0.9999... = 1?

[lemmingllama]

So this is just a fun thing to play around with math. Basically, I am saying that 0.999... is the same thing as 1. A lot of people do not believe that these two are equal since they think that 0.999... terminates eventually, which it doesnt. So, first let us prove with induction.

x = 0.999......

10x = 9.999......

10x-x = 9.999...... - 0.999....

9x = 9

x = 1

But if you look above, x = 0.999....

So that means that 0.999... is equal to one.

So how about another example, one that non-math people will understand. So you take a bottle of water and pour it into three glasses, so that the bottle is empty and there are equal amounts of water in each glass. You know that this is a third of the water in the bottle, basically 0.333.... You now have three sets of thirds.

So 0.333... + 0.333.... + 0.333... = 0.999...

But since this came from one bottle, then you know that this 0.999... must be a whole, basically 1.

There is also convergence theory, but anyone who understands that wont need me to spell it out for them. So thoughts?

[zx0]

Don't you learn this in school anymore? When I was in school this is on the syllabus at age 16. (UK GCSE Higher mathematics)

Recurring decimals are exact fractions. I do like it though

[lemmingllama]

Don't you learn this in school anymore? When I was in school this is on the syllabus at age 16. (UK GCSE Higher mathematics)

Recurring decimals are exact fractions. I do like it though

I dont think that it is taught anymore, and a lot of people dont seem to understand this concept. So spreading the word and all that

[Kasoivc]

Shrugs. I'm asian, I hate math. Therefore, anything beyond 2+2=Fish, Irritates the living hell out of me. I was told once that Algebra and Geometry are used to explain black holes.

Algebra = The Size and Depth of a Black Hole, or some kind of space related object.

Geometry = The Shape.

Either way, I hate math because it's not my cup of tea, and if it is for you, I tip my hat in your good direction because I wasn't born to commit myself to numbers.

As a result of 12 years of education, when I get recurring decimals, I just round up or down accordingly. I did have a thing for science, but in science, rounding gives your human made errors; apparently you either do it and explain yourself accordingly, or you don't and try to make a easy explanation of your actions - a much harder task.

[zx0]

I dont think that it is taught anymore, and a lot of people dont seem to understand this concept. So spreading the word and all that

And this is the reason the internet is full of crazy people who think 1 = 0, or that 6÷2(1+2)= 1.

I see these sort of polls on facebook all the time and it is so depressing to see people don't have a clue. Even my friends who I thought were fairly smart are shockingly poor at basic maths.

[lemmingllama]

And this is the reason the internet is full of crazy people who think 1 = 0, or that 6÷2(1+2)= 1.

I see these sort of polls on facebook all the time and it is so depressing to see people don't have a clue. Even my friends who I thought were fairly smart are shockingly poor at basic maths.

The 1 = 0 is mainly because some people seem to think that if you graph a line of x/x, then 0/0 must have a slope of 1 and equal 1. However they dont understand that there is an asymptote there. As for the other problem, it is because they read it as 6/2(1+2) rather than 6÷2(1+2)

[Dae314]

Ok so there are 3 math related thingies in this topic

1) 0.999... = 1. That's a topic that deals with infinity. Things are different when you have an infinite number of things. Deal with it.

2) 1=0 people who graph x/x and think there's a slope at (0,0) are idiots. No there's no asymptote there (wrong terminology asymptote) , the line just doesn't exist at that point. Divide by 0 fail.

3) That problem is a case that I like to use for prefix notation. With prefix notation there is no ambiguity about what order your operations should go in. For instance, the 2 different interpretations of the problem you gave would look like:

* / 6 2 + 1 2

/ 6 * 2 + 1 2

There's no need for parenthesis because there is no question about the order you evaluate those expressions. However, if I were to put parenthesis to make the operators look like functions in C++, it'd look like this (easier for you to read if you're not used to prefix notation):

*(/(6, 2), +(1, 2))

/(6, *(2, +(1,2)))

The basic idea of prefix notation is that you have a binary operator (+,-,*,/, etc.) which takes 2 arguments and does a function to the numbers (as opposed to a unary operator which takes 1 argument). For readability sake, we learned infix notation where the operator goes between the 2 arguments. However, prefix and postfix notations also exist where the operator goes before and after the arguments respectively. I like prefix because it resembles functions (which is what the operators represent), and it eliminates ambiguity so I wish it were taught .

