Is 1 = 0.9999......

[pyonir]

post 600 does not = 1.

 

[TuxDave]

Originally posted by: pyonir
post 600 does not = 1.

bah... you glory hog....

Post 600 + 0.999....

 

[bleeb]

0.9999.... != 1

 

[Haircut]

Originally posted by: Dufusyte

There are some values that we can't write out in full using decimal notation, this does not make them any different to other numbers.

Yes it does make them different:

- Interminal decimals cannot be written without resorting to suspicious looking superscripts and/or elipses

- Interminal decimals do not have a fixed value.

hint: if it had a fixed value you would be able to write it without resorting to suspicious notation.

Sometime notation is shorthand to save time, such as 10^100. But there are other types of notation that are actually quite insidious: they do not merely save time, but rather they mask an impossibility. Writing 0.9999... is *not* a time saver, as though the person would be able to write out the exact number if he had enough time and space. Rather, the elipsis insidiously masks the fact that 0.9999... is not a fixed value at all, which can never be written out. It is like a little shape-shifting critter that has snuck his way onto the page, and we should banish it forthwith.

OK, it's clear that you really don't know what you're talking about here, how does the fact that a number has a non-terminating decimal expansion mean that it does not have a fixed value.
If we look at the number line and pick a point at random on the line then chances are we will pick an irrational number. It is a fixed point on the line, but has a non-terminating decimal expansion.

See here for a definition of an irrational number
Link
Notice the bit where it says The decimal expansion of an irrational number is always nonterminating (it never ends) and nonrepeating (the digits display no repetitive pattern).

 

[Kyteland]

Originally posted by: MadRat

Originally posted by: RossGr
No, Pi is approximatly 22/7s .

Okay, RossGr, to make my point I want to compute Pi on a supercomputer. What is the equation?

PI is approximatly 355/113.

This is a very good equation to use when calulating PI on a computer.
PI/4=arctan(1/2)+arctan(1/3)
where arctan(x) = x - x3/3 + x5/5 - x7/7 + ...

more info here.

PS. I hope the "..." in the arctan(x) equation doen't throw you off. I know it is terribly complicated

 

[pyonir]

Originally posted by: TuxDave

Originally posted by: pyonir
post 600 does not = 1.

bah... you glory hog....

Post 600 + 0.999....

DUH!

 

[RossGr]

Interminal decimals do not have a fixed value.

This is simply not correct.

So you are telling us that .999.... wanders around? Does it ever get bigger then 1? Where does it go? How does a number wander? IF I write 2.00.... does 2 suddenly start to wander around?

Why if .999.... does not have a fixed value can it be bound above and below by arbitrally small numbers. My proof demonstrates that .999... is inside of ANY interval that you can make which is centered on 1, so where does .999.... go? It does not get outside of any interval surronding 1, how many numbers are that that if you subtract ANY number it is less then 1 AND if you add ANY number is greater then 1? The answer is that there is ONLY ! possible number that humber is 1 so ANY representation of a number which stays inside of that interval is equal to 1. This includes

.111...... Binary

.222...... base 3

.333.... base 4

Do we see a pattern? All of these are different representations of 1, your denying it does not make it false.

Your arbiirary defintions simply do not hold water, you have not given them any meaningful definition, simply wave your hands around and make claims. Show me how .999... wanders, what other values does it take? Lets be specific?

Since I can concieve of .999.... existing it cannot be swept under the rug, we have to deal with it. I have shown you how mathematicians deal with it and have even shown you how we have determined that it occupies a fixed point on the number line.

You would do well to quit living in denial and make an effort to understand this basic fact.

 

[bleeb]

1 != 0.9999...

 

[silverpig]

Really, you can't properly write "2" in decimal form. 2 is just a "special notation" for ...0002.000...

All decimal numbers are infinitely long strings of digits. Some people just decide that it's not a problem when the repeating digits are 0, and then go on to say that any other digit repeating isn't a valid number.

 

[TuxDave]

So did we win the 0.999... = 1 argument yet or not? If someone still has a counter argument, please repost it!

 

[Dufusyte]

RossGR,
as stated previously, the value of 0.99... lies between 0.99 and 1, and similarly the value of 0.9999... lies between 0.9999 and 1.

silverpig,
as for infinite 0's, they have no impact on the value of the number, therefore they are not written and do not need to be written.

 

[RossGr]

Originally posted by: Dufusyte
RossGR,
as stated previously, the value of 0.99... lies between 0.99 and 1, and similarly the value of 0.9999... lies between 0.9999 and 1.

silverpig,
as for infinite 0's, they have no impact on the value of the number, therefore they are not written and do not need to be written.

