Similarly, 1/3 = 0.3333…
So 3 times 1/3 = 0.9999… but also 3/3 = 1
Another nice one:
Let x = 0.9999… (multiply both sides by 10)
10x = 9.99999… (substitute 0.9999… = x)
10x = 9 + x (subtract x from both sides)
9x = 9 (divide both sides by 9)
x = 1
Not a proof, just wrong. In the “(substitute 0.9999… = x)” step, it was only done to one side, not both (the left side would’ve become 9.99999), therefore wrong.
For any a, b, c, if a = b and b = c, then a = c, right? The transitive property of equality.
For any a, b, x, if a = b, then x + a = x + b. The substitution property.
By combining both of these properties, for any a, b, x, y, if a = b and y = b + x, it follows that b + x = a + x and y = a + x.
In our example, a is x' (notice the ') and b is 0.999… (by definition). y is 10x' and x is 9. Let’s fill in the values.
If x' = 0.9999… (true by definition) and 10x = 0.999… + 9 (true by algebraic manipulation), then 0.999… + 9 = x' + 9 and 10x' = x' + 9.
if you are rearranging algebra you have to do the exact same thing on both sides
If you actually change any of the sides. Since, after substitution, the numeric value doesn’t change (literally the definition of equality), I don’t have to do anything – as I’m not rearranging. I’m merely presenting the same value in an equivalent manner. By contrast, when multiplying both sides by 10, since multiplication by 10 changes the concrete numeric value, I have to do it on both sides to maintain the equality relation (ditto for subtracting x'). But substitution never changes a numeric value – only rearranges what we already know.
(Edit)
Take the following simple system of equations.
5y = 3
x + y = 6
How would you solve it? Here’s how I would:
\begin{gather*} %% Ignore the LaTeX boilerplate, just so I could render it
\begin{cases}
y = \frac{3}{5} \\ % Isolate y by dividing both sides by 5
x = 6 - y % Subtract y from both sides
\end{cases} \\
x = 6 - \frac{3}{5} \\ % SUBSTITUTE 3/5 for y
x = 5.4 \\
(x, y) = (5.4, 0.6)
\end{gather*}
As I said, they didn’t substitute on both sides, only one, thus breaking the rules around rearranging algebra. Anything you do to one side you have to do to the other.
Nah that explanation is basically using an assumption to prove itself. You need to first prove that 1/3 does in fact equal .3333… which can be done using the ‘convoluted’ but not so convoluted proof
My degree is in mathematics. This is not how these notations are usually defined rigorously.
The most common way to do it starts from sequences of real numbers, then limits of sequences, then sequences of partial sums, then finally these notations turn out to just represent a special kind of limit of a sequence of partial sums.
If you want a bunch of details on this read further:
A sequence of real numbers can be thought of as an ordered nonterminating list of real numbers. For example: 1, 2, 3, … or 1/2, 1/3, 1/4, … or pi, 2, sqrt(2), 1000, 543212345, … or -1, 1, -1, 1, … Formally a sequence of real numbers is a function from the natural numbers to the real numbers.
A sequence of partial sums is just a sequence whose terms are defined via finite sums. For example: 1, 1+2, 1+2+3, … or 1/2, 1/2 + 1/4, 1/2 + 1/4 + 1/8, … or 1, 1 + 1/2, 1 + 1/2 + 1/3, … (do you see the pattern for each of these?)
The notion of a limit is sort of technical and can be found rigorously in any calculus book (such as Stewart’s Calculus) or any real analysis book (such as Rudin’s Principles of Mathematical Analysis) or many places online (such as Paul’s Online Math Notes). The main idea though is that sometimes sequences approximate certain values arbitrarily well. For example the sequence 1, 1/2, 1/3, 1/4, … gets as close to 0 as you like. Notice that no term of this sequence is actually 0. As another example notice the terms of the sequence 9/10, 9/10 + 9/100, 9/10 + 9/100 + 9/1000, … approximate the value 1 (try it on a calculator).
I want to stop here to make an important distinction. None of the above sequences are real numbers themselves because lists of numbers (or more formally functions from N to R) are not the same thing as individual real numbers.