[† Mute point]

Except if x = .9999

10x = 9.999............

10x-x = 8.999.................

9x = 8.9999......................

Thus, in the same equation X cannot possible equal 1. because...

if X=1

10x = 10

10x-x = 9

9x = 9

If the decimal is repeating, it is still less than 1, but the difference is infinitesimal and would not effect your result if you simply rounded up.

A .3 repeating decimal is better represented as a fraction 1/3, just as a .6 repeating is 2/3. However, 3/3 is not .9 repeating because when you add the final third there is a sufficient number to avoid the carry over that results in the repeating decimal. In fact, there is no decimal at that point because the result is now 1.

That being said, there is no whole number division that will have an end result of .9 repeating.

what you failed to calculate in your example, is that no matter how long it repeats, even if there is no end, when you multiply by 10, is that there is one less decimal digit. .9999 becomes 9.999. thus, you do not subtract .9 repeating into infinity from 9.9 repeating into infinity, you subtract .9 repeating into infinity from 9.9 repeating into (infinity-1).

[Dae314]

That being said, there is no whole number division that will have an end result of .9 repeating.

Actually... you have to remember that the set of natural numbers (which you refer to as whole numbers) is an infinite set of any number composed of any arbitrary amount of digits (0-9) in base 10 form that does not start with 0. By this logic, I can define any natural number in terms of the digits that make it up. For this example I'll use the notation #^n where # is a digit 0-9 and n is the amount of those digits repeated (e.g. 2^7 = 2222222, yes it's the same as power notation, but I'm redefining it so I can make this post easier to read). Now consider the number 3^<infinity>. This is a natural number because although it has an infinite number of digits, it is not a decimal.

Now, since we can't compare infinities (what I mean by this is that 3^<infinity> does not equal 3^<infinity> because those are 2 discrete values of infinity), I will say the variable i = <infinity> so that when I use i I'm referring to the same infinity (so 3^i = 3^i because i is the same infinity). Now we can say the rational number (9^i)/(1[0^i]) is equal to 0.999... repeating in decimal form. Note that 1[0^i] is not 1*[0^i] for this problem, but I'm placing a digit 1 in front of an i number of 0's (remember the notation I described above). Both of these are natural numbers, and two natural numbers used as the numerator and denominator of a fraction creates a rational number (you can actually do this with any 2 integers as long as the denominator isn't 0, but the set of natural numbers is a subset of the set of integers so it's all good).

The way you'd argue against this is to say that the set of natural numbers contains numbers made of an arbitrarily large number of digits, but does not contain any numbers made of an infinite number of digits. I'm not actually sure if that's the case, but either way would make sense to me .

[† Mute point]

A whole number is any number without a decimal or fraction. There is no combination of whole numbers that results in a .9 repeating decimal. No matter how high you go, the result is the same. all numbers, despite being capable of going on forever, are composed of a combination of the ten basic digits, 0,1,2,3,4,5,6,7,8,9. Knowing this, you can easily see that my statement is correct for yourself. no matter how high up you go, the results do not change. 11/33 is the same answer as 1/3, as is 111/333, 1111/3333. and so one down the line.

Many don't understand this concept, which is why you see so many using a calculator for basic math. and because this principal applies in basic math, by extension it also applies in higher math as well.

in your current equation, the number of zeros added after the 1 is equal to the number of decimal places in the answer. No matter how you see it, there is an end. Repeating decimals are repeating decimals, not because you are dividing by a factor of 10, but because the end result is the same when you continue to divide by the same number. 3 goes into 1 0 times with a remainder of one, 0.X 1 becomes 10 to determine the first decimal. 3 goes into ten 3 times with a remainder of 1, 0.3X. The cycle repeats. There can be no end, because there is always a remainder of 1 turning into 10 for the next decimal place.

[zx0]

Except if x = .9999

10x = 9.999............

10x-x = 8.999.................

9x = 8.9999......................

