0.999... = 1 (?)

[SIZE="7"]Go here.[/SIZE]

So, for the poll. After reading that, what do you think? Is 1 = 0.999..., or is it complete bullpomegranate?

 Spoiler

For me, long before that read, I knew 0.999... was equal to 1. Consider this a "social test" where I want to find out whether you obey to mathematical proofs or to your intuition.

I still say it's bull sh'it. A fraction of a number can't be the number. It's still missing a fragment needed to become that number. Sure, you can round up, but that doesn't mean 9/10ths = 1/1.

Nope, a number is anumber.
In the technical world, .99999... is very different from 1.

if 0.999... != 1 then the world of mathematical proofs will collapse.

I still say it's bull sh'it. A fraction of a number can't be the number. It's still missing a fragment needed to become that number. Sure, you can round up, but that doesn't mean 9/10ths = 1/1.

he's talking about 0.999999999999... repeated forever not 0.9

numbers that repeats at a pattern are rational numbers and can be written as a fraction of 2 whole numbers. By the technique of writing rational numbers as fraction that I learned when I was in gr 8. 0.999... is written as 9/9 which =1

They have some faulty premises.

"For any two different real numbers, you can pick a third number which is between them."

Who said that?

If the difference between two numbers is zero, then they are equal. For example, 5 - 5 = 0 because 5 = 5.

The difference between 1.0000... and 0.9999... is:

1.0000... - 0.9999... = 0.0000...
= 0

Therefore, they are equal.


Their arguement is bs. Sure, "..." might mean going on forever, but 1.0000... - 0.9999... = 0.0000(keep on going) until 1. This applies to the 3rd problem, too.

A sequence can only have one limit.

Observe that the limit of the sequence

0.9
0.99
0.999
0.9999
0.99999
...
is

0.9999...
That is, the sequence gets closer and closer to 0.9999..., in fact, infinitely close.

But the sequence also gets closer and closer to 1.0000..., in fact, infinitely close. So 1.0000... is a limit of this sequence too.

But a sequence can only have one limit, so 0.9999... and 1.0000... must be the same.

Isn't 0.99999... equal to 0.9999...?

The rest of his arguements are logical, but I still disagree due to my trig. teacher saying 0.999... approaches 1, but never gets there, and she had this entire complicated proof and crap.

I remember in Grade 10, my math teacher did a weird proof about this. I dunno what weird and crazy math theories and stuff can disprove it but I thought I'd share.

So, if x= 0.999~

10x = 9.999~

10x - x = 9.999~ - 0.999~

9x = 9.000

x = 1

o__o

Edit: LOL, didn't realize the link there provided the exact same thing. Stopped reading it after a few paragraphs

1 is an integer. 0.9999999 repeated isn't. Therefore 1 != 0.9999999999999. 0.9999999999999 is a number which is missing a fragment of its ability of becoming a whole number. Whether that be impossible to see the difference in, it's still not the same as the integer "1." It's still not a whole number.

They have some faulty premises.

"For any two different real numbers, you can pick a third number which is between them."

Who said that?
*facepalms that's the definition of real number

If the difference between two numbers is zero, then they are equal. For example, 5 - 5 = 0 because 5 = 5.

The difference between 1.0000... and 0.9999... is:

1.0000... - 0.9999... = 0.0000...
= 0

Therefore, they are equal.


Their arguement is bs. Sure, "..." might mean going on forever, but 1.0000... - 0.9999... = 0.0000(keep on going) until 1. This applies to the 3rd problem, too.
there won't be a "1" at a bottom, because the number goes on forever, forever means no end. If there is a 1, then the numbers does not go on forever, if you keep your reasoning, you'll fail calculus when you learn it

"0.9999... gets closer and closer to 1, but never reaches it."

Closer and closer? How can it be getting closer and closer? It's one number! Try the limit argument, above.

"0.9999... is a decimal representation of infinity, not a number."

Well, how come it's DEFINITELY bigger than 0.5 and smaller than 2? Just because something has infinitely many pieces doesn't mean it's infinite. Zeno figured this out 2500 years ago.


I disagree with the first quote, as it is one number, but... and with the second one, Xeno is just a crazy mathematician who's problems no one can really solve. One of them is like this:

Take the distance between you and an object. Cut it by half. Theoretically, you can keep on cutting in half into infinite. But....wouldn't this mean we can't be able to walk to that object, since it's infinite? So the universe is all screwed up and we're all gonna die.

There is obviously a flaw in his reasoning because he's only doing the limit up to the time that you and the object intersects, but not past it, that have nothing with this.

The rest of his arguements are logical, but I still disagree due to my trig. teacher saying 0.999... approaches 1, but never gets there, and she had this entire complicated proof and crap.May I tell you that high school teachers are not mathematicians?

1 is an integer. 0.9999999 repeated isn't. Therefore 1 != 0.9999999999999. 0.9999999999999 is a number which is missing a fragment of its ability of becoming a whole number. Whether that be impossible to see the difference in, it's still not the same as the integer "1." It's still not a whole number.

so 0.5+0.5 is also not a integer, "one" is not a integer because it must be "1".

They have some faulty premises.

"For any two different real numbers, you can pick a third number which is between them."

Who said that?

