1554

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15

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Topic

#1655
# 0.999... is NOT equal to 1

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Finished

The debate is finished. The distribution of the voting points and the winner are presented below.

Winner & statistics

After 5 votes and with 3 points ahead, the winner is...

garai

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- 4
- Time for argument
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- 5,000
- Voting period
- One month
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1507

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2

debates

50.0%

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Round 1

On my previous debate:

I definitely proved that 0.999 does = 1. The most compelling proof of that was:

X = 0.999...10x = 9.999...10x - x = 9.999... - 0.999...9x = 9X = 9/9 = 1

To prove that the conclusion is in fact wrong, i will repeat the same arithmetic on another number: ...999.0.

X = ...999.0

10x = ...9990.0

X - 10x = ...999.0 - ...9990.0

-9x = 9

X = 9/-9 = -1

Clearly something went wrong because even a single 9 before the decimal point is greater then 0, forget about an infinity of 9s. A negative number is an absurd conclusion, thus the argument that they are equal is reduced to absurdity.

This is interesting.

First of all, I want to thank you because you proved 0.999... = 1 for me. So there is no need to give some proof other than this.

But, in second "proof" you made mistake even before you started calculation. You said you will use same arithmetic on number ...999.0. And that is the mistake: number like this does not exist. Because if you said that number is ...999 then it is infinite number-there are infinite period of 9s after 0. and you can't just put 0 after infinite period of 9s otherwise it wouldn't be infinite... It isn't even a number. It's same as you wrote 0.999...0 - number like that does not exist.

Hope this proves my point.

Round 2

Your welcome, it wasn't something I was trying to hide.

However, it is contradictory to say that an infinite amount of 9s before the decimal is not valid, while accepting a number with infinite 9s after the decimal.

if ...999.0 does not exit, then neither does 0.999..., which invalidates the original proof of 0.999... = 1. Thank you very much. Please vote PRO! :)

It is contradictory to say that neither of our proofs are true and because of that one side or other should win. You tried to go off topic.

So, to review this shortly and clearly:

1) 0.999... IS periodic infinite number(check it anywhere, ask teacher in elementary school, if you aren't ashamed)

2) 0.999...0 (what you tried to calculate) IS NOT an infinite number like we said, because it has end-that's 0.

Don't vote for con, but VOTE for common sense. ;)

Thank you.

Round 3

2 correction:

1. Ending in zero doesnt make it any less equal to infinity. It has infinite 9s in front of it.

2. I was not trying to prove that ...999.0 = -1. I was trying to prove, as per topic, that 0.999... does NOT =1. Specifically by invalidating the method used to prove that it does and settle a long fought debate in mathematics.

1) You said, I quote: "Ending in zero doesnt make it any less equal to infinity." I am pretty sure that everyone reading this debate at the moment will agree with me that, if something has ENDING, then it is NOT INFINITE.

2) Objection to your second statement: You were trying to prove that 0.999... is not 1 BY proving that ...999.0 is =-1. So, if your method isnt valid and clear, conclusion from it cannot be any more valid or clear. :)

Round 4

1. In numbers, "ends in" refers to the smallest part of a number, the singles spot. Thus 1,000,009 "ends in" a 9. So having a zero in the last and smallest position doesnt change the infinite 9s before it. Hundreds, millions, billions, and on to infinity.

I trully do not understand the confusion here. Your argument revolving around "ends in" is essentially a play on words.

2. My opening statement, where i did the initial =-1 equation concluded with a declaration that both results are absurd. My arguments involved how similar the numbers and methods are, and I highlighted how if either one is invalid, both must be invalid.

Ill let #2 also be my closing statement.

As a bonus, i will show method to this madness with one more number:

X = ...999.999...

10x = ...999.999...

10x - x = ...999.999... - ...999.999...

9x = 0

X = 0

Which again seems absurd... until you realize that

...999.0 = -1

0.999... = 1

...999.999... = ...999.0 + 0.999...

