Poll: 0.999... = 1

[Athinira]

You haven't proven that yet, while there has been plenty of proof that shows that the two numbers are the same.

None of these proofs are acceptable, since they either rely on flawed representation of infinity or circular reasoning.

Actually, linear continuum as you mentioned yourself is a proof in itself. For two numbers x and y, where x < y, there exists a number z so that x < z < y.

Since z doesn't exist for x = 0.999... and y = 1, then that means that x < y is false and the two numbers are the same.

Above proof doesn't rely on circular reasoning. It relies on the standard mathematical rules for Real Numbers. What you might be thinking of is the Extended Real Number Line [http://en.wikipedia.org/wiki/Extended_real_number_line] which adds infinity and negative infinity to the Real Numbers system, meaning that infinitesimal numbers can exist there. But that still isn't the "real" Real Numbers system, which is why it's called the "extended" system. But in the Real Numbers system, infinitesimal values can't exist.

 

[grammarye]

I say it is flawed and incorrect when we work with infinity (either as a number or as a value), since when we approach the infinitely small or big there happens tons of stuff that basic calculus cannot deal with. The result is the common misconception (I'd almost call it delusion) that 0.99999... = 1.

I'm not sure delusion is fair. As I indicated, you can't have basic decimal maths as it stands today without the concept. Workaround? Corner case? I could work with any of those. Nevertheless, I would argue the burden is on the accuser of a theory to not merely poke holes, but to do better.

The definition of whether 0.99.. tends to 1 or is 1 is a mathematical argument with limited or zero usefulness to the modern world other than mathematicians, but one choice allows the overall system to keep going albeit perhaps not mathematically purely - the other doesn't.

If you were dividing 1/3, how would you represent it other than by writing 1/3? Note that our entire basis of computers, numerical interchange, banking, and a host of other things are based on the decimal system. As I said to Lyx, I have no problem with the idea that we haven't got a theory fully formed before using it, but I think there is a difference between saying '1/3 =/= 0.33...' and '1/3 = 0.33... because we say so and because it lets the rest of the system work until we replace decimal with something better'.

 

[Lyx]

So, here's the difficulty:

1/3 done as division, decimal by decimal leads you into an infinite loop of getting 0.3, then 0.33, and so on. Unless you're going to allege that at some point, if we did it enough, we'd get something other than 3, that must hold true.

If that is the case, and we accept that that division is what mathematicians everywhere will call division, and we also have the entirety of algebra agreed upon, such as 2a = b, then:

1 = 0.33... + 0.33... + 0.33...

It has to. You can't have a lossy process in your basic numeric operations. Division of one object into three must be reversed by taking those three objects and merging them back together. 'Something on top' is what exactly?

Okay, i'll repeat this again for the n-th time. I have no problem with that you are doing some additional operation (which in practice (non "conceptual") you are doing). To put it directly: My only problem with this is the symbol and term used. Call it "infinityAndSomething" and make a symbol for it, and i wont disagree anymore, simply because it then is no longer synonymous with infinity anymore. In fact, all you'd need to do is to call it what it actually is: infinity, rounded to the nearest number.

 

[AnnihilaSean]

This makes my face hurt.

 

[Davey Woo]

This theory was put here ages ago, and I'll say now what I did then.

0.999 = 0.999 and nothing else.
1 = 1 and nothing else.

When you say 0.999 = 1 you're just hiding the calculations in between that makes 0.999 equal to 1.

 

[Rubashov]

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...

