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Nope

Does 0.999 with nies that go on forever equal 1?

someone234

It does not. 1 is actually inferior because it has limitations.

anonymousdebater

No.

Let's imagine an infinite number line. What is the probability that if you select a random integer, it is 0. All integers are equally likely. It is 1 over infinity.

What is the probability of selecting any integer? 1 / infinity per integer times infinity (# integers), which equals 1.
However, if 1 / infinity = 0, then you get 0 times infinity, which equals 0.

1 / infinity =/= 0.

Therefore, 1 - 1 / infinity =/= 1 and 0.9999... does not equal 1.

Nope

anonymousdebater said:

No.

Let's imagine an infinite number line. What is the probability that if you select a random integer, it is 0. All integers are equally likely. It is 1 over infinity.

What is the probability of selecting any integer? 1 / infinity per integer times infinity (# integers), which equals 1.
However, if 1 / infinity = 0, then you get 0 times infinity, which equals 0.

1 / infinity =/= 0.

Therefore, 1 - 1 / infinity =/= 1 and 0.9999... does not equal 1.

X=0.9999....
10X=9.9999...
10X-X=9.9999....-0.9999....
9X=9
X=1
1=0.9999....

1/3=0.333...
3(1/3)=3(0.333...)
1=0.999...
1/9=0.1111
9/9=0.9999
9/9=3/3=1
: )

someone234

Nope said:

anonymousdebater said:

No.

Let's imagine an infinite number line. What is the probability that if you select a random integer, it is 0. All integers are equally likely. It is 1 over infinity.

What is the probability of selecting any integer? 1 / infinity per integer times infinity (# integers), which equals 1.
However, if 1 / infinity = 0, then you get 0 times infinity, which equals 0.

1 / infinity =/= 0.

Therefore, 1 - 1 / infinity =/= 1 and 0.9999... does not equal 1.

X=0.9999....
10X=9.9999...
10X-X=9.9999....-0.9999....
9X=9
X=1
1=0.9999....

1/3=0.333...
3(1/3)=3(0.333...)
1=0.999...
1/9=0.1111
9/9=0.9999
9/9=3/3=1
: )

This is where you went wrong. You never can say that equals anything without counting out the last 9.

Nope

someone234 said:

Nope said:

anonymousdebater said:

No.

Let's imagine an infinite number line. What is the probability that if you select a random integer, it is 0. All integers are equally likely. It is 1 over infinity.

What is the probability of selecting any integer? 1 / infinity per integer times infinity (# integers), which equals 1.
However, if 1 / infinity = 0, then you get 0 times infinity, which equals 0.

1 / infinity =/= 0.

Therefore, 1 - 1 / infinity =/= 1 and 0.9999... does not equal 1.

X=0.9999....
10X=9.9999...
10X-X=9.9999....-0.9999....
9X=9
X=1
1=0.9999....

1/3=0.333...
3(1/3)=3(0.333...)
1=0.999...
1/9=0.1111
9/9=0.9999
9/9=3/3=1
: )

This is where you went wrong. You never can say that equals anything without counting out the last 9.

There is no last 9. If there were it would not equal 1. It is infant. 1-0.999...=0.0000... because the number is infent that 1 at the end will never come. A 0 with an ifanet zeros behind it is just 0. You cannot count the nines because infanet is not a number. Why can't I say they are equal without counting the nines?

WilliamSchulz

.9999 is only ever equal to 1 when talking about infinity sequences, so if I have say 1 + 1/2 + 1/4 + 1/8 +1/16, eventually, after infinity, the number will be tantalizingly close to 2, which is the hint I am to receive, the problem is that that number is 1.99 inf. for the sequence. Thus, to equal the infinity sequence of 1, you would need 1/2 + 1/4 +1/8 etc, and .999999 is a stretching of the infinity that will never equal one but will be tantalizing close enough to round it to one anyways. 

Nope

WilliamSchulz
I am confused. First you said ".99999.. is only ever equal to 1 when taking about infinity sequence" But then you said ".999999 is a stretching of the infinity that will never equal one but will be tantalizing close enough to round it to one anyways.". Are you agreeing with me or not?

WilliamSchulz

It depends. In math, when concerning infinity sequences, the numbers should add up to one as the solution to the infinity sequence, but because the numbers stretch into fractions, it will never fully equal one. As mathematicians, we will label the number as one as the solution to the infinity sequence for purposes of Calculus, but in truth, .9999 will never truly equal one. 

Nope

WilliamSchulz
Are you saying it is impossible to write 1/3 as a number because 0.333... does not equal 1/3? What do you think 9.999...-0.999... is? Why does 0.999...+9=10*(0.999...) if 0.999... is not one?

WilliamSchulz

I think you are riding off of a strong point here, namely that we label 1/3 as .33 inf. and assume that 1/3 times three equals one, but wouldn't it just be .99999 inf? 