...999.999... = -1 + 1 = 0

At least the insane equations work together :)

You found the truth my friend, we all live in matrix and all world around you is a simulation. Maybe this could be a topic for some other debate...

Seriously now, I wanted to give myself some logical answer to this "problem" and I did some research. And even basic math with infinity is just...well,undefined. So, for example:

infinity+ 1= infinity, but subtracting infinity from infinity, well we could call it 'not-a-number-ness', because subtracting from infinity doesnt have a definite result,so your calculations fall short because of this. You tried to subtract infinity, and I respect that try, minds like you could change a world for better, just focus on something more practical. :))

After all, many big mathematician brains have been thinking about this long before us, so if it is consensus, it is consensus because of these things I mentioned above,and many others for sure.

We really live in a world full of absurds and anomalies, but thank you for high-quality debate.

Hope to see you in some next debates.

P.S. This could be named a 'tie' but let the votes prevail. :)

Yeah – and also, we encounter infinite series that converge on 1 in the real world all the time! It’s not like something we’ve never heard of.

You take a circle and divide its circumference by its diameter – there you have it, an infinite decimal.

You invest some money in an economy if inflation is fixed in the short run (e.g., a really really sticky economy) – and the spending exists in an infinite cycle. That’s 1 + C'(Y) + [C'(Y)]^2 + [C'(Y)]^3 ... and since C'(Y) belongs to (0,1), that’s identical to 1/(1-C'(Y)) (C'(Y) is the derivative of consumption with respect to real GDP).

You are not wrong. I get frustrated at these debates because, in most of them, even the "right" side of the debate seems to lack a fundamental understanding of mathematics.

It all boils down to what the decimal representation of a number even means. Each digit represents some multiple of a power of ten. The multiple is equal to the value of the digit at that place. The power of ten is dependent on that digits location relative to the decimal point, starting with 10^0 with the digit immediately to the left of the decimal point and increasing by one as you move left and decreasing by one as you move to the right. Thus any decimal representation can be expanded into a series as follows:

abc = a(10^2) + b(10^1) + c(10^0)

Example:

192 = 1(10^2) + 9(10^1) + 2(10^0) = 1(100) + 9(10) + 2(1) = 100 + 90 + 2 = 192

In either case of 0.999... or ...999.0 we have an infinite series. In the former case, the infinite series is equal to:

9/10 + 9/100 + 9/1000 + 9/10000 ...

Each term gets smaller and smaller and this series converges on a single value. Unsurprisingly, this series converges on (and is therefore equal to) 1. For the latter case however, we have:

9 + 90 + 900 + 9000 ...

Each term gets larger and larger and this series diverges. It means that its sum has no value (not even infinity). It just doesn't make mathematical sense to say that the sum of the series even exists.

Sorry, I’m not very well-versed in pure mathematics, but here’s my intuitive reaction to your argument (I may be wrong).

While the number 0.999... is clearly a number in the interval [0,1], the number ....999.0 seems to equal infinity if it extends indefinitely in the leftward direction. So your proof seems to rely on adding and subtracting infinities. I may be wrong here, though.

Vote Reported: RM // Mod action: Not remove

Vote is good

Man, how do I always miss these slam dunk debates.

“So... let me get this straight:

10-0.999...=0.000...1, correct? Ok.

The "..." in 0.000...1 represents an infinite number of 0s before the one. Since there is an infinite number of zeros, there will always be a 0 after the previous one. Thus, the 1 at the end will never be reached. Since the 1 will never be reached, 0.000...1 is the same as 0.000..., which is simply 0.

Going back to our equation, we now have 10-0.999...=0

Adding 0.999... on both sides, we get 10 = 0.999...

QED”

Where you see a 10, replace it with a 1. I had big brain fart lol

No prob!

Wow. Gonna take me a bit of time to sit down and digest that, but thanks, looks interesting.