I prefer:

b0.b1b2b3b4... = b0 + b1(1/10) + b2(1/10)^2 + b3(1/10)^3 + b4(1/10)^4 ...

if |r| < 1 then kr + kr^2 + kr^3 + ... = kr/(1-r)

So for 0.9...:

0.(9) = 9(1/10) + 9(1/10)^2 + 9(1/10)^3 + ... = (9(1/10))/(1-(1/10)) = 1

... you know what, I'm just gonna chalk this down to <link=http://en.wikipedia.org/wiki/Proof_by_intimidation>Proof by intimidation (cause this just confuses me)

tho that looks kinda like some of the proofs in <link=http://en.wikipedia.org/wiki/Invalid_proof>Mathematical fallacy, but i'm to lazy to dissect that whole equation just to tell you that 1 doesn't equal .999999

He's expressing 0.999... as a geometric series and showing that, because of what we know about geometric series, said series is equal to 1.

 

[PatrickXD]

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 isn't actually equal to 0.3 recurring. It's just the closest representation that we have. For that reason, decimals and not fractions are accepted at anything above GCSE maths in exams.

 

[Addendum_Forthcoming]

I suck at maths but technically wouldn't 10x (if x is .9~) - x = 8.9999999~?

If 10x = 9.999... and 9x = 8.999... then x = 1 since the difference between 8.9... and 9.9... is is 1, which still proves that 0.999... = 1

It's not though o.o

The difference is .9~,

.9~(x) + 8.9~ (9x) = 9.9~

.'. x = .9~

If X = 1 AND .9~ the difference would be .0~1 ... .9~ is not 1 n.n

Whilst the same can not be said of 3/3 = 1 ... and all numbers are divisible by themselves into real numbers.... 3/3 = 1 is still mathematically correct.

That which cannot be said about .9~/1 = 1, as .9~ =/= 1

 

[grammarye]

This is where i differ. I seek other things besides of "we can control stuff without understanding why". You know, things like... umm, understanding why. Explanation. A consistent worldview. And finally, sustainability (you can't do shit without concepts and understanding. So if the concepts and understanding doesn't matter, then this is like saying "Who cares about tomorrow? For now, the skyscraper hasn't come down yet."
...
I do not subscribe to the moral, that action and responsibility are isolated from each other. Among other things, it is logically false and untrue. A causal break to be precise. But it certainly is comfortable and popular

I would argue about this, but frankly, it's so off-topic and putting so many words into what I wrote that it's just not worth the effort. May I suggest simply that you consider that huge chip on your shoulder before attempting to find fault with others.

 

[Redingold]

Hey, let's try something.

Let n = 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + ... and so on, forever, involving all negative integer powers of 2.

Now, 2n = 1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + ... and so on.

This is 1 more than n.

2n = n + 1

Subtract n...

n = 1

So what you're saying is that n is actually infinitesimally smaller than 1, huh?

Would that make it equal to 0.999..., since that is also infinitesimally smaller than 1?

But I've just shown it's one.

What am I doing wrong? Where is the flaw that shows that when n + 1 = 2n, n =/= 1?

 

[slightly evil]

more or less...
but 0.9' is only aproximately equal to 1

 

[Athinira]

I suck at maths but technically wouldn't 10x (if x is .9~) - x = 8.9999999~?

If 10x = 9.999... and 9x = 8.999... then x = 1 since the difference between 8.9... and 9.9... is is 1, which still proves that 0.999... = 1

It's not though o.o

The difference is .9~,

.9~(x) + 8.9~ (9x) = 9.9~

.'. x = .9~

Actually no.

If 0.9r ISN't equal to 1, then ".9~(x) + 8.9~ (9x) = 9.9~" implies that the number 8 must exist somewhere in the end result (9.9r).

You bare basically saying that the difference between 8.9r and 9.9r is 0.9r. That implies that 0.9r equals 1, since 9.9r - 8.9r = 1.

I don't understand why people are still arguing this

Shouldn't common sense prevail soon?

Not if your "common sense" is wrong (or as a clever man once said: "Common sense is not so common.").

Math is a complicated branch of science and doesn't make sense if you don't take your time to understand it's rules. Specifically, you haven't taken your time to understand that one value can be represented in several (or rather, infinite) ways. 1 = 2/2 = 2 - 1 = (1/2) * 2. And yes, 1 = 0.999... as well.

