1 = 0.99... (RossGr)

[ZeroNine8]

Not that it in any way contributes to this discussion, but...
I never provided my argument as proof, merely raised a question that was answered in a section that I
didn't remember from previously reading it and missed this time. It's not like we're talking life and death here, its just a forum discussion. Upon re-reading, I realized what went on and that was the end of it. I realize that the internet is serious business and all, but as for my credentials, it really doesn't matter what they are. As it happens, I have a BS and MS in mechanical engineering as well as graduate level math courses and about 2 classes shy of a math minor. I don't claim to be a mathematical genius, nor do I use anything beyond 'practical' engineering mathematics from day to day. So, dad, can I play now or do I have to finish my broccoli first?

 

[Yozza]

Let x = 0.99999...
It can be proved easily that:
for all E > 0: 1 - E < x < 1 + E

The rest of the proof is by contradiction. Suppose x!=1. (x does not equal 1)
From the axioms of an ordered ring and the fact (1/2) exists in R, one can prove:

x < (1/2)(1 + x) < 1

Now, if we let E = 1 - (1/2)(1 + x) > 0
Then we have x < 1 - E.
This contradicts the statement (and proof by triviality ) that, for all E > 0: 1 - E < x

Therefore, our assumption that x!=1 was incorrect, and thus x = 1.

This method of proof predates limits, and is fully rigorous. It predates limits, and was essentially known to the ancient Greeks as the "Method of Exhaustion".

The theorem explicitly stated is:

If x and y are real numbers, and:
for all E > 0: y - E < x < y + E
Then we may conclude x = y.

and this is something that actually gets used in analysis; sometimes it is much easier to prove the hypotheses of the above theorem than to set up and prove the appropriate limit.

Incidentally, IMHO, everything is much more clear if we appeal to the actual formula for mapping decimals to reals:

0.9999.... = Sigma i=1...inf [9 * 10^-i] (read as "sum of 9 * 10^-i, for i=1 to i=infinity")

Since we know the sum of this infinite geometric series to be:
a + aq + aq^2 + aq^3 + ... = a/(1 - q)

Thus, it follows that
0.999... = 0.9 + 0.09 + 0.009 + ... = (9/10) * [1 + (1/10) + (1/10)^2 + (1/10)^3 + ... ]

which, by the sum of an infinite series, gives
(9 * 10^-1) / (1 - 10^-1) = 1

I don't see how anyone can argue against something as fundamental as this in mathematics. If you disagree with it, you obviously haven't grasped basic concepts of mathematics and Number Theory.

Edit - formatting, minor tweaks

 

[RossGr]

Originally posted by: ZeroNine8
Not that it in any way contributes to this discussion, but...
I never provided my argument as proof, merely raised a question that was answered in a section that I
didn't remember from previously reading it and missed this time. It's not like we're talking life and death here, its just a forum discussion. Upon re-reading, I realized what went on and that was the end of it. I realize that the internet is serious business and all, but as for my credentials, it really doesn't matter what they are. As it happens, I have a BS and MS in mechanical engineering as well as graduate level math courses and about 2 classes shy of a math minor. I don't claim to be a mathematical genius, nor do I use anything beyond 'practical' engineering mathematics from day to day. So, dad, can I play now or do I have to finish my broccoli first?

Not to worry. I find it refreshing that someone on the web will say. "Opps"


All to frequently many continue to argue from believes while refusing to see logical proofs.


An engineer..... Should'a figured....

 

[rjain]

Heh.

 

[Lynx516]

Sorry for being very harsh on you ZeroNine. Just got a bit annoyed with people having a go at perfectly good proofs.

 

[f95toli]

IRossGr: I am not questioning your proof but like som other people I am not sure about the implication. What you have proven is equality not identity, right?
It's been a few years since I studied number theory but as far as I can rememer you can show that two real numbers are equal but you can never prove that they are identical

The theorem as written by Yozza is:

If <I>x</I> and <I>y</I> are real numbers, and:
for all E > 0: y - E < x < y + E
Then we may conclude x = y

but that theorem only states that x is equal to y (x=y), not that x is identical to y (and yes, there is a difference, that much I do remember)

 

[The Green Bean]

my head hurts

 

[Matthias99]

AFAIK, equality for real numbers is symmetric -- that is, it's defined such that if x=y, y=x as well. It's also transitive (if x=y, and y=z, then x=z) and reflective (x=x for all x).