If you want me to show you the problem with the equation in this debate using this math, let me know ;)

Dang I feel like a Math Professor now lol

Since 0.999… is a repeating decimal, it can be expressed as an infinite sum:

0.999… = 0.9 + 0.09 + 0.009 + 0.0009 + …

S∞ = The entire infinite sum → S∞ = 0.999...

A = The first term → A = 0.9

R = The constant multiple (this can be derived from dividing the second term from the first)

R = (9/100)/(9/10) ← Dividing two fractions is the same as multiplying the first fraction by the reciprocal of the second.

R = (9/100)*(10/9)

R = 90/900 ← Simplify

R = 1/10

Since |1/10| < 1, we will use the second case of the equation. We will now plug in our numbers into the equation.

0.999… = 0.9/(1 - 1/10)

0.999… = (9/10)/(10/10 - 1/10)

0.999… = (9/10)/(9/10) ← Any number divided by itself is 1

0.999… = 1

QED

For the second case, remember that x^n will increase (gets closer to infinity) as n increases (gets closer to infinity). In this case, N = ∞. So, R^N = R^∞ = ∞. With that, we get this:

S∞ = A(1 - R^∞)/(1 - R), |R| > 1

S∞ = A(1 - ∞)/(some negative number)

S∞ = A(-∞ + 1)/(some negative number)

At this point, it is important to note that infinity plus/minus any finite number is still infinity. Similarly, negative infinity plus/minus any finite number is still negative infinity. Thus:

S∞ = A(-∞ + 1)/(some negative number)

S∞ = A(-∞)/(some negative number)

S∞ = -A(∞)/-(some positive number) ← remove the negative from the negative number

S∞ = A(∞)/(some positive number)

S∞ = ∞/(some positive number) ← Infinity/(Any finite number) = Infinity

S∞ = ∞

For this case, we can see that the infinite geometric series equation will equal to infinity. This leaves us with the third case:

S∞ = A(1 - R^∞)/(1 - R), |R| < 1

For the third case, remember that x^n will decrease (gets closer to zero) as n increases (gets closer to infinity). In this case, N = ∞. So, R^N = R^∞ = 0. With that, we get this:

S∞ = A(1 - R^∞)/(1 - R), |R| < 1

S∞ = A(1 - 0)/(1 - R)

S∞ = A(1)/(1 - R) ← Remember: A(1) = A

S∞ = A/(1 - R)

With that said, there are two possible cases for this equation:

S∞ = A(1 - R^∞)/(1 - R), |R| > 1 → S∞ = ∞

S∞ = A/(1 - R), |R| < 1

I will now use this equation to prove that 0.999… = 1.

Now that we have the equation for the geometric series, we need to establish what an infinite geometric series is.

An infinite geometric series is one which has an infinite amount of terms in its sequence. Since N is the number of terms in the sequence, and an infinite geometric series has an infinite amount of terms in its sequence (by definition), N = ∞. Let the entire infinite geometric series be S∞.

Retrieving the equation for the geometric series, and plugging ∞ into N, we get this:

S∞ = A(1 - R^∞)/(1 - R)

At this point, it is important to note the nature of exponents. Any number (x) to the power of something (n) will get larger or smaller when n gets larger, depending on what x is. The following will be the case for positive and negative numbers, so we can put an absolute value sign (| |) around x. This will make x an absolute value number. What an absolute value sign does is that it turns a negative number into a positive one. Positive numbers will not be affected.

If |x| > 1, then x^n will increase as n increases. (eg, 2^2 = 4, 2^3 = 8, etc.)

If |x| < 1, then x^n will decrease as n increases. (eg, (½)^2 = ¼, (½)^3 = ⅛)

If |x| = 1, then x^n will remain the same.