Continuing with the discussion of sequences approximating numbers, when a sequence, call it A, approximates some number L, we say “A converges”. If we want to also specify the particular number that A converges to we say “A converges to L”. We give the number L a special name called “the limit of the sequence A”.
Notice in particular L is just some special real number. L may or may not be a term of A. We have several examples of sequences above with limits that are not themselves terms of the sequence. The sequence 0, 0, 0, … has as its limit the number 0 and every term of this sequence is also 0. The sequence 0, 1, 0, 0, … where only the second term is 1, has limit 0 and some but not all of its terms are 0.
Suppose we define a sequence a1, a2, a3, … where each of the an numbers is one of the numbers from 0, 1, 2, 3, 4, 5, 6, 7, 8 or 9. It can be shown that any sequence of the form a1/10, a1/10 + a2/100, a1/10 + a2/100 + a3/1000, … converges (it is too technical for me to show this here but this is explained briefly in Rudin ch 1 or Hrbacek/Jech’s Introduction To Set Theory).
As an example if each of the an values is 1 our sequence of partial sums above simplifies to 0.1,0.11,0.111,… if the an sequence is 0, 2, 0, 2, … our sequence of partial sums is 0.0, 0.02, 0.020, 0.0202, …
We define the notation 0 . a1 a2 a3 … to be the limit of the sequence of partial sums a1/10, a1/10 + a2/100, a1/10 + a2/100 + a3/1000, … where the an values are all chosen as mentioned above. This limit always exists as specified above also.
In particular 0 . a1 a2 a3 … is just some number and it may or may not be distinct from any term in the sequence of sums we used to define it.
When each of the an values is the same number it is possible to compute this sum explicitly. See here (where a=an, r=1/10 and subtract 1 if necessary to account for the given series having 1 as its first term).
So by definition the particular case where each an is 9 gives us our definition for 0.999…
To recap: the value of 0.999… is essentially just whatever value the (simplified) sequence of partial sums 0.9, 0.99, 0.999, … converges to. This is not necessarily the value of any one particular term of the sequence. It is the value (informally) that the sequence is approximating. The value that the sequence 0.9, 0.99, 0.999, … is approximating can be proved to be 1. So 0.999… = 1, essentially by definition.
Sure, when you start decoupling the numbers from their actual values. The only thing this proves is that the fraction-to-decimal conversion is inaccurate. Your floating points (and for that matter, our mathematical model) don’t have enough precision to appropriately model what the value of 7/9 actually is. The variation is negligible though, and that’s the core of this, is the variation off what it actually is is so small as to be insignificant and, really undefinable to us - but that doesn’t actually matter in practice, so we just ignore it or convert it. But at the end of the day 0.999… does not equal 1. A number which is not 1 is not equal to 1. That would be absurd. We’re just bad at converting fractions in our current mathematical understanding.
Edit: wow, this has proven HIGHLY unpopular, probably because it’s apparently incorrect. See below for about a dozen people educating me on math I’ve never heard of. The “intuitive” explanation on the Wikipedia page for this makes zero sense to me largely because I don’t understand how and why a repeating decimal can be considered a real number. But I’ll leave that to the math nerds and shut my mouth on the subject.
The rigorous explanation for why 0.999…=1 is that 0.999… represents a geometric series of the form 9/10+9/10^2+… by definition, i.e. this is what that notation literally means. The sum of this series follows by taking the limit of the corresponding partial sums of this series (see here) which happens to evaluate to 1 in the particular case of 0.999… this step is by definition of a convergent infinite series.
If they aren’t equal, there should be a number in between that separates them. Between 0.1 and 0.2 i can come up with 0.15. Between 0.1 and 0.15 is 0.125. You can keep going, but if the numbers are equal, there is nothing in between. There’s no gap between 0.1 and 0.1, so they are equal.
What number comes between 0.999… and 1?
(I used to think it was imprecise representations too, but this is how it made sense to me :)
My brother. You are scared of infinities. Look up the infinite hotel problem. I will lay it out for you if you are interested.
Image you are incharge of a hotel and it has infinite rooms. Currently your hotel is at full capacity… Meaning all rooms are occupied.
A new guest arrives. What do you do? Surely your hotel is full and you can’t take him in… Right?
WRONG!!!
You tell the resident of room 1 to move to room 2, you tell the resident of room 2 to move to room 3 and so on… You tell the resident of room n to move to room n+1.
