Poll: 0.999... = 1

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Rabid Toilet

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emeraldrafael said:
Rabid Toilet said:
Firstly, you cant because 0.9...! =/= 1.
Logical fallacy, you're saying 0.99... =/= 1 because 0.99... =/= 1.

Secondly 0^2 =/= 1. 2^0 = 1. But thats oka.
Have you not taken algebra? Solve the equation I presented and you get two values of x.
0^2 =/= 1 because 0 =/= 1. However, 0^2 = 0 and 1^2 = 1. Just because x has different values doesn't mean you can plug different ones into the same equation at the same time.
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.
Do you also not understand limits? It's the existence of limits that proves .99... = 1.

lim x -> infinity of 1/(10^x) <--- the theoretical distance between .99... and 1
lim = 0 <--- the distance between .99... and 1

If there is no distance between two numbers, they are the same number.
 

acer840

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Rabid Toilet said:
acer840 said:
Rabid Toilet said:
acer840 said:
havass said:
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.
That's because it's my equation that is in the first quote box.
 

IMakeIce

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If you go to college and take a Discrete Math course, you're likely to be introduced to this proof. I had seen it in high school and didn't really believe it because of the "logical" problems with it. Try to turn your intuition off for a moment; if you can't do that, you really don't have much hope in higher level math anyway.

My professor explained it after he showed the proof in a way I hadn't heard before, but helped me understand a lot better.

.(9) does, in fact = 1, here's the thing

No one would ever dispute the fact that 1/3 = .(3) however, you're talking about two numbers living in totally different worlds. 1/3 lives in Q (rational numbers), .(3) lives in R (real numbers). These are two ways, in different number sets to express the same idea (after all, numbers are simply ideas). Likewise, we have 1 (which lives in N (natural numbers), or Z, Q, or R for that matter) and .(9) which lives in R (real numbers).

1/3 + 1/3 + 1/3 does, in fact, = 1 You aren't going to dispute that, because it is fact.
.(3) + .(3) + .(3) does, in fact, =.(9) You aren't going to dispute that, because it is fact.

You can always substitute x for x. 1/3 + 1/3 + 1/3 = 1/3 + 1/3 + 1/3 = .(3) + .(3) + .(3)... 1 = .(9)

You're thinking about this in absolute terms when numbers are not absolute. We have different ways of representing many numbers. Because they are in different number sets we perceive them differently; they are not different in the slightest.
 

Ravek

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Rabid Toilet said:
I had the same reaction, but dammit, it's just so tempting to prove them wrong.
But as you should have noticed by now, logic doesn't work on these people.
 

Coldie

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emeraldrafael said:
Secondly 0^2 =/= 1. 2^0 = 1. But thats oka.
You misunderstand how variables work. The variable has only one value at a time, but can have an infinite amount of values. So in this case, the solutions for x[sup]2[/sup] = x are, indeed, x = 0, 1.

0[sup]2[/sup] = 0
1[sup]2[/sup] = 1

QED.

SomethingAmazing said:
No.

.9999 is lesser than 1. Very very slightly lesser, but still lesser.
Would you kindly provide this 'very very slight' difference between 1 and 0.(9)? I'm very very much interested in seeing it.

I won't be surprised at all if said difference is 0.
 

blankedboy

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BlacklightVirus said:
PoisonUnagi said:
BlacklightVirus said:
havass said:
crudus said:
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.
No matter how many threes you have, it still won't quite be a third. Sorry.
 

Rabid Toilet

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acer840 said:
That's because it's my equation that is in the first quote box.
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...
The original equation from the first page.
 

Sprntr_Zomby

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Ok for simplicities sake and to end this arguement
x=.99999 (the amount of nines doesn't matter it is true for any number of nines, I'm just truncating at a smaller amount of nines.)
10x=9.9999 (this is the step that everyone gets wrong as multipling (.9*.1^n)! by 10 gets (9*.1^n)!, not (9*.1^(n+1))!. (9*.1^(n+1))! is 9.99999 when n+5)
thus 9.9999-.99999=8.99991=9x
8.99991/9=9x/9,8.99991/9=.99999
QED x=.99999.

Try this with a calculator for any amount of nines and you'll see that my math is right.
 