What's common sense to you is laughable in the eyes of math Ph.d.'s. It's commonly accepted that 0.999... equals 1 amongst educated mathmaticians, even if it sounds like bollocks to people who doesn't understand this branch of math properly.

 

[Liam Moriarty]

2003 called, it wants its math meme back.

This has nothing to do with any meme. I'm not some 4chan idiot. I want to see how many people reject the concept.

Rule 50

OT: I guess this makes sense, but I'm going to show my geometry teacher this just to see what she says.

 

[Trivun]

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.

I'm in my third year of a Maths degree and that's one of several proofs I was shown in my first year. My professors agree that it's correct and a valid proof, so all the so-called 'Math majors' crudus refers to are wrong.

On topic, I know multiple proofs for this and the statement is indeed true. Anyone who disagrees, feel free to quote me, and come back with a completely logical and mathematically correct disproof, only then will I take you seriously .

 

[Piflik]

Hey, let's try something.

Let n = 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + ... and so on, forever, involving all negative integer powers of 2.

Now, 2n = 1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + ... and so on.

This is 1 more than n.

2n = n + 1

Subtract n...

n = 1

So what you're saying is that n is actually infinitesimally smaller than 1, huh?

Would that make it equal to 0.999..., since that is also infinitesimally smaller than 1?

But I've just shown it's one.

What am I doing wrong? Where is the flaw that shows that when n + 1 = 2n, n =/= 1?

Again your geometric series...I said it before and will say it again. A limit is not a value. It is a limit and will never reach that value.

The limit of the sum 1/(2^n) for n -> infinity is 1. That much is true, but the function will approach this value asymptotically and never reach it.

 

[Ishnuvalok]

End of thread.

 

[grammarye]

I don't understand why people are still arguing this

Shouldn't common sense prevail soon?

1 = 1

0.999... = 0.999...

Red = Red

Blue = Blue

Right = Right

Left = Left

No matter how many times a 9 gets added to that decimal place, it won't equal 1.

So what value should you multiple by 3 to get precisely 1?

 

[Piflik]

I don't understand why people are still arguing this

Shouldn't common sense prevail soon?

1 = 1

0.999... = 0.999...

Red = Red

Blue = Blue

Right = Right

Left = Left

No matter how many times a 9 gets added to that decimal place, it won't equal 1.

So what value should you multiple by 3 to get precisely 1?

1/3

 

[grammarye]

...

No matter how many times a 9 gets added to that decimal place, it won't equal 1.

So what value should you multiple by 3 to get precisely 1?

1/3

Now do it without using a tautology, which proves nothing.

 

[Maze1125]

0.333... * 3 =/= 1, since 0.33333... =/= 1/3

0.333... is a flawed representation of 1/3. No matter how far you go, will always be an infinitesimal difference between 1/3 and 0.333...

Let a[sub]n[/sub] be the nth term of the sequence: 0.4, 0.34, 0.334, 0.3334, ...
Let b[sub]n[/sub] be the nth term of the sequence: 1/3, 1/3, 1/3, 1/3, ...
Let c[sub]n[/sub] be the nth term of the sequence: 0.3, 0.33, 0.333, 0.3333, ...

Now, the limits of a[sub]n[/sub] and c[sub]n[/sub], as n tends to infinity, are both obviously 0.333...
The limit of b[sub]n[/sub] is also clearly 1/3.
But, for all n, a[sub]n[/sub] > b[sub]n[/sub] > c[sub]n[/sub], and the limits of a[sub]n[/sub] and c[sub]n[/sub] are equal. So, by the squeeze theorem, the limit of b[sub]n[/sub] is equal to the limits of a[sub]n[/sub] and c[sub]n[/sub].

So, the limits of b[sub]n[/sub] are both 1/3 and 0.333..., and limits are unique. Therefore 0.333... = 1/3
QED