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Moosey should read the thread. We proved it with Algebra

1/9 = .1111111111111111111111111...
8/9 = .8888888888888888888888888...

If every eight and one add on to makes nines, I don't understand how it gets to 1. All of my math teachers are new this year, so they are kind of crappy in discussing this kind of stuff.

It's really weird.. 5000 years ago, none of this stuff even mattered.. it's like now the world is made up of pictures we call numbers. I don't even know what I'm rambling about .

No it's not. At sometime at the end, after multiplying by 10, you'll get an extra ...9. This is impossible; where does this extra value come from?


Taking the limit is not the same thing as rounding. A limit simply gives what happens to f(x) as x get's closer to that value. If we have the complex equation:
{ y = x for (-infinity, 0) and (0, infinity)
{ y = 10 for 0

Then, the limit of x --> 0 is 0; but it's not actually 0.

The fact is .999..., while close to 1, is not equal to 1.



I agree, .9999... does in fact approach 1. Does this make it equal to? Nope.

Ya, my AP Calculas teacher spent 30 minutes explaining this to our class because people kept asking questions. It IS true, mathematical forumulas all point out that .9 repeating equals 1.

I merely state that it is all a matter of opinion. Because numbers can be infinite, .9999 add infinite will always be .9999 add infinite. However if you simply round it off it will only then be 1.

This was a trick, although a rather interesting one. Cheeze is right.

To explain this appropriately.

Ok, now if you look at the multiplication of 0.9999*10, you'll obviously notice that the decimal place moves over one space. But this is not all that happens when mutiplying by ten.
When you multiply a number by 10, it adds a 0 to the end of the number. In any normal number this 0 makes absolutely no difference, except in problems like this.
The fact that there is a zero at the end of an infinite number makes the answer different.

GIVEN: x = 0.999...
10x = 9.999...0
10x - x = 9.999...0 - 0.999...
9x = 9.000...1
x = 0.999...

It's not correct, but you have to think rather abstractly to get the right answer. (Adding a zero to the end of an infinitly long number. )

There are infinite nines. Either way, assuming that they all move up a place, then we have a zero at the end. A zero at the end is essentially dropped, just like a zero before any real number.

0100 = 100
.990 = .99

By definition, it wouldn't be .999... anymore then.

Depending on who you ask, to break this down, is that .9999 will either equal 1 or .9999 will always be .9999

Exactly. It's going to one. It's not exactly one. That's why when you take the limit, it is exactly 1!

Do you know the reason for the decimal? Because the number is less than a whole... Which means they are unequal numbers. Duh.

Hmm... I see why you would think 9/9 equal .99999999999 due to the patern but it equals 1.

I mean, if you have 9 pies and 9 people, each person is going to get a pie, there won't be .0000000000000000000000000000001 left over.

I'm to lazy to do any of the algebra work here but anything divided by itself = 1

When you simplify it, it simplifies to 1.

X = .999

10X = 9.990

10X-X= 8.991

8.991\9 = .999

The limit of .999~ is 1. But that doesn't make .999~ equal to one.

Rantent by sayin you add a zero when multiplying with ten would also mean adding a zero to the X as well. So its still proves the equation is right.

You cant add a zero to x. X is constant it cant be modified. By using that logic.


X = 10

10x = 100

x = 100??

Well, it really depends what 1 you are talking about.
1.000... ≠ 0.999...
However:
1 = 0.999...
This is due to rules of significant digits.

I never expressed .999... as a limit. I've always used .999...

I don't want it approaching any number in terms of limits. Thus, I'm not taking a limit and it does not equal 1.

Besides, are you going to tell me 1/8 is also a limit approaching some rational number? Or pi? Both of these functions can be expressed in terms of limits.

weeee topics with no answer. O boy!!