With this in mind, we can address the equation at hand. An absolute value number increasing can be thought of as getting closer to infinity, while an absolute value number decreasing can be thought of as getting closer to zero. There are three possible cases for the equation:

S∞ = A(1 - R^∞)/(1 - R), |R| = 1

S∞ = A(1 - R^∞)/(1 - R), |R| > 1

S∞ = A(1 - R^∞)/(1 - R), |R| < 1

For the first case, we get this:

S∞ = A(1 - R^∞)/(1 - R), |R| = 1 → S∞ = A(1 - 1^∞)/(1 - 1) → S∞ = A(1 - 1^∞)/0

There are two problems with this. Firstly, we cannot divide by zero. Secondly, 1^∞ is indeterminate. Thus, we can outright eliminate this case.

[continued in next comment]

Since any number subtracted by itself is 0, we are left with:

S - SR = A - AR^N + 0 + 0 + 0 + ...(all of the terms in here are 0)... + 0 + 0

Simplified, we get this:

S(1) - SR = A(1) - AR^N

Since any number times 1 is itself, the number can be restated as itself times 1.

We can factor out the S and the A to get this:

S(1 - R) = A(1 - R^N)

Since we are trying to find S, we divide both sides by 1-R. As a result, we get:

S = A(1 - R^N)/(1 - R)

And this is the equation for the geometric series.

[Continued in next comment]

This goes on for the amount of terms that are in the sequence. Let's call the number of terms in the sequence "N", and the sequence as a whole as "s".(Again, what we call these by is completely arbitrary)

So, a geometric sequence can be shown as:

s = [A, AR, AR^2, ..., AR^(N-3), AR^(N-2), AR^(N-1)]

The reason that the final number in the sequence is AR^(N-1) and not AR^N is because on the 1st term, N = 0 (R^0 = 1, and A*1 is simply A).

Since a series is simply the sum of a sequence, a geometric series (let's call the variable for the entire series "S") can be shown as:

S = A + AR + AR^2 + ... + AR^(N-3) + AR^(N-2) + AR^(N-1)

This is our first equation for the proof.

If we multiply R by both sides, we get:

SR = R * [A + AR + AR^2 + ... + AR^(N-3) + AR^(N-2) + AR^(N-1)]

Remember that the exponent simply shows the number of times that something is being multiplied by. So, if we multiply something raised to the power of n (n is simply the exponent of that something) by that same something, then it is the same as adding 1 to n. With that in mind, we get:

SR = AR + AR^2 + AR^3 + ... + AR^(N-3+1) + AR^(N-2+1) + AR^(N-1+1)

= AR + AR^2 + AR^3 + ... + AR^(N-2) + AR^(N-1) + AR^N

This is our second equation.

Now, we subtract the second equation from the first (First equation is S, second equation is SR):

S - SR = [A + AR + AR^2 + ... + AR^(N-2) + AR^(N-1)] - [AR + AR^2 + AR^3 + ... + AR^(N-2) + AR^(N-1) + AR^N]

The minus sign in front of the first series can be distributed into each number within the series:

S - SR = [A + AR + AR^2 + ... + AR^(N-2) + AR^(N-1)] + [-AR - AR^2 - AR^3 - ... - AR^(N-2) - AR^(N-1) - AR^N]

We can get rid of the brackets:

S - SR = A + AR + AR^2 + ... + AR^(N-2) + AR^(N-1) - AR - AR^2 - AR^3 - ... - AR^(N-2) - AR^(N-1) - AR^N

Then we can rearrange the equation:

S - SR = A - AR^N + AR - AR + AR^2 - AR^2 + AR^3 - AR^3 + ... - ... + AR^(N-2) - AR^(N-2) + AR^(N-1) - AR^(N-1)

[continued in next comment]

I'll show you the infinite geometric series proof that I forgot to explain in full last time.

Before I go on to proving that 0.999...=1, I need to establish what an infinite geometric series is. Before I can do that, we need to establish what a geometric series is. Before that, I need to define what a series is.

A series is simply the sum of a sequence. A sequence is just a set of numbers. For example, 15 is the sum of the series [1,2,3,4,5] (the square brackets are just the formal way of showing a set). If you add all the numbers of the sequence together, you will get 15.