Now you have room 1 empty
But sir… How did I create an extra room?
You didn’t. The question is the same as asking yourself that is there a number for which n+1 doesn’t exist. The answer is no… I can always add 1.
Infinity doesn’t behave like other numbers since it isn’t technically a number.
So when you write 0.99999… You are playing with things that aren’t normal. Maths has come with fuckall ways to deal with stuff like this.
Well you may say, this is absurd… There is nothing in reality that behaves this way. Well yes and no. You know how the building blocks of our universe obey quantum mechanics? The equations contain lots of infinities but only at intermediate steps. You have to “renormalise” them to make them go away. Nature apparently has infinities but likes to hide the from us.
The infinity problem is so fucked up. You know the reason physics people are unable to quantize gravity? Surely they can do the same thing to gravity as they did to say electromagnetic force? NOPE. Gravitation doesn’t normalise. You get left with infinities in your final answer.
Anyways. Keep on learning, the world has a lot of information and it’s amazing. And the only thing that makes us human is the ability to learn and grow from it. I wish you all the very best.
One way to think about this is that we represent numbers in different ways. For example, 1 can be 1.0, or a single hash mark, or a dot, or 1/1, or 10/10. All of them point to some platonic ideal world version of the concept of the number 1.
What we have here is two different representations of the same number that are in a similar representation. 1 and 0.999… both point to the same concept.
I strongly agree with you, and while the people replying aren’t wrong, they’re arguing for something that I don’t think you said.
1/3 ≈ 0.333… in the same way that approximating a circle with polygons of increasing side number has a limit of a circle, but will never yeild a circle with just geometry.
0.999… ≈ 1 in the same way that shuffling infinite people around an infinite hotel leaves infinite free rooms, but if you try to do the paperwork, no one will ever get anywhere.
Decimals require you to check the end of the number to see if you can round up, but there never will be an end. Thus we need higher mathematics to avoid the halting problem. People get taught how decimals work, find this bug, and then instead of being told how decimals are broken, get told how they’re wrong for using the tools they’ve been taught.
If we just accept that decimals fail with infinite steps, the transition to new tools would be so much easier, and reflect the same transition into new tools in other sciences. Like Bohr’s Atom, Newton’s Gravity, Linnaean Taxonomy, or Comte’s Positivism.
That does very accurately sum up my understanding of the matter, thanks. I haven’t been adding on to any of the other conversation in order to avoid putting my foot in my mouth further, but you’ve pretty much hit the nail on the head here. And the higher mathematics required to solve this halting problem are beyond me.
No, it equals 0.999…
2/9 = 0.222… 7/9 = 0.777…
0.222… + 0.777… = 0.999… 2/9 + 7/9 = 1
0.999… = 1
No, it equals 1.
Similarly, 1/3 = 0.3333…
So 3 times 1/3 = 0.9999… but also 3/3 = 1
Another nice one:
Let x = 0.9999… (multiply both sides by 10)
10x = 9.99999… (substitute 0.9999… = x)
10x = 9 + x (subtract x from both sides)
9x = 9 (divide both sides by 9)
x = 1
My favorite thing about this argument is that not only are you right, but you can prove it with math.
Not a proof, just wrong. In the “(substitute 0.9999… = x)” step, it was only done to one side, not both (the left side would’ve become 9.99999), therefore wrong.
The substitution property of equality is a part of its definition; you can substitute anywhere.
And if you are rearranging algebra you have to do the exact same thing on both sides, always
And if you don’t then you can no longer claim they are still equal.
For any
a,b,c, ifa = bandb = c, thena = c, right? The transitive property of equality.For any
a,b,x, ifa = b, thenx + a = x + b. The substitution property.By combining both of these properties, for any
a,b,x,y, ifa = bandy = b + x, it follows thatb + x = a + xandy = a + x.In our example,
aisx'(notice the') andbis0.999…(by definition).yis10x'andxis9. Let’s fill in the values.If
x' = 0.9999…(true by definition) and10x = 0.999… + 9(true by algebraic manipulation), then0.999… + 9 = x' + 9and10x' = x' + 9.If you actually change any of the sides. Since, after substitution, the numeric value doesn’t change (literally the definition of equality), I don’t have to do anything – as I’m not rearranging. I’m merely presenting the same value in an equivalent manner. By contrast, when multiplying both sides by 10, since multiplication by 10 changes the concrete numeric value, I have to do it on both sides to maintain the equality relation (ditto for subtracting
x'). But substitution never changes a numeric value – only rearranges what we already know.(Edit)
Take the following simple system of equations.