Both can be right, let's assume that 1 = .9999999999999....etc, repeating, repeating...

what do you get out of winning this? proving that all of our mathmatics are pure useless? I dunno, I'll stop posting here before i'm reported for spam ;P

Limits - It defines the value y of a function f(x) to which the function's value of a given number aproaches to, but is not equal to.

0.999... does not equal 1 and will never equal to 1.

A limit will only say the value to which a function get close too, in some cases it will never equal to the limit value because there is a hole in the function.

I can prove that 0.999... does not equal 1 in every single mathematics.

Algebra - If a number is subtracted from another, and the result is 0, then the two numbers are equal.

1 - 0.999... = x
0.000...1 = x

1 and 0.999... are not equal.

Geometry - If a right triangle has 2 sides with the same lenght, then the third side will equal to the same lenght of the first sides times the square root of 2.

1[sup]2[/sup] + 1[sup]2[/sup] = x[sup]2[/sup]
2[sup]1/2[/sup] = x


0.999...[sup]2[/sup] + 0.999...[sup]2[/sup] = x[sup]2[/sup]
[2(0.999...8000...1)][sup]1/2[/sup] = x
[1.999...6000...1][sup]1/2[/sup] = x

The answers are not the same, therefore the sides of the triangles are different.

Trigonometry - I suck at Trig and can't think of soemthing

Calculus - Definite Integrals

[sup]1[/sup]
∫f(x)dx
[sub]0.999...[/sub]

Let u=g(x) which is part of f(x).

[sup]g(1)[/sup]
∫[u(du/f'(x))]
[sub]g(0.999...)[/sub]

Conclusions:

1) z will equal 1 and 0.999, y will equal infinite.

lim[sub](n-->y)[/sub][g(z)][sup]n[/sup] = infinite

So as y grows and grows, you will see a greater difference between g(1) and g(0.999).

2) If 1 and 0.999... were equal, then g(1) and g(0.999...) would be equal and therefore, giving the answer of 0 because the fundamental theorem of Calculus states that:

[sup]b[/sup]
∫f(x)dx = F(b) - F(a) when F(x) is a primitive of f(x).
[sub]a[/sub]

Those who have studied Calculus know that the answer would be 0, it doesn't matter how the primitive function come out, but we do know that if the integral is solved using the given theorem, it would give an answer with high probability of being lim[sub](x-->n)[/sub] = 0 meaning that it is not 0, it approaches the value, there will also be an assymptote because it will never equal to 0 because if it was 0, there wouldn't be an integral. Which brings me to this variation of the fundamental theorem of Calculus.

[sup]a[/sup]
∫f(x)dx = 0
[sub]a[/sub]

3) 1/0.999... = 1.000...1

4) Moose's Calculator Comment

5) The following asumption is wrong:


If 0.999... = 1 then you could say that 1 = 0.999... by using the transitive property and by using the same method, you should arrive at the same conclusion, which you don't, you arrive at 1=1 and therefore breaking the transitive law which states that if a=b then b=a and in this case, a=b but b≠a therefore making a=b null, meaning a≠b.

Edited: Fixed some of the Calculus Conclusions to make them more accurate.

this topic is silly. guys this is proven in science. X = 1. GET IT?
because (0.9 dot above 9) goes on forever, there can be no 000000000001 you can put in. Silly things. If that 9 goes on forever, there is no difference between a 1 and a (0.9 dot above 9.) It's mathematically proven. plus DKTB and few others proved it true anyways.
think of it logically. and easily. 0.9999999999999999... goes on. and never stops.
the whole point is, the gap is too small to be considered different. it's like a perfect another way to say 1. but longer. it's like human, you wouldn't see a difference between 2 same boxes but in atomic scale, they are different. so. in logic its not different. in science it's not different at all. really. and. since they go on forever... theres no 0 in the end, and all it will do is bring up a number one digit that will not do anything down there because its "FOREVER"

Lithium, I think you're right...but beer_keg just totally pwned the thread ;P

i bet beer keg wouldn't be able to tell the difference if it was in boxes. not numbers