Poll: 0.999... = 1

[zfactor]

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...

Uh, wait, the second part will have one less 9 after the decimal point than the first part.

No. That's not how infinity works. Both 10x and x have infinite decimal places, so it doesn't make sense to talk about one having one less decimal place than the other.

He multiplied by 10 thus shifting the decimal place over to the right one space. X had infinity decimal places and 10x has infinity - 1 decimal places. The infinite decimal places applies to x, not to the polynomial 10x. At least that's how I see it, the answer to the entire question relies on how you work with infinity...

This would be a valid argument if it weren't for the fact that one less than infinity is still infinite. In fact, fifty trillion less than infinity is still infinite. Any finite number less than infinity is still infinite. If this doesn't apply to the concept you call infinity, then what you're talking about isn't infinity at all in the mathematical sense; it's an arbitrary finite value that you've decided to call infinity.

-sigh-

I'm just going to stop now because the answer is it depends.

 

[Pyode]

1/3 = .3333...

therefore 3/3 = .9999... = 1

This was what I immediately though of when I read the thread title. I was really surprised to see those other examples that prove it as well.

If that doesn't make sense to you, imagine two ... let's make them space ships, travelling along the same axis in the same direction at the exact same speed which will never alter. If Rocket 1 is 10 metres in front of rocket 2, and they both start at the same time, after an infinite amount of time has passed, the distance they are from rocket 2's starting point is infinity, but rocket 1 is 10 metres in front of rocket 2. Therefore, both ships have travelled an infinite distance, but the distance between the origin and rocket 1 is a greater degree of infinity (by a distance of 10 metres) than rocket 2.

That's a really interesting way to look at it.

 

[Rabid Toilet]

Again, true, but those are the same based on fractions vs decimals. This is a whole number and a fraction we're talking about. .9(infite) would be called a fraction. Called a fraction because it is a fraction of a number. And therefore not a whole number, which one is.

Can you show an example where a fraction, a decimal and a whole number are all the same number?

Easily.

1/3 = .333...
3 * (1/3) = .999...
3 * (1/3) = 3/3
3/3 = 1

.999... = 3/3 = 1

Except 1/3 is not .333333[infinity]. It is .33333[infinity][1/3]. By your example, 3 * [1/3] would be .9999999[infinity]. If you tack on the [1/3] at the end of the infinity decimal, there is an extra .0000[infinity]1 which makes the whole thing equal exactly 1.

But, balls, there doesn't seem to be any way to prove this without circular logic (i reasoned something extra needed to be there because .999999 =/= 1)... I really need to stop getting in on these physics/math forums... I'll make my brain explode at some point.

YOU SEE? THIS IS WHAT HAPPENS WHEN I POST IN THESE THREADS. NOW I'M GOING TO BE HERE ALL DAY RESPONDING. THANKS A LOT!

There is no .000[infinity]1 because there is no end to the zeroes. If the zeroes never end, you cannot place a 1 a the end, because it doesn't exist. That's what infinity means.

 

[crudus]

are you refering to canceling out (a - b) by dividing both sides by it? so, (a - b) / (a - b), and (a - b) = 0, so it would be 0 / 0, or simplified, 0, which is a mathematically correct term

We are indeed referring to the canceling out of the (a-b). However (a-b)/(a-b) never happens. 2a2/(a-b) happens.

Some of the math in this thread is flawed... jeez. Double check, please.

.9999... is not equal to 1.
Why?

Still being in high school, and with my chemistry background, I would have to say that .9 etc. is not equal to 1 because of its significant figures. You can round up when the math is over, but it will never be correct. .9999999 will always be <1, no matter how close it gets. In the math world, this is just how it has to be.

In real life, it would get so close to 1 that it wouldn't really matter.
But in theoretical conversations, it will always be <1.

Don't worry. Math starts to blow your mind-hole once you hit your second or third year of calculus. Although it starts with understanding infinity which is surprisingly difficult.

Spoiler: here is a nice proof for you which should give you a taste

What it is saying is to take 9 and divide it by 10 and add it to 9/100, then add it to 9/1000 and so on. If you notice you eventually get .9999..., and it ends up equaling 1 in the end.

 

[Rabid Toilet]

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...

However 0.999... < 1 as 1x10 = 10
0.999...x10 = 9.999...
Therefore 0.999...(not equal)1

SEE HERE I AM STILL RESPONDING.