AS I SAID .999.... is inside of an ARBITARAILY small interval surrounding 1 evidently you cannot conceive of just what this implies. why you picked 4 nines is not clear, is that some magic number why not pick 100,000 nines? This btw is the DEFINITION of equality in the real numbers. Perhaps you do understand what that means? You might check the definition of definition perhaps you need basic english instruction.
Clearly you are talking about some other number system. Must be the same one matrat uses, must be interesting to have a number line full of holes. Tell me are the holes every where? or just where you choose to place them; If .999... is not 1 and there is nothing between them there must be a hole which contains.... What?


You still have not told me how to deal with the fractional representation which has an ifinitlly repeating pattern, to say it cannot exist is nonsence, I just created it, it exists in my mind and now if you can open your narrow mind it will exist in yours, now how do I deal with it? Where does it go?

When will you answer a question?

 

[silverpig]

Originally posted by: Dufusyte
RossGR,
as stated previously, the value of 0.99... lies between 0.99 and 1, and similarly the value of 0.9999... lies between 0.9999 and 1.

silverpig,
as for infinite 0's, they have no impact on the value of the number, therefore they are not written and do not need to be written.

You may want to clarify what you mean here then:

"- Interminal decimals do not have a fixed value."

2.000... whether you deem the zeros necessary or not, is an interminal decimal.


As to your reply to RossGR, you seem close to understanding exactly why 0.999... does equal 1.

I will attempt to show Ross's argument in a slightly different way.

I'm sure you can easily see that:

0.9 < 0.999... < 1.1

Now, let's "tighten the noose" a little bit:

0.99 < 0.999... < 1.01

0.999999 < 0.999... < 1.000001


As you can see, 0.999... will always fit inside the inequalities, no matter how small you make the division. The only number that can exist in the middle of the inequalites no matter how tight you make the constriction is 1.

 

[MikeA]

Enough of this .9999 = 1 or .9999 != 1 Time for a little physics. This whole thing reminded me of a debate I once got into with my physics teacher. If an object is falling from rest towards Earth infinitely far from Earth, at what speed will it hit the Earth's surface? And another question, what's infinity + infinity?

 

[silverpig]

infinity + infinity = infinity

Oh, and as to your physics problem: You can't define a position infinitely far away from another. Even if you could, the object would never hit the earth as it started out infinitely far away.

Why don't we all work on finding the non-500 page solution to Fermat's last theorem? I'll even go so far as to do the first step: Writing out the problem.

For any (n, x, y, z ) | n, x, y, z E I, n > 2, solve:

x^n + y^n = z^n

 

[MikeA]

Well, that's exactly what I said. And this is how he replied. "If the object is launched at ~11.2km/s it will stop at infinity, so if it starts falling from infinity towards earth it will hit the surface at ~11.2km/s. " So should of I received 100% on that test, or not?

Oh and infinity + infinity != infinity. Does infinity - infinity = 0?

 

[silverpig]

Originally posted by: MikeA
Well, that's exactly what I said. And this is how he replied. "If the object is launched at ~11.2km/s it will stop at infinity, so if it starts falling from infinity towards earth it will hit the surface at ~11.2km/s. " So should of I received 100% on that test, or not?

Oh and infinity + infinity != infinity. Does infinity - infinity = 0?

I'm guessing the problem is about escape velocity. If the object is given exact escape velocity, it won't stop. Tell him you'd see the heat death of the universe before the object hit the earth again

Oh, and taken from here:

It makes sense to define:

infinity + r = r + infinity = infinity
(-infinity) + r = r + (-infinity) = -infinity
infinity + infinity = infinity
(-infinity) + (-infinity) = -infinity
infinity - r = infinity
(-infinity) - r = -infinity
r - infinity = -infinity
r - (-infinity) = infinity
infinity - (-infinity) = infinity
(-infinity) - infinity = -infinity
infinity * r = r * infinity = infinity for r > 0
(-infinity) * r = r * (-infinity) = -infinity for r > 0
infinity * r = r * infinity = -infinity for r < 0
(-infinity) * r = r * (-infinity) = infinity for r < 0
infinity * infinity = (-infinity) * (-infinity) = infinity
infinity * (-infinity) = (-infinity) * infinity = -infinity
infinity / r = infinity for r > 0
(-infinity) / r = -infinity for r > 0
infinity / r = -infinity for r < 0
(-infinity) / r = infinity for r < 0
r / infinity = 0
r / (-infinity) = 0

 

[MadRat]

Originally posted by: here

You can try to make up a good set of rules, but it always leads to nonsense, so to avoid all the trouble we just say that it doesn't make sense to divide by zero.