 

[RossGr]

Originally posted by: f95toli
IRossGr: I am not questioning your proof but like som other people I am not sure about the implication. What you have proven is equality not identity, right?
It's been a few years since I studied number theory but as far as I can rememer you can show that two real numbers are equal but you can never prove that they are identical

The theorem as written by Yozza is:

If <I>x</I> and <I>y</I> are real numbers, and:
for all E > 0: y - E < x < y + E
Then we may conclude x = y

but that theorem only states that x is equal to y (x=y), not that x is identical to y (and yes, there is a difference, that much I do remember)

I have Proven equality as defined on the real line (Thank you Yozza for providing the final proof!). I am not familiar with the mathematical definition of identiy. I am more of an applied mathematician, never did any number theory.

 

[DrPizza]

OMG, more intelligent people posting in a .999... = 1 thread who aren't basing their view on "in my opinion..."

I've been lobbying the mods for a "forum rules" for Off Topic. I think I'm going to ask them to include "rule 7: .999repeating = 1. Do not bother posting about it."

 

[DrPizza]

Originally posted by: uart

Originally posted by: DrPizza

Originally posted by: uart

Originally posted by: nrg99
what about 1=0?

Sorry, the closest I can get to that is 0^0 = 1. Is that whacky enough for you?

0^0 is undefined. Simple explanation: x^12 divided by x^9 = x^3.... subtract the coefficients. x^12 divided by x^12 = x^ (12-12) = x^0 = 1. However, if x=0, then the initial fraction involves dividing by 0. Thus, 0^12 divided by 0^12 does NOT equal 1. It is division by 0 which is undefined. The entire rule for "x^0 = 1" includes "x != 0"

0^0 as a limit is indeterminate (and depending on what the limit is of, 1 is a possible answer).

Yes I know that strictly speaking 0^0 is undefined, I was just trying to be controversial.

x^0 = 1 and is well defined for all x other than 0.

0^x = 0 for all x>0 and is Inf (or undefined) for all x<0. I can remember a Maths Lecturer once told me that some authors "define 0^0" as being 1 for convenience, but that this was a little controversial and not universally accepted.

BTW, Both Windows calculator and Maple7 return 0^0=1.

I just ran into a problem that can be verified graphically or through a table of values...

Limit as x-> infinity of [x^((ln a)/(1 + ln x))] leads to the indeterminate form limit = 0^0. Now, if 0^0 DID = 1 then this wouldn't be indeterminate. Applying L'Hopital's Rule, the limit = a.
Fortunately, my TI-83 plus says "ERROMAIN" I wonder if Maple 8 still returns 1.

 

[rjain]

Since 0.0 isn't necessarily 0 in floating point math, we are best off assuming that it's not 0, in which case raising it to the (approximately) zeroth power gives us (approximately) 1.

 

[uart]

Originally posted by: DrPizza I just ran into a problem that can be verified graphically or through a table of values...

Limit as x-> infinity of [x^((ln a)/(1 + ln x))] leads to the indeterminate form limit = 0^0. Now, if 0^0 DID = 1 then this wouldn't be indeterminate. Applying L'Hopital's Rule, the limit = a.
Fortunately, my TI-83 plus says "ERROMAIN" I wonder if Maple 8 still returns 1.

Hang on, I'm pretty sure that the Limit as x-> infinity of [x^((ln a)/(1 + ln x))] really is a.

Let f = x^((ln a)/(1 + ln x))]

then ln(f) = ln(a) * ln(x) / ( 1 + ln(x) )

So Lim x-> Infinity of ln(f) is ln(a)

Is this not enpugh to prove that the limit as x->Inf of f(x) is a ?

 

[RossGr]

0.000...............1

How do you define this representation?

In formal mathematics an ellipsis appearing in the middle of a number represents a finite number of omitted digits, your multiple periods has no meaning what so ever. If you are attempting to say there are a infinite number of zeros followed by a one, then you are being self contradictory. How can you have an infinite number (ie unending) followed by something? As soon as you place the 1 you have terminated your string of zeros, therefore it is not infinite.

You are correct for any FINITE string of 9's there is a corresponding string of zeros followed by a 1 which sum to 1. This fact is the heart of my proof. Did you take the time to read clear through it?

 

[uart]

Originally posted by: RossGr

0.000...............1

How do you define this representation? ....

Ross, who are you quoting there? This thread is getting too long and I cant see who you're referring to.

 

[RossGr]

I swear I clicked on the last page, a post showed up that I thought was new. So I just quoted it, thinking it was the newest post. No, I didn't look at the date time stamp. Looks like it was some old post in the middle of the thread! Sorry, probably should have just deleted my reply when I realized my error.

 

[DrPizza]

I got thrown off by that .0000...1 thing too... I was wondering what you were talking about. Then, Uart quoted me and repeated what I said (except showed some steps), confusing me more... I thought I was in the twilight zone or something.