A geometric series is a series in which each subsequent number in the series' sequence is multiplied by some constant. There will be the first number in the sequence, and then that number will be multiplied by something to get the next number, which will be multiplied again by that something to get the next, and so on and so forth. Let's define the first number as "A" and the constant which "A" is being multiplied by as "R". These variables are completely arbitrary, and we can call them whatever we want. Given these two variables, we can now define our geometric sequence in a mathematical sense.

The first number of the sequence is just A, by definition.

The second is A*R, since a geometric sequence is, by definition, in which each subsequent number in it is the previous one multiplied by a constant. This can be rewritten as AR.

The third is AR*R, since we are multiplying the last term by the same constant (R). Since R*R can be rewritten as R^2, this term can be rewritten as AR^2. One important thing to remember is that the exponent of something simply denotes how many times that variable is multiplied by.

The fourth term is AR^3, since the previous term is AR^2 and we are multiplying it by R again.

[continued in next comment]

Subtraction is a valid operation.

I did 10x - x = 10x -x in the proper order

10x is the ...9990.0

X is ....999.0

10x - x was equal to ...9990.0 (aka 10x) - ...999.0 (aka x)

It was all in the proper order.

Does that negate your entire vote? And how do i bring this comment to the proper mods attention if it does?

So... let me get this straight:

10-0.999...=0.000...1, correct? Ok.

The "..." in 0.000...1 represents an infinite number of 0s before the one. Since there is an infinite number of zeros, there will always be a 0 after the previous one. Thus, the 1 at the end will never be reached. Since the 1 will never be reached, 0.000...1 is the same as 0.000..., which is simply 0.

Going back to our equation, we now have 10-0.999...=0

Adding 0.999... on both sides, we get 10 = 0.999...

QED

Im unsure of that conclusion, but i am sure about your error. So i will be arguing a bit of devils advocate based of my previous mindset.

If one were to try to manually calculate all the 9s... even given infinite time no they will never reach that final 9 that makes it a one...

But the number itself is not growing 1 number at a time. It is in its entirety now. All infinite 9s are there inside of its definition. And *at the point of infinity*, when all infinity of that number is valued at once. It is equal to 1.

Its like an asymptotes in math class. It approaches the line, but never reaches the line as far as you can see... but if you extend that line to infinity, it does meet the line at that infinite point.

Then you will admit that 0.999... never becomes 1.000... because it forever can never reach it.

Red pill time.

The issue isnt whether its a 9, a 1, or a 0. The issue is with the concept of "last". There is no "last". When faced with an infinity, last has no meaning.

There is no last. There is no after. Thats the reality of infinity. Your sentences dont make any sense in the context of infinity.

The only way to negate this concept is to show how mathematically absurd it is with similar examples, as i tried.

Do you think because the last 9 of 0.999...9 happens to be the same digit as the others means it's more attainable than the '1' at the end of 0.000...1?

I dont understand what your referingto with the "just as" in that first sentence, but i think the key is in my last question.

Can there be a last, or an after, in an infinite series?

Just as much as the infinite 0's followed by a 1 that is the difference has the '1'.

So, I'd totally concede the '1' is never reached, and neither is the '9' because recurring numbers basically can't truly exist in their base. Base-10 doesn't allow for 1/3, it also doesn't allow for whatever sum you use to get 0.999... to ever be equal to 1 unless you abusively play around with algebra in a one-way-conversion pseudo trick that revolves around the fact that 0.9999.... is a fake value.

Does a series of infinite 9s have a "last" 9?

Then it never reaches the last 9.

There is no after the infinity. The most you can say is "at the point of infinity". Before that point 0.999... "approaches" 1, but the number isnt actually changing with time or approaching anything. If you take the number as a whole, "at the point of infinity" it supposedly becomes 1.

Im not sure of that conclusion, but i am sure of your "after infinity" flaw. 0.999...x does not exist, there is no "x", it will always be "..." with no end. Whether the x is another 9 or a different number, there is no final digit. Ever. Your notation doesnt make sense.