Rabid Toilet

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PoisonUnagi said:
BlacklightVirus said:
PoisonUnagi said:
BlacklightVirus said:
havass said:
crudus said:
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.
No matter how many threes you have, it still won't quite be a third. Sorry.
If you have an infinite number of threes, it does indeed equal a third.
 

crazy_egyptian

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Ok i completely agree with this point that 0.999... = 1 and i have had it proven to me by one of my maths professors in a way that has been said earlier in the thread, the one that states that if 1 =/= 0.999..., what is the number in between?, but my point here is, i don't see a poll in this thread.
 

Ravek

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IMakeIce said:
No one would ever dispute the fact that 1/3 = .3333 however, you're talking about two numbers living in totally different worlds. 1/3 lives in Q (rational numbers), .(3) lives in R (real numbers). These are two ways, in different number sets to express the same idea (after all, numbers are simply ideas). Likewise, we have 1 (which lives in N (natural numbers), or Z, Q, or R for that matter) and .(9) which lives in R (real numbers).
A tempting fairytale, but nonsense. If x is in S, then x = y implies y is also in S. 0.33(3) is also a rational number, because it's the exact same thing as 1/3. By the same reasoning, 1/3 is also a real number. Of course, all rational numbers are also real numbers, so that was obvious anyway.

Your general point is okay though: these numbers are entities that operate under strict predefined rules. If you don't follow the rules when manipulating them, you're going to end up with nonsense.
 

Coldie

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PoisonUnagi said:
No matter how many threes you have, it still won't quite be a third. Sorry.
When you have an infinite amount of threes, it will be exactly a third. But I'm not sorry, that's just how math works.
 

zoulza

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Sprntr_Zomby said:
Ok for simplicities sake and to end this arguement
x=.99999 (the amount of nines doesn't matter it is true for any number of nines, I'm just truncating at a smaller amount of nines.)
10x=9.9999 (this is the step that everyone gets wrong as multipling (.9*.1^n)! by 10 gets (9*.1^n)!, not (9*.1^(n+1))!. (9*.1^(n+1))! is 9.99999 when n+5)
thus 9.9999-.99999=8.99991=9x
8.99991/9=9x/9,8.99991/9=.99999
QED x=.99999.

Try this with a calculator for any amount of nines and you'll see that my math is right.
It's true for any finite number of nines, but not when the number of nines is infinite.

Infinity does not work that way.
 

Rabid Toilet

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Sprntr_Zomby said:
Ok for simplicities sake and to end this arguement
x=.99999 (the amount of nines doesn't matter it is true for any number of nines, I'm just truncating at a smaller amount of nines.)
10x=9.9999 (this is the step that everyone gets wrong as multipling (.9*.1^n)! by 10 gets (9*.1^n)!, not (9*.1^(n+1))!. (9*.1^(n+1))! is 9.99999 when n+5)
thus 9.9999-.99999=8.99991=9x
8.99991/9=9x/9,8.99991/9=.99999
QED x=.99999.

Try this with a calculator for any amount of nines and you'll see that my math is right.
I bolded the problem for you. You can't do any of this with a calculator, because a calculator can only deal with finite numbers of decimals. We're talking about the nines going on for infinity, in which case there is no infinity - 1, because that is still infinity, which can't be represented with a calculator.
 
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PoisonUnagi said:
BlacklightVirus said:
PoisonUnagi said:
BlacklightVirus said:
havass said:
crudus said:
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.
No matter how many threes you have, it still won't quite be a third. Sorry.
I think you're trolling. 1/3 will always be less than 1.(3). That's a mathematical fact.
 

Rabid Toilet

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BlacklightVirus said:
PoisonUnagi said:
BlacklightVirus said:
PoisonUnagi said:
BlacklightVirus said:
havass said:
crudus said:
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.
No matter how many threes you have, it still won't quite be a third. Sorry.
I think you're trolling. 1/3 will always be less than 1.(3). That's a mathematical fact.
I think he meant to say 0.33..., not 1.33...

I don't think anyone would argue that 1/3 > 1.333..., but arguing that adding threes to the end of 0.33 will always be less than 1/3 makes more sense.
 

smithy_2045

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PoisonUnagi said:
BlacklightVirus said:
PoisonUnagi said:
BlacklightVirus said:
havass said:
crudus said:
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.
No matter how many threes you have, it still won't quite be a third. Sorry.
1.333... = 4/3 > 1/3 = 0.333...

SomethingAmazing said:
BlacklightVirus said:
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.
Well then your own damn numbers are confusing you. Because .9999... is still less than 1.
No it isn't.
 

gl1koz3

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Well, the definition is acceptable, because infinitely close is direct proximity.

Not much to say.