How would you solve it? Here’s how I would:
Here’s how Microsoft Math Solver would do it.
They multiplied both sides by 10.
0.9999… times 10 is 9.9999…
X times 10 is 10x.
10x is 9.9999999…
As I said, they didn’t substitute on both sides, only one, thus breaking the rules around rearranging algebra. Anything you do to one side you have to do to the other.
Except it doesn’t. The math is wrong. Do the exact same formula, but use .5555… instead of .9999…
Guess it turns out .5555… is also 1.
Lol you can’t do math apparently, take a logic course sometime
Let x=0.555…
10x=5.555…
10x=5+x
9x=5
x=5/9=0.555…
Oh honey
you have to do this now
That’s the best explanation of this I’ve ever seen, thank you!
That’s more convoluted than the 1/3, 2/3, 3/3 thing.
3/3 = 0.99999…
3/3 = 1
If somebody still wants to argue after that, don’t bother.
Nah that explanation is basically using an assumption to prove itself. You need to first prove that 1/3 does in fact equal .3333… which can be done using the ‘convoluted’ but not so convoluted proof
If you can’t do it without fractions or a … then it can’t be done.
We’ve found a time traveller from ancient Greece…
Edit: sorry. I mean we’ve found a time traveller from ancient Mesopotamia.
1/3=0.333…
2/3=0.666…
3/3=0.999…=1
Fractions and base 10 are two different systems. You’re only approximating what 1/3 is when you write out 0.3333…
The … is because you can’t actually make it correct in base 10.
The fractions are still in base 10 lmfao literally what the fuck are you talking about and where are getting this from?
You keep getting basic shit wrong, and it makes you look dumb. Stop talking and go read a wiki.
What exactly do you think notations like 0.999… and 0.333… mean?
That it repeats forever, to no end. Because it can never actually be correct, just that the difference becomes insignificant.
My degree is in mathematics. This is not how these notations are usually defined rigorously.
The most common way to do it starts from sequences of real numbers, then limits of sequences, then sequences of partial sums, then finally these notations turn out to just represent a special kind of limit of a sequence of partial sums.
If you want a bunch of details on this read further:
A sequence of real numbers can be thought of as an ordered nonterminating list of real numbers. For example: 1, 2, 3, … or 1/2, 1/3, 1/4, … or pi, 2, sqrt(2), 1000, 543212345, … or -1, 1, -1, 1, … Formally a sequence of real numbers is a function from the natural numbers to the real numbers.
A sequence of partial sums is just a sequence whose terms are defined via finite sums. For example: 1, 1+2, 1+2+3, … or 1/2, 1/2 + 1/4, 1/2 + 1/4 + 1/8, … or 1, 1 + 1/2, 1 + 1/2 + 1/3, … (do you see the pattern for each of these?)
The notion of a limit is sort of technical and can be found rigorously in any calculus book (such as Stewart’s Calculus) or any real analysis book (such as Rudin’s Principles of Mathematical Analysis) or many places online (such as Paul’s Online Math Notes). The main idea though is that sometimes sequences approximate certain values arbitrarily well. For example the sequence 1, 1/2, 1/3, 1/4, … gets as close to 0 as you like. Notice that no term of this sequence is actually 0. As another example notice the terms of the sequence 9/10, 9/10 + 9/100, 9/10 + 9/100 + 9/1000, … approximate the value 1 (try it on a calculator).
I want to stop here to make an important distinction. None of the above sequences are real numbers themselves because lists of numbers (or more formally functions from N to R) are not the same thing as individual real numbers.
Continuing with the discussion of sequences approximating numbers, when a sequence, call it A, approximates some number L, we say “A converges”. If we want to also specify the particular number that A converges to we say “A converges to L”. We give the number L a special name called “the limit of the sequence A”.
