Poll: 0.999... = 1

[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

 

[Aerialfrogg]

It is generally accepted among mathematicians that 0.999... = 1.
One "rebellious" college students and people who don't know anything about math in the first place question this FACT.

But hey, how about another question:

0/0 =
0, 1 or &#8734;?

UNDEFINED. Although, I think it should be &#8734;
If 0! = 1 then we should be able to divide by zero.

 

[Piflik]

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

Refer to my previous posts...Limits are not values. Both sequences will approach 1/3 asymptotically and never reach it. Ever.

If you want to argue then answer the following question: If the limit for 1/n for n -> 0 = infinity, what is 1/0?