So after the infinity it never adds on the 1 to become 1 from the 9, yes?

There is no "after" with infinity. There will be 9s, none of them are after the infinity. There will never be a 1, or any number not part of the repeating pattern.

The problem with your logic is your insistence on an "after the infinity". Thats a ludicrous statement by itself.

1 after infinite 0's is directly comparable to a 9 after infinite 9's. Just as unattainable.

Last time we pursued this discussion, you refused to agree to disagree. I was convinced by a math argument that may or may not have been valid, but your logical argument is illogical to me. 0.000...1 is nonsensical because you are trying to put a 1 after infinite zeros.... except there is no after infinity.... i dont want to force a conversation that will end in a rage quit, and i have no interest in a circular conversation. If it becomes clear neither of us is convincing the other, and we've heard all of each others arguments, lets agree to disagree.

Likewise, there is no 9 at the end of 999..., there are always more 9s after that one. There will.never however be a different number once the ... kicks in

To do this angle, why not point out that the '9' at the end of the 0.999... is no more or less attainable than the '1' at the end of the 0.000...1?

This blackmails the opponent to admit the difference is valid or that 0.999... is a fake value.

3*0.333..., according to my proposition, is incalculable because you are trying to calculate a number with infinite digits. You can end an equation with 0.333..., but you cant start a computation with it. Thats what i was trying to demonstrate.

It's wrong. 3*0.333... never makes 1 because it needs a 'third' at the end of the 3's.

It has convinced people, just not on this website posting in public.

But your proofs are not killer, are quite illogical. I cant understand how you can label a proof as "killer" when it hasn't convinced anyone.

Your argument about 0.333... is undercut by your claim that you cant start at 1 and get to 0.999..., because you can start at 1/3 and end at 0.333... as well as the reverse. You simply misunderstand infinity.

I doubt i will convince you, we can agree to disagree... but why would i use your "killer" proofs when they never worked for you in your debates?

The point i was trying to make is you cant calculate with these ludacris numbers. You can get them in a conclusion if you start with normal numbers like 1/3, but you can continue the maths, or start the maths with infinite digit incalculable numbers. But if you divide 1 by 3, you will forever get another 3 to the end of the number. There will never be a 4, no 5, no 05. Those dont even make any sense at all.

Glad to see that this debate made some serious polemics here, looking forward to see you in the next one!

Schucks

As I stated before, if ...9999.0 is infinite (as seems to be the case with the ... to the left of the 9s), then you’re left with ♾-10♾. 10♾ is still infinity, so you’re left with ♾- ♾. This is indeterminate, meaning that there is no definite or definable variable. You can’t calculate an answer from an indeterminate equation. Since your entire argument hinges on the basis that ...9999.0-...9990.0=9 (essentially, ♾-♾=9), and this is shown not to be the case, the argument falls apart.

I may be growing conflicted due to some follow up research. Can you elaborate on how that fact defeats my argument?

Good on you for debating from multiple perspectives, but the error I pointed out (if ...9999.0 is infinite) pretty much defeats the entire argument.

You may be right with this number, but i initially simply assumed it was just a number. Many debates here go crazy with formalities, i just want to dive in :)

And by that, I mean explaining what ...9999.0 means

I’m just saying, you might wanna clarify your terms before proceeding

Perhaps. Once again this is an ongoing debate. Maybe towards the later rounds depending on my opponents perfornance.

If ...9999.0 is suppose to be infinity (as the “...” to the left of the 9s implies), then you end up with ♾-♾, which is indeterminate.

That seems like something that may come up in the next round, ill save my answer til then :p

Question:

What number is ...9999.0?

Double contrary, my bad. Im on mobile, forgive any typos and doubling :/

Yes 0.999... = 1 is the consensus.

Just to be sure i understood you clearly, you said that it is "contrary to the contrary",it is like minus and minus give plus, so basically you are telling that 0.999... = 1 is consensus and that it is difficult for you to prove the otherwise? :))