Thus, in the same equation X cannot possible equal 1. because...

if X=1

10x = 10

10x-x = 9

9x = 9

If the decimal is repeating, it is still less than 1, but the difference is infinitesimal and would not effect your result if you simply rounded up.

Your first example is incorrect.

x = .9999...

10x = 9.999...

10x-x = 9

9x = 9

x = 1

The two prove that 0.9... = 1. It is exactly equal to 1. There is no rounding involved.

what you failed to calculate in your example, is that no matter how long it repeats, even if there is no end, when you multiply by 10, is that there is one less decimal digit. .9999 becomes 9.999. thus, you do not subtract .9 repeating into infinity from 9.9 repeating into infinity, you subtract .9 repeating into infinity from 9.9 repeating into (infinity-1).

Again you are wrong. Multiplying it by 10 does not make one less digit! You misunderstand what infinity is.

This actually remind me of a puzzle I expect most of you have heard before, but will put here for fun

A hotel has an infinite number of rooms. The hotel is full with an infinite number of guests.

Q1. If a person arrives at the hotel looking for a room, how can can we fit them in?

Q2. If an infinite number of guests arrive (at the already full hotel) how can we give each guest a room?

[Dae314]

You're also misunderstanding infinity in my example.

Using the same variables I defined in my previous post, I'll breakdown what's happening. You usually can't divide 2 infinitely long integers because they're infinitely long so your answer would have no end . However because we're dividing by a power of 10 we're able to abstract out from dividing each individual digit and just say that for each 0 in the denominator, the location of the decimal point moves over 1 place. With 9^i, the decimal is after an infinite number of 9's. If we were some how able to count up what infinity of 9's there are and put that many 0's after a 1 in the denominator we'd end up with a decimal point in front of an infinite number of 9's. So 9^i divided by 1[0^i] creates 0.9999999... because i refers to the same infinity. There must be an infinite number of 9's after the decimal point because the numerator consists of an infinite number of 9's.

Remember that while both of these numbers have an infinite number of digits in them, neither of them are infinity so we're not breaking any rules by dividing infinity over infinity.

[zx0]

3) That problem is a case that I like to use for prefix notation. With prefix notation there is no ambiguity about what order your operations should go in. For instance, the 2 different interpretations of the problem you gave would look like:

* / 6 2 + 1 2

/ 6 * 2 + 1 2

There is no ambiguity here. The rules are clear.

Brackets

Indices

Division and Multiplication

Addition and Subtraction

Division and Multiplication are done from left to right, as are Addition and Subtraction.

The ONLY answer to the question 6÷2(1+2) is 9.

Brackets first: 6÷2(3)

Then multiplication and division from LEFT TO RIGHT; division first: 3(3)

Finally the multiplication: 9

A couple of other things people have said, someone used the term "infinity -1".

infinity minus 1 = infinity

And I believe your argument, Dae314, is misleading. You can't have an infinite number of digits and than say it's not infinity. Your answer as defined would be 0.[9^i], which would not be the same as 0.9 recurring.

This thread was not a maths trick or anything, 0.9 recurring is equal to 1.

0.9 recurring = S(infinity) = a / (1 - r), where a = 0.9 and r = 0.1

S(infinity) = 0.9 / (0.9) = 1

If you want to understand fully why this is you will have to go and learn some basic maths (basic as in building blocks, not easy!). Completeness axiom; http://www.emathzone.com/tutorials/real-analysis/completeness-axiom.html

Also every recurring decimal can be written as an exact fraction using the method in the first post.

[Dae314]

Yes, I admitted earlier that my idea can be argued using your argument that an infinite number of digits may not be considered an integer anymore since integers allow arbitrarily large numbers of digits, but not necessarily an infinite number of digits. Also, you may have missed when I put i = infinity, so I did mean 0.[9^infinity] or 0.999... repeating forever. I use i there so I can refer to the same infinity . Even if you didn't get that though you're right that I'm not proving things in a very mathematically sound style, but that's what makes things more interesting in non-formal discussion :3.

If you assert that my 1[0^i] is equal to infinity then my equation turns into 0 because I'm dividing by infinity (note, I'm talking limits here).

