.9 recurring.

Author: zedvictor4

Posts

Total: 10
zedvictor4
zedvictor4's avatar
Debates: 22
Posts: 11,259
3
3
6
zedvictor4's avatar
zedvictor4
3
3
6
Will always be one part short of a whole.

1/10, 1/100, 1/1000, 1/10000 etc. etc. etc.


Interestingly:

Decimally one cannot cut a cake into three equal slices.

Whereas fractionally one can cut a cake into three equal slices.

Thirds as it were.


So if you are sharing a cake with two other people, always demand a third, rather than a decimal slice.

Or you might be .000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001 etc,  short of a calorie.
SirAnonymous
SirAnonymous's avatar
Debates: 3
Posts: 4,140
3
7
10
SirAnonymous's avatar
SirAnonymous
3
7
10
-->
@zedvictor4
Actually, while this seems intuitively true, it isn't. There's actually a mathematical proof here, which I will present to show off my utter nerdiness. Before I begin, I need to establish two things: the density of rational and irrational numbers. Real numbers are divided into two sets: rational and irrational. These sets have properties called the density of rational/irrational numbers. This property means that:

For every (ir)rational numbers A and B such that A<B, there exists some (ir)rational number C such that A<C<B.

In plain English, if you have any two real numbers, whether rational or irrational, there will be a third number that is between them. For example, if we take 2.00 and 2.01, there are numbers like 2.005 that are between them.

Now suppose we have two real numbers A and B and that there is no number C such that A<C<B. Assume that A<B. Since A<B, by the density of rational/irrational numbers, there exists some number C such that A<C<B. This is a contradiction: we cannot have A<B and have no number C between them. Either A=B and there is no C, or A<B and there is a C.

The second thing to know is the mathematical definition of infinity. It means to increase without bound (or decrease, for negative infinity). That is, there is no point at which we can stop and say that infinity won't get any larger.

Now take 0.9 repeating and 1. By definition, 0.9 repeating has an infinite number of 9s after the decimal point. By definition, infinite means increasing without bound. Now suppose that there exists some C such that 0.9 repeating < C<1. C cannot be (1 - 0.1). It cannot be (1 - 0.01), or (1 - 0.0001). In fact, no matter how many 0s I add between the 1 and the decimal point,  0.9 repeating will be greater than C. Now suppose that we take the number (1 - 0.000...1), with an infinite number of 0s between the decimal point and the 1. That would fit the bill, right? Wrong. Notice what just happened. The 1 is in the infinity + 1 position after the decimal point. However, infinity + 1 is not a real number;  that contradicts the definition of infinity. There cannot be a real number larger than infinity because infinity increases without bound. Consequently, 0.000...1, and thereby (1 - 0.000...1), is not a real number. Thus, there is no number C between 0.9 repeating and 1. Therefore, the only option is that A=B, proving that 0.9 repeating = 1.


zedvictor4
zedvictor4's avatar
Debates: 22
Posts: 11,259
3
3
6
zedvictor4's avatar
zedvictor4
3
3
6
-->
@SirAnonymous
Is mathematics always representative of reality?

So can parts diminish infinitely so that there is always one part remaining.....Provocatively we can call it the GOD PART if you like.

Or are you saying that ultimately parts cease to exist......Therefore potential is in fact, finite.

To know the answer might explain something.


Though I suppose that fractionally a part is always one part.

So what we are questioning, is can a part be fractionally reduced to a finite point of non-existence.

If not then .9 recurring will never equal 1, because as I stated previously there will always be one part.


AH!......Is that what you are saying?


SirAnonymous
SirAnonymous's avatar
Debates: 3
Posts: 4,140
3
7
10
SirAnonymous's avatar
SirAnonymous
3
7
10
-->
@zedvictor4
Is mathematics always representative of reality?
Not sure. This is a case of mathematics representing mathematics, though.
AH!......Is that what you are saying?
Kind of. Mathematically, an infinitely small fraction is just equal to 0. I can have 0.00000000000000000000000000000000000000000000000000000001 be a real number. It is a finite quantity. However, 0.000...1 is not. So if you have 0.9 recurring, there is no finite point remaining.