What happens if you add apples to oranges? It just doesn't make sense, so the easiest thing is just to say that it doesn't make sense, or, as a mathematician would say, "it is undefined."

Maybe that's the best way to look at it. When, in mathematics, you see a statement like "operation XYZ is undefined", you should translate it in your head to "operation XYZ doesn't make sense."
- Dr. Tom

Dr. Tom's above statement on adding apples to oranges seems to contradict Dr. Rob.

 

[silverpig]

No they don't. 0.999... and 1 are single points on the real number line. Infinity is different.

 

[MadRat]

Is this statement below what you thought I was referring? It wasn't.

Originally posted by: silverpig
infinity + infinity = infinity

I was referring to Dr. Rob and his equations referring to infinity.

I will say, though, that "infinity" + "infinity" <> "infinity". Rather it would logically equal (2 * "infinity"), but that too is not the case. The problem is one simply cannot multiply infinity for exactly the opposite reason one cannot divide with zero. Zero (null) and infinity are subjective values that do not follow the same rules as other numbers.

 

[ElFenix]

Originally posted by: pyonir
post 600 does not = 1.

that was reply 600, the 600th post was the one before yours

 

[ElFenix]

Originally posted by: MadRat

Originally posted by: ElFenix

Originally posted by: MadRat

Originally posted by: bigredguy
If .999...=1
then couldn't 1=1.00..001?

These math geniuses deny the existence of any number ending with .000...1 because they short-sidedly deny the existence of any value at the infinite position other than 0.

thats not the infinith position! thats a finite position!

Again, you confuse a position with a value. The position can equal any number. In no way does the position affect the value,

in no way does the position affect the value? so 1.000000001 = 1.0000000000000000001?

 

[RossGr]

Why do ya'll insist on writing .00....001 ?

The best way to designate some arbitaraly small decimal number is by writting 10^-N, where N can be any positive integer. This precisily covers All possible numbers, which have commonly written in this thread, as .00...001. In addition it alleviates any possible confusion over a mystic digit in the "infinite possition". All digits in a real number are located in fixed positons which are enumerated by the N. With this notation we can know exactly how many zeros (N-1) there are between the decimal point and the 1. By not specifing the N, which is what the notation .000...0001 implies, does not negate its existance.

My proof shows that if you add 10^-N to .999... the result is larger then 1 for all N>0. It also shows that if you subtract 10^-N from 1 the result is LESS then .999... For All N>0. These 2 inequalities demonstrate that 1 and .999... occupy the same point on the number line, thus are equal.

 

[MadRat]

Originally posted by: ElFenix

in no way does the position affect the value? so 1.000000001 = 1.0000000000000000001?

1 + 10^-9 <> 1 + 10^-19

The value of the decimal using an integer of "1" at 10^-9 position is 10^10's greater than the value of the decimal using an integer of "1" at the 10^-19 position. But the value of integer at either position is still a "1". Whether the integer is at 10^-9 or 10^(-infinite) position, it is still an integer of "1" at either position.

 

[MadRat]

Originally posted by: RossGr
Why do ya'll insist on writing .00....001 ?

The best way to designate some arbitaraly small decimal number is by writting 10^-N, where N can be any positive integer. This precisily covers All possible numbers, which have commonly written in this thread, as .00...001. In addition it alleviates any possible confusion over a mystic digit in the "infinite possition". All digits in a real number are located in fixed positons which are enumerated by the N. With this notation we can know exactly how many zeros (N-1) there are between the decimal point and the 1. By not specifing the N, which is what the notation .000...0001 implies, does not negate its existance.

Errr, nobody denied the existence of any place on the number line other than you specifically saying .000...1 cannot exist. I just said you have the problem that it either exists to disprove your point or it does not exist, but I never made the call. The problem is comparing apples to orangles, subjective values <> objective values.

Originally posted by: RossGr
My proof shows that if you add 10^-N to .999... the result is larger then 1 for all N>0.

And I've already pointed out that you cannot add to a sum that equals infinite, regardless if you use it to represent an infinitely small (10^-infinity) or large (10^infinity) decimal. You want to use a rule of thumb to reason out infinitely small numbers when infinite doesn't follow the same rules as other numbers. Your argument is akin to someone insisting that division by zero is possible. Hence the conclusion of your proof has a gaping wound:

Originally posted by: RossGr
It also shows that if you subtract 10^-N from 1 the result is LESS then .999... For All N>0. These 2 inequalities demonstrate that 1 and .999... occupy the same point on the number line, thus are equal.