stay away from it, and stop embarassing yourselves.

 

[emeraldrafael]

Snip

Firstly, you cant because 0.9...! =/= 1.

Secondly 0^2 =/= 1. 2^0 = 1. But thats oka.

Thirdly, no. And infinite number means you go back infinitely. It doesnt change the concept of what the number's value equals cause all numbers have a set value. Put it in volume if you want, 0.9...! will never fill the same as 1.

And again, go do it on a calculator that has a limit. Do it then show me what you get, and I'll admit I'm wrong. Show me this graphically. Until then, based on the concept of limits, its impossible.

 

[Rabid Toilet]

God, not this thread again. People who don't understand maths should just stay away from it, and stop embarassing yourselves.

I had the same reaction, but dammit, it's just so tempting to prove them wrong.

 

[quhouv vhqwiufv qwiu]

Every math major I have talked to and showed that to has described that as "shady".

I myself have my doubts about it, but I just can't find anything wrong in any step of the proof! Every step is perfectly logical.

There is an even simpler proof.

1/3 = 0.333...

1/3 + 1/3 + 1/3 = 3/3 = 1

But 0.333... + 0.333... + 0.333 = 0.999...

Hence 0.999 = 1

But 1/3 is greater than 1.33333...
So you've failed this round.

1/3 > 1.333...? No... just no.

 

[Coldie]

You apparently have no idea how logic/mathematical proofs work so here I go:
claiming x=0.999... is an assumption, 1=0.999... is what he intended to prove.
the reason his proof is wrong is because he failed to prove the assumption.

Understanding it requires college level math.

The assumption that is to be proven here is that '1 = 0.(9)'. The proof starts with a definition of a variable, x, as a value of '0.(9)'. You don't prove that x = 0.(9), that's what you just defined so it's true automatically. By definition. To prove the above assumption, you need to derive it, '1 = 0.(9)' from a statement known to be true, by applying whatever mathematical manipulations are allowed/possible within the limits of the System you're currently working in (e.g., if your system doesn't have multiplication or addition, the proof would have to be much more complex, if at all possible).

Your understanding of math, proofs, and your assumptions about my level of education are flawed. Or perhaps you're just trolling.

 

[2xDouble]

Allright. I've heard enough. Let me use a little bit of different logic:

A) Infinitesimal = mathematically insignificant. (definition)

B) Infinitesimal =/= 0. (definition)

C) Mathematically insignificant = ignored by Math (definition)

D) Mathematically insignificant =/= nonexistent. (definition)

if B AND D,
therefore: E) Infinitesimal AND Mathematically insignificant =/= nonexistent AND =/= 0.
if A AND E,
therefore: E[sub]2[/sub])Infinitesimal =/= nonexistent AND =/= 0

if C AND D,
therefore: F) ignored by Math =/= nonexistent

G) (not)nonexistant = existing (definition)

because E[sub]2[/sub], F and G:
Things exist within Math that Math cannot describe.

 

[quhouv vhqwiufv qwiu]

No,I am pretty damn sure that .999999... is .99999...

You're right, .9999... does in fact equal .9999...

But .9999... equals 1.

No.

.9999 is lesser than 1. Very very slightly lesser, but still lesser.

This is the problem which people not educated in math always find. As I have already said the reals do not permit numbers infinitesimally small.