0.9999 = 1... It truly does!

Conan Edogawa

By Conan Edogawa,
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Conan Edogawa

Conan Edogawa 837

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I am going to introduce a counterintuitive principle.. that 0.9999 repeating equals 1!

This has been known to mathematicians for centuries, but to the average student it remains puzzling to see how this is true, since they might think it only approaches 1.

Here are some proofs.

With basic point set topology, we can easily prove that it actually is 1.

Theorem: Let S = {0.9, 0.99, 0.999, ...}. Then 1 ∈ S.

Proof: First, note that S is closed, since its complement R\S = (-∞, 0.9) U (0.99, 0.999) U ... is a union of open sets, and thus open. Also, 1 is an accumulation point (limit point) of S, since any open ball B(1, r) contains some point in S that is not equal to 1 (I have not seen anyone dispute that 0.9 + 0.09 + 0.009 + ... approaches 1, and this is basically what this sentence says). But S is closed, and thus contains all of its accumulation points. Then 1 ∈ S.

It should be clear that 0.999... is the element of S that is equal to 1.

It also should be said that the equality of 0.9999 and 1 depend heavily on the absence of non-zero infinitesimals. 

Algebraic proof 1(Below proof contains a non-rigorous definition.)

I am going to introduce a counterintuitive principle.. that 0.9999 repeating equals 1!

This has been known to mathematicians for centuries, but to the average student it remains puzzling to see how this is true, since they might think it only approaches 1.

Here are some proofs.

With basic point set topology, we can easily prove that it actually is 1.

Theorem: Let S = {0.9, 0.99, 0.999, ...}. Then 1 ∈ S.

Proof: First, note that S is closed, since its complement R\S = (-∞, 0.9) U (0.99, 0.999) U ... is a union of open sets, and thus open. Also, 1 is an accumulation point (limit point) of S, since any open ball B(1, r) contains some point in S that is not equal to 1 (I have not seen anyone dispute that 0.9 + 0.09 + 0.009 + ... approaches 1, and this is basically what this sentence says). But S is closed, and thus contains all of its accumulation points. Then 1 ∈ S.

It should be clear that 0.999... is the element of S that is equal to 1.

It also should be said that the equality of 0.9999 and 1 depend heavily on the absence of non-zero infinitesimals. 

Algebraic proof 1(Below proof contains a non-rigorous definition.)

{\begin{aligned}x&=0.999\ldots \\10x&=9.999\ldots &{\text{multiply by }}10\\10x&=9+0.999\ldots \\10x&=9+x&{\text{definition of }}x\\9x&=9&{\text{subtract }}x\\x&=1&{\text{divide by }}9\end{aligned}}

Another proof relies on the expansion of 0.9 repeating into a series.

For 0.99999... one can rely on the geometric convergence principle.

ar+ar^{2}+ar^{3}+\cdots ={\frac {ar}{1-r}}. ..

Because the common ratio is .1, we can

0.999\ldots =9\left({\tfrac {1}{10}}\right)+9\left({\tfrac {1}{10}}\right)^{2}+9\left({\tfrac {1}{10}}\right)^{3}+\cdots ={\frac {9\left({\tfrac {1}{10}}\right)}{1-{\tfrac {1}{10}}}}=1.\,

make short work of such an intimidating number.

Hopefully this has somewhat enlightened you into real analysis. 

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Conan Edogawa

Conan Edogawa 837

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2 hours ago, Master Flap said:

I have a simpler to understand proof.

1/3 = 0.33333...

3 * 1/3 = 3 * 0.33333.... = 0.9999......

But 3 * 1/3 = 1

Therefore 1 = 0.99999....

A fine proof, but it sheds little light on decimals and the numbers they represent, where the question underlies whether two decimals are said to be equal at all.

Also, these proofs are not rigorous as they do not contain an analytical definition of 0.999..., which completely invalidates the use of such a proof.

Edited by Field of Dreams
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Shawnic

Shawnic 2,977

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1 minute ago, Field of Dreams said:

A fine proof, but it sheds little light on decimals and the numbers they represent, where the question underlies whether two decimals are said to be equal at all.

Also, these proofs are not rigorous as they do not contain an analytical definition of 0.999...

Yeah, I agree. I find the concept of 0.999... difficult to fully grasp. 

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Grant

Grant 646

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7 hours ago, Master Flap said:

I have a simpler to understand proof.

1/3 = 0.33333...

3 * 1/3 = 3 * 0.33333.... = 0.9999......

But 3 * 1/3 = 1

Therefore 1 = 0.99999....

I was waiting for Master Flap to reply to this. :P

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Cytochrome

Cytochrome 156

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I have something that can benefit all of these posts..

Let use a Dedekind cut approach. This is indeed a comprehensive proof, though its elegance is hidden inside this bountiful proof..

Define every real number as the infinite set of those less than x. For example, e is the infinite set of all numbers less than e.

So the number 0.999.. is the set such that x < 0, x < 0.9... x < .999.

{\begin{aligned}1-\left({\tfrac {1}{10}}\right)^{n}\end{aligned}}.

Every element of 0.9999.. is less than 1.

This implies 

{\begin{aligned}{\tfrac {a}{b}}<1-\left({\tfrac {1}{10}}\right)^{b}\end{aligned}}.

Which then in turn implies that 0.9999.. = 1 because they have the same numbers in their sets.

Q.E.D

Edited by Luminary Spoon
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