Notice in particular L is just some special real number. L may or may not be a term of A. We have several examples of sequences above with limits that are not themselves terms of the sequence. The sequence 0, 0, 0, … has as its limit the number 0 and every term of this sequence is also 0. The sequence 0, 1, 0, 0, … where only the second term is 1, has limit 0 and some but not all of its terms are 0.
Suppose we define a sequence a1, a2, a3, … where each of the an numbers is one of the numbers from 0, 1, 2, 3, 4, 5, 6, 7, 8 or 9. It can be shown that any sequence of the form a1/10, a1/10 + a2/100, a1/10 + a2/100 + a3/1000, … converges (it is too technical for me to show this here but this is explained briefly in Rudin ch 1 or Hrbacek/Jech’s Introduction To Set Theory).
As an example if each of the an values is 1 our sequence of partial sums above simplifies to 0.1,0.11,0.111,… if the an sequence is 0, 2, 0, 2, … our sequence of partial sums is 0.0, 0.02, 0.020, 0.0202, …
We define the notation 0 . a1 a2 a3 … to be the limit of the sequence of partial sums a1/10, a1/10 + a2/100, a1/10 + a2/100 + a3/1000, … where the an values are all chosen as mentioned above. This limit always exists as specified above also.
In particular 0 . a1 a2 a3 … is just some number and it may or may not be distinct from any term in the sequence of sums we used to define it.
When each of the an values is the same number it is possible to compute this sum explicitly. See here (where a=an, r=1/10 and subtract 1 if necessary to account for the given series having 1 as its first term).
So by definition the particular case where each an is 9 gives us our definition for 0.999…
To recap: the value of 0.999… is essentially just whatever value the (simplified) sequence of partial sums 0.9, 0.99, 0.999, … converges to. This is not necessarily the value of any one particular term of the sequence. It is the value (informally) that the sequence is approximating. The value that the sequence 0.9, 0.99, 0.999, … is approximating can be proved to be 1. So 0.999… = 1, essentially by definition.
That’s is a very precise and very good answer, but im still at a loss as to how all the .9,.99,.999,.9999 eventually just becomes 1.
Numbers are not their notations.
Sure, let’s do it in base 3. 3 in base 3 is 10, and 3^(-1) is 10^(-1), so:
1/3 in base 10 = 1/10 in base 3
0.3… in base 10 = 0.1 in base 3
Multiply by 3 on both sides:
3 × 0.3… in base 10 = 10 × 0.1 in base 3
0.9… in base 10 = 1 in base 3.
But 1 in base 3 is also 1 in base 10, so:
0.9… in base 10 = 1 in base 10
You’re having to use … to make your conversion again. If you need to to an irrational number to make your equation correct, it isn’t really correct.
Do you know what an irrational number is?
Sure, when you start decoupling the numbers from their actual values. The only thing this proves is that the fraction-to-decimal conversion is inaccurate. Your floating points (and for that matter, our mathematical model) don’t have enough precision to appropriately model what the value of 7/9 actually is. The variation is negligible though, and that’s the core of this, is the variation off what it actually is is so small as to be insignificant and, really undefinable to us - but that doesn’t actually matter in practice, so we just ignore it or convert it. But at the end of the day 0.999… does not equal 1. A number which is not 1 is not equal to 1. That would be absurd. We’re just bad at converting fractions in our current mathematical understanding.
Edit: wow, this has proven HIGHLY unpopular, probably because it’s apparently incorrect. See below for about a dozen people educating me on math I’ve never heard of. The “intuitive” explanation on the Wikipedia page for this makes zero sense to me largely because I don’t understand how and why a repeating decimal can be considered a real number. But I’ll leave that to the math nerds and shut my mouth on the subject.
You are just wrong.
The rigorous explanation for why 0.999…=1 is that 0.999… represents a geometric series of the form 9/10+9/10^2+… by definition, i.e. this is what that notation literally means. The sum of this series follows by taking the limit of the corresponding partial sums of this series (see here) which happens to evaluate to 1 in the particular case of 0.999… this step is by definition of a convergent infinite series.
It still equals 1, you can prove it without using fractions:
x = 0.999…
10x = 9.999…
10x = 9 + 0.999…
10x = 9 + x
9x = 9
x = 1
There’s even a Wikipedia page on the subject
I hate this because you have to subtract .99999… from 10. Which is just the same as saying 10 - .99999… = 9
Which is the whole controversy but you made it complicated.