It comes from the property that real numbers can be compared. If two numbers are different, that is, a =/= b, then either a < b or b < a, correct? This also means that there exist a number c, which makes a + c = b, or b + c = a. The number c can then be written as c = a - b (or b - a), if you know both a and b. If you divide c by 2, then a < a + c/2 < b, which proves there is a number between.

1 is an integer. 0.9999999 repeated isn't. Therefore 1 != 0.9999999999999. 0.9999999999999 is a number which is missing a fragment of its ability of becoming a whole number. Whether that be impossible to see the difference in, it's still not the same as the integer "1." It's still not a whole number.

We're not talking about different sets of numbers. True enough, 1 is an integer, but 1 is a real number as well. (1 can be written as 1.0, 1.000, 1.000... etc)

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They have some faulty premises.

"For any two different real numbers, you can pick a third number which is between them."

Who said that?

If the difference between two numbers is zero, then they are equal. For example, 5 - 5 = 0 because 5 = 5.

The difference between 1.0000... and 0.9999... is:

1.0000... - 0.9999... = 0.0000...
= 0

Therefore, they are equal.


Their arguement is bs. Sure, "..." might mean going on forever, but 1.0000... - 0.9999... = 0.0000(keep on going) until 1. This applies to the 3rd problem, too.

A sequence can only have one limit.

Observe that the limit of the sequence

0.9
0.99
0.999
0.9999
0.99999
...
is

0.9999...
That is, the sequence gets closer and closer to 0.9999..., in fact, infinitely close.

But the sequence also gets closer and closer to 1.0000..., in fact, infinitely close. So 1.0000... is a limit of this sequence too.

But a sequence can only have one limit, so 0.9999... and 1.0000... must be the same.

Isn't 0.99999... equal to 0.9999...?

The rest of his arguements are logical, but I still disagree due to my trig. teacher saying 0.999... approaches 1, but never gets there, and she had this entire complicated proof and crap.

Basically exactly what I thought. At some point 0.0000000...1 has to happen.
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http://en.wikipedia.org/wiki/0.999...

*facepalms that's the definition of real number

Not really. That number in between can be imaginary, anyways.

there won't be a "1" at a bottom, because the number goes on forever, forever means no end. If there is a 1, then the numbers does not go on forever, if you keep your reasoning, you'll fail calculus when you learn it

I probably will.

There is obviously a flaw in his reasoning because he's only doing the limit up to the time that you and the object intersects, but not past it, that have nothing with this.

Already deleted, but not because of this, but because I said the exact same thing the creator of the website said, just proving it for different reasons....

May I tell you that high school teachers are not mathematicians?

Say hi to my teacher.

Edit: Yeah, I'll admit my intuition is very, very, strong. However, I blame that on all of those Sherlock Holmes, Detective Conan, and Agatha Christie novels that I read.

Hi there guys, Shouri here, math major. Just took Real Analysis in which this topic came up on about the first week. The complicated version for why .9999.... = 1 deals with something called cauchy sequences and equivalence classes.

When we only deal with the rational numbers (numbers of the form a/b where a and b are integers and b=/=0 ), things like .999=/=1. But in order to build the real numbers, we needed numbers that could possibly have an infinite number of decimal places. We defined number such as pi in terms of a cauchy sequence of rational numbers. For instance, the cauchy sequence of pi is {3, 3.1, 3.14,....} We then said that the number pi is the limit of that sequence. But the thing is that there is more than one sequence which can get you to pi. For instance, another cauchy sequence for pi is {3.2, 3.15, 3.142, 3.1416, 3.14160, 3.141593,....}. We can call these two sequences equivalent since the differences in the nth terms for these sequences approaches zero.

Now we look at two cauchy sequences, one for 0.9999..... and one for 1:

{0.9, 0.99, 0.999, 0.9999, 0.99999,.....} Call this sequence S and call the nth term sn (term 3 would be s3 = 0.999)
and
{1.0, 1.0, 1.0,...} Call this sequence R, and the nth term rn.

Notedly cauchy sequences are allowed to be just one number repeated in which case, they represent that very number.


The differences in the sequence are:

sn-rn ={.1, .01, .001,....} which is

nth term = 1/(10^n)

You then take the limit of this sequence as n approaches infinity which is zero. Thus the two sequences R and S are part of the same equivalence class.

If two numbers x and y have cauchy sequences in the same equivalence class then x=y.

Thus .9999......=1

Not really. That number in between can be imaginary, anyways.
no, the number in between is real
I probably will.


Already deleted, but not because of this, but because I said the exact same thing the creator of the website said, just proving it for different reasons....

1. Whatever you say, mate.

2. Your point of posting your teacher?

Not really. That number in between can be imaginary, anyways.

There is NO imaginary number between two real numbers. The imaginary numbers are not found on the REAL number line.

There is NO imaginary number between two real numbers. The imaginary numbers are not found on the REAL number line.

So I basically concede. I wish I wrote down my teacher's proof, though.

2. Your point of posting your teacher?

Harvard university?
5 awards?
17 publications?

what about your teacher?

Harvard university?
5 awards?
17 publications?

what about your teacher?

I said, what was the point of posting your teacher? I posted mine to prove that mine was also a mathematician.

My teacher has 2 patents in some cellphone camera formula device and is a millionaire =/.

point is your teacher is a minority compared to mine and the majority of mathematicians out there. (unless he didn't say it, and you just made it up)