43 days later

Ramshutu
Ramshutu's avatar
Debates: 43
Posts: 2,768
6
9
10
Ramshutu's avatar
Ramshutu
6
9
10
-->
@zedvictor4
Easily provable that 0.9999r = 1.

X = 0.999r

Multiple both sides by 1

10x = 9.999r

Subtract x from both sides 
10x - x= 9.9999r - 0.999999r

9x = 9

Which, after running all the calculations and consulting with mathematicians, yields:

X = 1

Meaning 0.999r = 1
zedvictor4
zedvictor4's avatar
Debates: 22
Posts: 11,259
3
3
6
zedvictor4's avatar
zedvictor4
3
3
6
-->
@Ramshutu
For sure.
 

The magic of X.


Still doesn't alter the fact that 0.999r will never achieve a whole.


Nonetheless another way of looking at it is:

X = .999r

ergo .999r =.999r

Multiply both sides by 1

.999r = .999r (Not sure where you got 9.99r from)

If you subtract X from both sides (Given that the value of X is .999r)

We are left with 

0 = 0
Ramshutu
Ramshutu's avatar
Debates: 43
Posts: 2,768
6
9
10
Ramshutu's avatar
Ramshutu
6
9
10
-->
@zedvictor4
Still doesn't alter the fact that 0.999r will never achieve a whole.
Theres two things you’re tripping up on here. Firstly is decimal notation (Because 0.999r = 1/3 * 3 = 1) - your seeing the 0.999 but missing the nuance of what recurring means, because decimals cannot properly express what it means, it’s a bit confusing.

The second issue you’re tripping up on, is the nature of the recurring, and infinities.

If you took 0 followed by a billion 9s, it would never reach 1. A trillion - closer, but doesn’t reach 1. A googleplex of 9s - close but still no.

Any finite number of 9s you can have, there is always a number between it and one - as the number has an end you can still add a 9 to.

You can never reach 1 with finite numbers, but 0.99r is not a finite number. It is closer to 1 than any finite number of 9s that you can ever specify because you cannot add anything to it to make it closer to 1 -there is no actual real number between 0.99r and 1.

You’re treating 0.99r like it’s a finite number, but it’s an infinite number, and that infinite nature is what allows it to reach 1:

Infinities cancel nevers.

I will never reach the end of an infinitely long road - unless I walk for an infinite time. 

So it’s best to stop thinking of 0.99r in terms of finite numbers.


























badger
badger's avatar
Debates: 0
Posts: 2,087
3
3
3
badger's avatar
badger
3
3
3
-->
@zedvictor4
Not sure where you got 9.99r from
He meant to write multiply by 10, zed. It's actually a neat little proof. Multiply by 10 because this shifts the decimal place, but shifting the decimal places makes no difference to the right hand side due to it's being infinitely recurring. 

So .99r * 10 = 9.99r. 
9.99r - the original .99r = 9. 

He could have done the same with multiplying by 100 or 1000 or 10,000 or whatever.

It's actually a neat little proof. 
Ramshutu
Ramshutu's avatar
Debates: 43
Posts: 2,768
6
9
10
Ramshutu's avatar
Ramshutu
6
9
10
-->
@zedvictor4
Sorry I didn’t realize I had included a 1 instead of a 10 when i said multiple both sides by 10.
zedvictor4
zedvictor4's avatar
Debates: 22
Posts: 11,259
3
3
6
zedvictor4's avatar
zedvictor4
3
3
6
-->
@Ramshutu
A Zedku for Ramshutu.


Everything is thought.

Which imagines an infinite quantity.

Therefore,

One can also imagine 

.1

The missing tenth.

Nonetheless,

If the missing tenth is not worthy of consideration.

Is it OK if I have it.