It would be better just to have them do the long subtraction
If they don’t get it and keep trying to show you how you are wrong they will at least be out of your hair until forever.
You don’t subtract from 10, but from 10x0.999… I mean your statement is also true but it just proves the point further.
No, you do subtract from 9.999999…
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Do that same math, but use .5555… instead of .9999…
Have you tried it? You get 0.555… which kinda proves the point does it not?
???
Not sure what you’re aiming for. It proves that the setup works, I suppose.
x = 0.555…
10x = 5.555…
10x = 5 + 0.555…
10x = 5+x
9x = 5
x = 5/9
5/9 = 0.555…
So it shows that this approach will indeed provide a result for x that matches what x is supposed to be.
Hopefully it helped?
If they aren’t equal, there should be a number in between that separates them. Between 0.1 and 0.2 i can come up with 0.15. Between 0.1 and 0.15 is 0.125. You can keep going, but if the numbers are equal, there is nothing in between. There’s no gap between 0.1 and 0.1, so they are equal.
What number comes between 0.999… and 1?
(I used to think it was imprecise representations too, but this is how it made sense to me :)
My brother. You are scared of infinities. Look up the infinite hotel problem. I will lay it out for you if you are interested.
Image you are incharge of a hotel and it has infinite rooms. Currently your hotel is at full capacity… Meaning all rooms are occupied. A new guest arrives. What do you do? Surely your hotel is full and you can’t take him in… Right? WRONG!!! You tell the resident of room 1 to move to room 2, you tell the resident of room 2 to move to room 3 and so on… You tell the resident of room n to move to room n+1. Now you have room 1 empty
But sir… How did I create an extra room? You didn’t. The question is the same as asking yourself that is there a number for which n+1 doesn’t exist. The answer is no… I can always add 1.
Infinity doesn’t behave like other numbers since it isn’t technically a number.
So when you write 0.99999… You are playing with things that aren’t normal. Maths has come with fuckall ways to deal with stuff like this.
Well you may say, this is absurd… There is nothing in reality that behaves this way. Well yes and no. You know how the building blocks of our universe obey quantum mechanics? The equations contain lots of infinities but only at intermediate steps. You have to “renormalise” them to make them go away. Nature apparently has infinities but likes to hide the from us.
The infinity problem is so fucked up. You know the reason physics people are unable to quantize gravity? Surely they can do the same thing to gravity as they did to say electromagnetic force? NOPE. Gravitation doesn’t normalise. You get left with infinities in your final answer.
Anyways. Keep on learning, the world has a lot of information and it’s amazing. And the only thing that makes us human is the ability to learn and grow from it. I wish you all the very best.
It’s a correct proof.
One way to think about this is that we represent numbers in different ways. For example, 1 can be 1.0, or a single hash mark, or a dot, or 1/1, or 10/10. All of them point to some platonic ideal world version of the concept of the number 1.
What we have here is two different representations of the same number that are in a similar representation. 1 and 0.999… both point to the same concept.
I strongly agree with you, and while the people replying aren’t wrong, they’re arguing for something that I don’t think you said.
1/3 ≈ 0.333… in the same way that approximating a circle with polygons of increasing side number has a limit of a circle, but will never yeild a circle with just geometry.
0.999… ≈ 1 in the same way that shuffling infinite people around an infinite hotel leaves infinite free rooms, but if you try to do the paperwork, no one will ever get anywhere.
Decimals require you to check the end of the number to see if you can round up, but there never will be an end. Thus we need higher mathematics to avoid the halting problem. People get taught how decimals work, find this bug, and then instead of being told how decimals are broken, get told how they’re wrong for using the tools they’ve been taught.
If we just accept that decimals fail with infinite steps, the transition to new tools would be so much easier, and reflect the same transition into new tools in other sciences. Like Bohr’s Atom, Newton’s Gravity, Linnaean Taxonomy, or Comte’s Positivism.
That does very accurately sum up my understanding of the matter, thanks. I haven’t been adding on to any of the other conversation in order to avoid putting my foot in my mouth further, but you’ve pretty much hit the nail on the head here. And the higher mathematics required to solve this halting problem are beyond me.
THAT’S EXACTLY WHAT I SAID.
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