I guess the response to this would simply be that 1/3 might be .333 inf. and somewhere in all of it, there is a 4. Again, based on rounding rules, we are willing to overlook .00000001 to round the number to one, even if it isn't. 

Ampersand

WilliamSchulz said:

I think you are riding off of a strong point here, namely that we label 1/3 as .33 inf. and assume that 1/3 times three equals one, but wouldn't it just be .99999 inf? 

I guess the response to this would simply be that 1/3 might be .333 inf. and somewhere in all of it, there is a 4. Again, based on rounding rules, we are willing to overlook .00000001 to round the number to one, even if it isn't. 

There isn't a 4 in there though, because the 3s go on to infinity. There is no end to the threes so there can be no four.

anonymousdebater

@Nope ;
"A 0 with an ifanet zeros behind it is just 0."

0 with infinite zeros behind it is 1 over infinity. But is that zero?
 

Fascism

Yes because 

1/3 = 0.333... 
So (1/3)*3 = 0.999... = 1. 

someone234

@Fascism 1/(3/3) = 1

1/3*3 = 0.9999...

0.99999... =/= 1


You can't just round it and say it's a truth.

1/(3/3) = 1/1 but 1/3*3 is only equal to it if you round it away from its true value to a lesser amount of figures.

Fascism

@someone234 ;
1/(3/3) = (1/1)/(3/3)
(1/1)/(3/3) = 1/1 = 1
(1/1)*(3/3) = 1/3 * 3 = 0.999...
Therefore, 1/(3/3) = 1/3*3, and 1 = 0.999... This isn't rounding. I only used addition, multiplication, and division. 

BaconToes

But think of it like this using @Nope's previous equation:
X=0.9999....
10X=9.9999...
10X-X=9.9999....-0.9999....But
9X=9
X=1
1=0.9999....

But what if we did 1-0.999..., wouldn't the answer be 0.0...1? Not 0

Nope

BaconToes
1 is at the end of the the decimal. However because 0.999... is infante so must 0.000.... There is no end to the decimal. If there is no end then there can be no one. Having a infanet amount of zeros with no end to place the one is equal to zero with an ifanet amount of zeros which is zero. 

Ampersand

BaconToes said:

But think of it like this using @Nope's previous equation:
X=0.9999....
10X=9.9999...
10X-X=9.9999....-0.9999....But
9X=9
X=1
1=0.9999....

But what if we did 1-0.999..., wouldn't the answer be 0.0...1? Not 0

No, it wouldn't. You can't simultaneously say that there are both infinite zeroes and finite zeroes because that's contradictory. You assume the former because there have to be infinite zeroes if there were infinite nines and you specify the latter because eventually the 0s have to stop (hence not infinite) so you can have the 1 at the end.

EmeryPearson

Ah, do to the mathematical proofs that can be offered on this, I have been persuaded. .999... Does in fact equate to 1.

MayCaesar

I love questions like this! Let me try to give a mathematical, and yet easy to understand, explanation of this fact.

First, in mathematics we use the concept of limit. The limit is the number to which a given sequence of numbers converges.
For example, the sequence 1, 0.1, 0.01, 0.001, ... converges to 0, that is it gets infinitely close to 0, hence its limit is 0. The sequence 1, 1+1/2, 1+1/2+1/4, 1+1/2+1/4+1/8, ... converges to 2, hence its limit is 2
Strict definition of limit is a number such that the difference between it and the sequence gets smaller than any possible value, starting at some point. Does it get closer to some value than 0.000000000000001? Closer than 0.00000000000000000000000000000001? Closer than...? If the answer to all these questions is "yes", then this value is the limit.

Now, let us come back to the original question: does 0.999... equal 1? To answer this, we must first understand what 0.999... exactly means. By definition, it is nothing more than the limit of the following sequence:
0.9, 0.99, 0.999, 0.9999, ...
What is the limit of that sequence? It is easy to see that it is 1. Indeed, 1 - 0.9 = 0.1; 1 - 0.99 = 0.01; 1 - 0.999 = 0.001 - it is just the same sequence as in the first example above, and it converges to 0. So the difference between 1 and this sequence gets infinitely small, hence the limit of this sequence is 1.

Hence, 0.999... = 1. There is no trick here: 0.999... is just a smart way of writing down a limit of a sequence converging to 1.

---

Note, however, that not all repeating decimals can be written as a simple number. For example, 0.444... cannot be written as one short number. However, it can be written as a fraction of 4/9. It is easy to see from the following:

0.444.../0.999... = 4/9
0.999... = 1
0.444.../1 = 4/9
0.444... = 4/9

drodgers

Yes for all practical purposes.

For precision, it does not.

AlexOland

Yes. And this is an important idea in mathematics. One could argue that calculus is based upon this simple idea. There is an infinitely small difference between 0.999.. and 1. 