RE ambiguity. There is ambiguity otherwise why do you need to define all those order of operation rules in your post? If there were no ambiguity, you wouldn't need to give any special rules regarding the order you're expected to do things in. You might argue that there's the special rules with prefix notation that the operator must come first and the two operands must come second, but that's necessary for any type of notation definition . The order of operations is set by the way you write the problem, not by the rules people assigned to disambiguate ambiguous situations.

[zx0]

Those are the basic rules of mathematics, there is no ambiguity because all mathematicians know them and apply them in the same way. I only stated them in my post because it seemed that people were unaware of them.

https://en.wikipedia.org/wiki/Order_of_operations

There have to be these rules otherwise you could claim that 2 + 4 * 2 = 12 instead of 10!!!

[Dae314]

Those are the basic rules of mathematics, there is no ambiguity because all mathematicians know them and apply them in the same way. I only stated them in my post because it seemed that people were unaware of them.

https://en.wikipedia.org/wiki/Order_of_operations

There have to be these rules otherwise you could claim that 2 + 4 * 2 = 12 instead of 10!!!

Or you just don't make rules and say:

+ 2 * 4 2 = 10

* + 2 4 2 = 12

.

[Malyssa Rahl]

Sarah was correct on several points here..

The first being how truly repeating decimals work, and that there is no fraction or division that ends as 0.9 repeating. Which means that when you set it up, even if you are considering it an infinite, it cannot be one. There will always be an end, and when you reach that end and multiply it by 10, the decimal point moves one step to the right, and thus you lose a single decimal place in favor of gaining a whole number space. You cannot change this simple fact simply by wishing it to be different, or trying to confuse the facts with symbols and big words.

[lemmingllama]

Ok, because some people are now arguing about 0.999... not equaling 1, let me show you convergence theorum.

ar + ar^2 + ar^3 + ..... = ar/(1-r)

9(1/10) + 9(1/10)^2 + 9(1/10)^3 + .... = (9(1/10) / 1-(1/10))

=1

This shows that 0.999... = 1

[Malyssa Rahl]

Once again you spout symbols to attempt to prove a point that cannot be true. In plain words, explain how a .9 repeating answer can come about through the natural mathematical laws, not through applying infinity to a number that doesn't come out to be an infinite by any mathematical equation not including infinity. If you can't do this, your point is invalid, as the counter points have been made in terms anyone can understand, and clearly refute everything you've tried to show.

[Dae314]

Once again you spout symbols to attempt to prove a point that cannot be true. In plain words, explain how a .9 repeating answer can come about through the natural mathematical laws, not through applying infinity to a number that doesn't come out to be an infinite by any mathematical equation not including infinity. If you can't do this, your point is invalid, as the counter points have been made in terms anyone can understand, and clearly refute everything you've tried to show.

Just because you or other people cannot understand what he's saying doesn't mean it's invalid. Now if nobody but him understands what he's saying (even other mathematicians) then he's got to do some work to either figure out what he did wrong or get people to understand where he's coming from xD. You are not a leading mathematician though, so just because you don't recognize something doesn't mean it's invalid.

What lemming put out there was an infinite summation. In otherwords, it's an sum of an infinite number of values. In this case, it's the summation of 9*(1/10)^n for n = 1 to infinity. This is a perfectly valid expression (actually it's not since the way you'd evaluate that is by using limits, but using a limit here is just being proper, you really don't have to if you know that's where a limit belongs anyway). If you don't know what summations and limits are, take some more calculus or look at the result from wolfram alpha. It's not some mysterious force that nobody understands, it's just a bit of calculus. His original post is somewhat hard to validate and my answer is hard to validate, but this expression is easily proven.

In case you're not totally convinced by an infinite sum, know this: infinite summations are used in a ton of places in math. It's not isolated to proving that 0.99999... = 1. One really well known area that I can think of right now is Fourier Series which use summations of sines and consines to approximate any function. As you take a Fourier series to infinity, the series matches the function completely. Fourier series are used frequently in the fields of signal and image processing.

And yes, I was aware of this series (in fact, I found it the first time I looked up this problem when I came to this thread). It's just more fun (as I said earlier) to attempt unconventional solutions, especially in non-formal contexts .