Your argument is a logical fallicy. You're saying .99... doesn't equal 1 because .99... doesn't equal 1.

9.99... = 9 + .99...
9.99... = 9 + 1 (since .99... = 1, as he showed in his proof)
9.99... = 10

 

[zfactor]

x = 0.9999999999 (a finite number of 9s, for proof of concept purposes [or you can use infinity])
10x = 9.999999999 (this is one less 9 than the number of 9s in x [or infinity - 1])

So the 10x - x would actually equal 8.9999999991, not 9. [or 8.(infinity - 1 nines)1] And no, that fraction shouldn't theoretically explode because the extra 1 on the end is the decimal place provided by subtracting 1 from infinity (providing you can do that, and in my brain, yes you can).

You're doing it wrong. Infinity and infinity-1 are in fact the same quantity. That's kind of the point of infinity. You can't say there's an infinite number of nines and then a 1 on the end because there is no end!

As such, if x = 0.9999..., then 10x - x does equal 9.

Oh hi there, you joined just to point out my flawed ideas about infinity?

I respect that, good sir!

But, I still disagree with you. I say infinity - 1 is not infinity because it is one less than infinity. That's what let's me stick the 1 on the end, there is still one more spot availible. Now infinity + 1 is infinity. That I won't argue.

Yet alas, I need sleep now...

 

[Denamic]

Also, 1/3 does not actually equal .333333333 in infinity.
That's kind of an approximative place-holdery value because the actual value is between numbers in base 10.
1/3 equals 1/3.

 

[Agayek]

Why is it invalid, though?

Because 10x-x isn't 9. It's 8.999999....9991.

 

[Rubashov]

Whenever someone puts that up to me, I point to this shirt:

Spoiler

It's funny because that shirt divided by 0.

Actually, it didn't :-/ That proof works.

The shirt divides both sides by (a - b). Because a = b, (a - b) = 0. Thus, the proof is invalid.

are you refering to canceling out (a - b) by dividing both sides by it? so, (a - b) / (a - b), and (a - b) = 0, so it would be 0 / 0, or simplified, 0, which is a mathematically correct term

What? 0/0 is not 0. It's not anything, because it divides by zero, and allowing division by zero produces contradictions like the one shown on the shirt.

 

[AnOriginalConcept]

Here we go. Another definitive proof, unless you'd like to contest the rules of summation.

 

[zfactor]

Except 1/3 is not .333333[infinity]. It is .33333[infinity][1/3]. By your example, 3 * [1/3] would be .9999999[infinity]. If you tack on the [1/3] at the end of the infinity decimal, there is an extra .0000[infinity]1 which makes the whole thing equal exactly 1.

But, balls, there doesn't seem to be any way to prove this without circular logic (i reasoned something extra needed to be there because .999999 =/= 1)... I really need to stop getting in on these physics/math forums... I'll make my brain explode at some point.

YOU SEE? THIS IS WHAT HAPPENS WHEN I POST IN THESE THREADS. NOW I'M GOING TO BE HERE ALL DAY RESPONDING. THANKS A LOT!

There is no .000[infinity]1 because there is no end to the zeroes. If the zeroes never end, you cannot place a 1 a the end, because it doesn't exist. That's what infinity means.

Well, the same thing is happening to me and I just noticed an error in my post...

It should be

there is an extra .0000[infinity - 1]1 which makes the whole thing equal exactly 1.

That works in my brain, don't know about yours.

NOW STOP REPLYING TO ME SO I CAN GET SOME FUCKING SLEEP!

 

[Scooter789]

I hate you for starting this topic. This has been beaten to death on the internet for ages. Stop it. Everyone stop it. Talk about something else that matters.

I'm not posting my side of this. I will not fuel this fire.

VIVA LA RESISTANCE or whatever.

 

[Rabid Toilet]

Why is it invalid, though?

Because 10x-x isn't 9. It's 8.999999....9991.

There is no 1 at the end, because there are an infinite number of 9's. Therefore, there is no end at which to place a 1.

 

[Coldie]

But I would say no unless it says round up, simply cause no two numbers can be the same.

Oh the answer to that is simple. No two different numbers can have the same canonical representation, yes. That's precisely why (9) (that is, an infinite amount of nines) is not allowed in a canonical notation, as it would give any number two representations, e.g. 571.25 = 571.24(9). However, a number can have any amount of [non-canon] representations, so 0.(9), 0.99999..., 1, 8-7, 243/243 all represent the same number.

