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0.999...= 1

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Re: 0.999...= 1

#131

Post by Scarblac »

tom-thistime wrote: post #121
undecisive wrote: post #115

There is no proof that will ever satisfy a person dead-set against this. Ever since I brought this home from school as a child, my whole family ribbed me mercilessly for it. If you tell a person that 3/6 = 1/2, they'll believe you - because they have been taught from an early age that fractions can have multiple "representations" for the same underlying amount. People mistakenly believe that decimal numbers don't hav…

(STATEMENT OF PERSONAL IGNORANCE [SOPI]: Anyone who actually understands this stuff please correct my mistakes below. Thanks.) In the real numbers, which are not always simple or intuitive, 0.99... = 1. That's true and I seem to understand the proof. But the real numbers aren't the only system that might be sitting behind "0.99..." and "1" when I write those symbols down and talk intuitively to people in my family. T…

A number is just a number, it doesn't approach anything. A series can approach something, but a number can't. In any system where 0.99... is valid notation for a number, it doesn't approach anything.

Re: 0.999...= 1

#133

Post by CogitoCogito »

judofyr wrote: post #108
CogitoCogito wrote: post #94

Earlier quoted context omitted.

> Personally I've always thought "proofs" using "arithmetic" are right, but kind of stated backwards. I've never considered them right at all. By saying something like 0.9... x 10 = 9.9... and then saying that 9.9... - 0.9... = 9 you're basically just a priori defining 0.9... to be 1. In other words you're basically just defining 0.9... as a symbol to be some number x which has the property that 10x - x = 9. So you'r…

You only need to define 0.9… as 9/10 + 9/100 + 9/1000 + …. Without knowing how that series converges you can then use the two expressions mentioned to conclude that it has to be equivalent to 1.

> You only need to define 0.9… as 9/10 + 9/100 + 9/1000 + …. Without knowing how that series converges you can then use the two expressions mentioned to conclude that it has to be equivalent to 1.

Sure you can provide a hand-wavy argument and try to give some intuition if you'd like. That doesn't make it any sort of logical proof though. I guess it depends on what you're after.

Re: 0.999...= 1

#135

Post by [deleted] »

traderjane wrote: post #119
undecisive wrote: post #115

There is no proof that will ever satisfy a person dead-set against this. Ever since I brought this home from school as a child, my whole family ribbed me mercilessly for it. If you tell a person that 3/6 = 1/2, they'll believe you - because they have been taught from an early age that fractions can have multiple "representations" for the same underlying amount. People mistakenly believe that decimal numbers don't hav…

You must go up to something like limits to make ... meaningful.

[deleted]

Re: 0.999...= 1

#136

Post by karatinversion »

ristos wrote: post #112

I don't think that 0.999... = 1 is actually provable. I think this and all of calculus is actually axiomatic, which has the following axiom: Given ε = 1/∞ then: ε = 0 Am I wrong in thinking this way? It seems as though there's no way to actually truly prove that an infinite series converging towards zero actually hits zero (from a constructivist pov)

That's not an axiom we take; the study of real numbers takes some algebraic and order axioms, and a completeness axiom:

Any non-empty set of real numbers with an upper bound has a least upper bound.

We can't prove your statement

Given ε = 1/∞ then: ε = 0

because it is not well-defined, but we can prove this:

If ε ≥ 0 and, for every natural number n, ε For suppose there exists an ε which is a counter-example, i.e. ε > 0 and ε S = { x : x a real number, x > 0 and x is non-empty, and has an upper bound (e.g. 1). So it has a least upper bound, say y. In particular, y is an upper bound, so 2y is not in S. It is > 0, so there must exist a natural number N for which 2y >= 1/N. But then y/2 > 1/4N, and 4N is also a natural number. So for any element x of S, x This is a contradiction, so the claim is proved.

Re: 0.999...= 1

#137

Post by lmkg »

undecisive wrote: post #115

There is no proof that will ever satisfy a person dead-set against this. Ever since I brought this home from school as a child, my whole family ribbed me mercilessly for it. If you tell a person that 3/6 = 1/2, they'll believe you - because they have been taught from an early age that fractions can have multiple "representations" for the same underlying amount. People mistakenly believe that decimal numbers don't hav…

It's more fundamental: People seem to have the intuition that the decimal representation of a number is a number. I don't know if it's because decimals resemble the ntural numbers or what, but decimals seem to have a primacy for people that fractions do not. The idea that there's a gap between the symbol for a thing and the thing itself is the stumbling block.

Re: 0.999...= 1

#138

Post by kaetemi »

saagarjha wrote: post #15
ltbarcly3 wrote: post #7

This is 'more intuitive' if you think about it this way: If any two real numbers are not equal, then you can take the average and get a third number that is half way between them. Conversely, if the average of two numbers is equal to either of the numbers, then the two numbers are equal. (this isn't a proof, just a way to convince yourself of this) What's the average of .9999... and 1?

0.999…5 obviously.

Okay, now multiply that by 2.

Re: 0.999...= 1

#139

Post by baryphonic »

knzhou wrote: post #44

Personally I've always thought "proofs" using "arithmetic" are right, but kind of stated backwards. The point is that in elementary school arithmetic, you define addition, multiplication, subtraction, division, decimals, and equality, but you never define "...". Until you've defined "...", it's just a meaningless sequence of marks on paper. You can't prove anything about it using arithmetic, or otherwise. What the "a…

The way '...' is used here is perfectly consistent with being defined as a geometric series where the ratio between successive elements is 1/10 and the start term is the final digit. Geometric series always converge when the absolute value of the ratio of less than 1.

I should note that when I learned about rational & irrational numbers in elementary school (I think third or fourth grade), we used a "bar" notation where we'd put a bar over the last digits in a decimal expression that repeated forever (i.e. it corresponded exactly to a geometric series with r = (1 / 10)^k where k is the number of digits under the bar, though we didn't know about that at the time). Our teachers explained that the difference between a rational and irrational number was that there would be no pattern you could ever find in an irrational number that would allow us to use the bar, which is surprisingly accurate for grade school arithmetic.

Re: 0.999...= 1

#140

Post by tom-thistime »

Scarblac wrote: post #131
tom-thistime wrote: post #121

Earlier quoted context omitted.

(STATEMENT OF PERSONAL IGNORANCE [SOPI]: Anyone who actually understands this stuff please correct my mistakes below. Thanks.) In the real numbers, which are not always simple or intuitive, 0.99... = 1. That's true and I seem to understand the proof. But the real numbers aren't the only system that might be sitting behind "0.99..." and "1" when I write those symbols down and talk intuitively to people in my family. T…

A number is just a number, it doesn't approach anything. A series can approach something, but a number can't. In any system where 0.99... is valid notation for a number, it doesn't approach anything.

Sure. Instead of "0.99..." please substitute lim n->inf sum(1..n)(9 times 10^-n).

The point I'm making is that the "obvious truth" 0.99... = 1 that we're all talking about depends on the assumption that we're working in the real numbers.

I claim that the real numbers are not something intuitively obvious to every sufficiently intelligent person; instead they are kind of weird and technical. I go on to claim, though I'm more unsure of this, that the real numbers are not even the only way to make calculus work.

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