 As the fathers of calculus has proved, if something is infinitely small it does not exist. Therefore the difference between 0.999.. and 1 is non-existent.

Topaet

The misunderstanding that 0.999... does not equal 1 stems from a misunderstanding of the concept of actual infinity and mistaking it for potential infinity. It is correct, that if there was a "0." followed by a limited amount of 9s, it would never equal 1. However, if there is an unlimited amount of 9s following the "0.", then it does equal 1. To prove this I will present a few mathematical proofs that 0.999... equals 1:

1. Reductio ad absurdum:
If 1 and 0.999... are not equal, then we should be able to find a distinct number in between them (their average). There is, however, no number in between 0.999... and 1. Therefore, this is a reductio ad absurdum argument that proves that 0.999... can't not equal 1 and thus 0.999... equals 1.
---------------------------------------------------------------------------------------
2. The decimal expansion for 1/3:
1/3 = 0.333...
0.333... * 3= 1/3 * 3 = 0.999... = 3/3 = 1

---------------------------------------------------------------------------------------
3. Subtracting off the infinite sequence:
x = 0.999... |*10
10x = 9.999...
10x = 9 + 0.999...
10x = 9 + x | -x
9x = 9 |"9
x = 1

--------------------------------------------------------------------------------------
4. Arithmetic proof:
1-0.999... = 0.000...
Since infinitely small numbers do not exist (Norton and Baldwin, 2018), 0.000... equals 0 and because any number subtracted from an equal value is zero, 0.999... equals 1.
--------------------------------------------------------------------------------------
In conclusion, there are more proofs that 0.999... equals 1 but the four that I have presented should be sufficiently convincing and thus I have affirmed the resolution that 0.999... does equal 1.

Topaet

The common misunderstanding that 0.999... does not equal 1 stems from a misunderstanding of the concept of actual infinity and mistaking it for potential infinity. It is correct, that if there was a "0." followed by a limited amount of 9s, it would never equal 1. However, if there is an unlimited amount of 9s following the "0.", then it does equal 1. To prove this I will present a few mathematical proofs that 0.999... equals 1:

1. Reductio ad absurdum:
If 1 and 0.999... are not equal, then we should be able to find a distinct number in between them (their average). There is, however, no number in between 0.999... and 1. Therefore, this is a reductio ad absurdum argument that proves that 0.999... can not not equal 1 and thus 0.999... equals 1.
---------------------------------------------------------------------------------------
2. The decimal expansion for 1/3:
1/3 = 0.333...
0.333... * 3= 1/3 * 3 = 0.999... = 3/3 = 1

Therefore 0.999... equals 1.
---------------------------------------------------------------------------------------
3. Subtracting off the infinite sequence:
x = 0.999... |*10
10x = 9.999...
10x = 9 + 0.999...
10x = 9 + x | -x
9x = 9 |"9
x = 1

Therefore 0.999... equals 1.
--------------------------------------------------------------------------------------
4. Arithmetic proof:
1-0.999... = 0.000...
Since any number subtracted from an equal value is zero, 0.999... equals 1.
--------------------------------------------------------------------------------------
In conclusion, there are more proofs that 0.999... equals 1 but the four that I have presented should be sufficiently convincing and thus I have affirmed the resolution that 0.999... does equal 1.

Topaet

The common misunderstanding that 0.999... does not equal 1 stems from a misunderstanding of the concept of actual infinity and mistaking it for potential infinity. It is correct, that if there was a "0." followed by a limited amount of 9s, it would never equal 1. However, if there is an unlimited amount of 9s following the "0.", then it does equal 1. To prove this I will present a few mathematical proofs that 0.999... equals 1:

1. Reductio ad absurdum:
If 1 and 0.999... are not equal, then we should be able to find a distinct number in between them (their average). There is, however, no number in between 0.999... and 1. Therefore, this is a reductio ad absurdum argument that proves that 0.999... can not not equal 1 and thus 0.999... equals 1.
---------------------------------------------------------------------------------------
2. The decimal expansion for 1/3:
1/3 = 0.333...
0.333... * 3= 1/3 * 3 = 0.999... = 3/3 = 1

Therefore 0.999... equals 1.
---------------------------------------------------------------------------------------
3. Subtracting off the infinite sequence:
x = 0.999... |*10
10x = 9.999...
10x = 9 + 0.999...
10x = 9 + x | -x
9x = 9 |"9
x = 1

Therefore 0.999... equals 1.
--------------------------------------------------------------------------------------
4. Arithmetic proof:
1-0.999... = 0.000...
Since any number subtracted from an equal value is zero, 0.999... equals 1.
--------------------------------------------------------------------------------------
In conclusion, there are more proofs that 0.999... equals 1 but the four that I have presented should be sufficiently convincing and thus I have affirmed the resolution that 0.999... does equal 1.