Yeah, I was taking the [infinity - 1 =/= infinity] approach, because it makes sense to me. But that's probably because I think like a computer and I treat infinity as the maximum value any variable can hold. Therefore [infinity - 1 =/= infinity] but [infinity + 1 = infinity], because if you can go higher than infinity, that new value is infinity.

You approach is incorrect, no matter how much sense it does or does not make. Infinity is not any single value, it is just 'infinity', even and odd at the same time. Any arithmetic operation with one infinite operand will produce infinity as the result (except paired with 0 or x/infinity). In arithmetics, 1 + infinity = infinity + 1 = infinity = infinity - 1 = -1 + infinity. So, no matter how you approach it, 0.(9) is the exact same number as 1, or x.y(9) = x.[y+1]

Infinity math gets more complicated fun with Ordinal Numbers, but that's a part of Set Theory, look it up for more confusion entertainment.

 

[Rabid Toilet]

Except 1/3 is not .333333[infinity]. It is .33333[infinity][1/3]. By your example, 3 * [1/3] would be .9999999[infinity]. If you tack on the [1/3] at the end of the infinity decimal, there is an extra .0000[infinity]1 which makes the whole thing equal exactly 1.

But, balls, there doesn't seem to be any way to prove this without circular logic (i reasoned something extra needed to be there because .999999 =/= 1)... I really need to stop getting in on these physics/math forums... I'll make my brain explode at some point.

YOU SEE? THIS IS WHAT HAPPENS WHEN I POST IN THESE THREADS. NOW I'M GOING TO BE HERE ALL DAY RESPONDING. THANKS A LOT!

There is no .000[infinity]1 because there is no end to the zeroes. If the zeroes never end, you cannot place a 1 a the end, because it doesn't exist. That's what infinity means.

Well, the same thing is happening to me and I just noticed an error in my post...

It should be

there is an extra .0000[infinity - 1]1 which makes the whole thing equal exactly 1.

That works in my brain, don't know about yours.

NOW STOP REPLYING TO ME SO I CAN GET SOME FUCKING SLEEP!

IF I HAVE TO BE HERE REPLYING TO PEOPLE THEN SO DO YOU!

Infinity - 1 = Infinity

This is because infinity is not a defined number, it is a concept. No matter how many you take away from infinity, it still goes on forever, because it is by it's very definition, infinite.

 

[Rubashov]

Why is it invalid, though?

Because 10x-x isn't 9. It's 8.999999....9991.

That doesn't make sense. You're saying that 10x-x is a number with an infinite number of decimal places occupied by nines, but the last decimal place is occupied by a one. Which means that you're essentially saying that 10x-x has both an infinite number of decimal places and a finite number of decimal places. That's a contradiction.

 

[crudus]

Why is it invalid, though?

Because 10x-x isn't 9. It's 8.999999....9991.

I see it now. I finally understand where the circular logic is coming from. I knew that proof was shady!

p.s. .999... still equals one. That is just a bad proof.

nevermind. I had a lapse in judgment there.

SEE HERE I AM STILL RESPONDING.

Don't worry. I am still here as well. We can make it through this.

 

[BENZOOKA]

0.999... is not 1
It is as simple as that. No matter how you twist and try to prove it.

 

[Dwarfman]

Toooo...much...higher math...brain...caaaann't

 

[Rabid Toilet]

Why is it invalid, though?

Because 10x-x isn't 9. It's 8.999999....9991.

I see it now. I finally understand where the circular logic is coming from. I knew that proof was shady!

p.s. .999... still equals one. That is just a bad proof.

SEE HERE I AM STILL RESPONDING.

Don't worry. I am still here as well. We can make it through this.

Actually, the reason the proof is shady is because its first step is x = .99...

Basically, you can't start a proof with something that hasn't been proven to be true, because you can only do something to both sides of an equation if the two sides are equal.

For example:

1 = 2
Multiply both sides by zero
0 = 0
Thus, the first equation is also true.

Obviously, this is false, because 1 does not equal 2. If you start out with a true equation, however, such as 0 = 0, there is no way to get to 1 = 2 because it is not a true statement.

Edit:
Okay, yeah, a lot of people have pointed out that x = .99... is technically true because you're just defining a variable. I'm still pretty sure the reason is something like that though.