Why Does (1/3)*3 = 1 and Not .9 Repeating?

[ldzcableguy]

How can 1/3 multiplied by 3 give us 1 if 1/3 is the repeating decimal .3 to an infinite number of decimal places? If you multiplied 3 * .3 with an infinite number of repeating 3's wouldn't you get .9 with an infinite number of 9's repeating? Why do we say that (1/3) = .3 repeating and when you multiply that times 3 you get 1 instead of .9 repeating?

 

[BoulderHead]

Yikes!

 

[Tom McCurdy]

because .999999999999999999999999999999999 repated equals one when you apply calculus

look up zeno's paradoxs it has similar ideas

 

[Math Is Hard]

Yikes!

heh heh heh! here we go again! ... poor Hurkyl... poor Matt.. poor Tom...
I feel sorry for them already.

 

[Tom McCurdy]

heh heh heh! here we go again! ... poor Hurkyl... poor Matt.. poor Tom...
I feel sorry for them already.

well at least people are thinking

 

[Gokul43201]

How can 1/3 multiplied by 3 give us 1 if 1/3 is the repeating decimal .3 to an infinite number of decimal places? If you multiplied 3 * .3 with an infinite number of repeating 3's wouldn't you get .9 with an infinite number of 9's repeating? Why do we say that (1/3) = .3 repeating and when you multiply that times 3 you get 1 instead of .9 repeating?

Because the real number represented by the repeating decimal 0.999... is defined to be identical to the number represented by 1.

Read this or try this thread.

 

[Zurtex]

Just so you guys know I actually managed to convince some people on another forum that 0.9 recurring = 1 who were arguing otherwise.

 

[Tide]

How can 1/3 multiplied by 3 give us 1 if 1/3 is the repeating decimal .3 to an infinite number of decimal places? If you multiplied 3 * .3 with an infinite number of repeating 3's wouldn't you get .9 with an infinite number of 9's repeating? Why do we say that (1/3) = .3 repeating and when you multiply that times 3 you get 1 instead of .9 repeating?

Can you find any number between 0.\bar 3 and \frac{1}{3} ? If not, then they must be the same! :-)

 

[BoulderHead]

heh heh heh! here we go again! ... poor Hurkyl... poor Matt.. poor Tom...
I feel sorry for them already.

Haha, my thoughts exactly!

I will say it can be an indicator someone has been using their coconut to try and make sense of things but, well, not always (often turns ugly for no good reason).

 

[Janitor]

Where has Organic been of late?

 

[ldzcableguy]

because .999999999999999999999999999999999 repated equals one when you apply calculus

look up zeno's paradoxs it has similar ideas

What is the calculus behind it?

 

[Gokul43201]

I believe what Tom had in mind was the limit of the infinite sum "0.9 + 0.09 + 0.009 + ..." which is just a geometric progression, and its sum to an infinite number of terms is simply 0.9/(1- 0.1) = 1.

 

[ReyChiquito]

i like this one

0.33333\bar{3}=a

3.33333\bar{3}=10a

10a-a=9a=3

a=\frac{3}{9}=\frac{1}{3}

0.333\bar{3}=\frac{1}{3}

 

[matt grime]

One thing I've never understood is why people argue against the fact that it is even remotely possible for two decimals to represent the same real number yet are perfectly happy to accept there are an infinite number of rational representations for some element in Q {1/2, 2/4. 3/6, ...} Surely two different ones for only a few numbers must be a fantastic improvement over infinitely many for all.

 

[Hurkyl]

My best guess is that they don't internalize 1/2 as being a number, but rather an arithmetic expression. The thing that bugs me is how many think of 0.999... as some strange sort of varying number.

 

[russ_watters]

Because of this, I don't understand why this should be that hard of an issue:

Can you find any number between 0.\bar 3 and \frac{1}{3} ? If not, then they must be the same! :-)

Some people have tried to get around it by pulling new numbers out of their a--air (0.000...1), but if you can't answer the question using the real number system, game over. I'm currently in page 9 of a similar thread at BadAstronomy, where the answer to that question was "an infinitessimally small number." It makes me wonder whether this is an honest argument.

The one (sorta) legitimate concern I've seen is from engineers who think its a matter of precision: you have to stop somewhere to round it off and that's how you get 1.

 

[matt grime]

It cannot be an infinitesimal number, and non-zero, in the sense of non-standard analysis, since 1/3 - 0.33... is real, and an infinitesimal, hence zero - even in nonstandard analysis 0.9.. and 1 are the same.

 

[omicron]

We should have a sticky for this. Or maybe we should have a faq forum for the different topics. Btw you can try here

 

[check]

We should have a sticky for this. Or maybe we should have a faq forum for the different topics.

Nah, people would still continue to post this 5 times a day. I recommend just changing the url to 0.9repeating=1.end_of.story_so.please.stop_asking.com and replace the PF banner with something similar.

 

[nnnnnnnn]

9 = 9.9... - .9... = 10*(.9...) - 1(.9...) = (10 - 1)*(.9...) = 9*(.9...)

-> 9 = 9*(.9...)
-> 1 = .9...

Using this same method can be used to prove the geometric series (I think its the geometric series) which is where it ties into calculus...

 

[okidream]

How can 1/3 multiplied by 3 give us 1 if 1/3 is the repeating decimal .3 to an infinite number of decimal places? If you multiplied 3 * .3 with an infinite number of repeating 3's wouldn't you get .9 with an infinite number of 9's repeating? Why do we say that (1/3) = .3 repeating and when you multiply that times 3 you get 1 instead of .9 repeating?

Hi. If you don't mind, I like to share my opinion on this.

This type of thing is new, as before the decimal system was invented, there was not such a problem. This is a problem with the decimal system and not a problem with fractions, as in ancient times, I guess.

As an illustration: 3 is odd and hence dividing it into itself i.e to 3 components, (1/3 each) it should never be perfect in reality, except in your own imagination, (which is the fraction 1/3...which reveals the beauty of maths). Such are the case with all the other odds, except for the special odd - 5 and its selective multiples. (I'm still figuring out what is so special about this, like the fact we're born with 5 fingers and toes and 4 limbs+head).

To be honest, I have not seen any usefulness of decimal point system in number theory - it could be very useful to precision engineering, etc, elsewhere though.

My conclusion is: 1/3 is never 0.333..., but 0.3333... is 1/3.
(Just as a square is a rectangle but a rectangle is not a square, kind of arg)

 

[matt grime]

"My conclusion is: 1/3 is never 0.333..., but 0.3333... is 1/3.
(Just as a square is a rectangle but a rectangle is not a square, kind of arg)"

Words almost fail me.

If you can't figure out why 1/5 has a nice decimal base 10, then try understandin why 1/3 = 0.1 base 3.

 

[Cosmo16]

Here's a proof that I think is sound. Tell me if I'm wrong

x=.9 repeating
10x=9.9 repeating
10x-x=9.9 repeating - .9 repeating
9x=9
x=1

 

[matt grime]

Presumes that the arithmetic operations have been defined on infinite strings.

 

[StatusX]

its all limits. in that proof, using a finite string of 9s, the error gets smaller and smaller as the length grows, so in the infinite limit, it is exact. Every proof depends on the idea of a limit and the fact that all these symbols are intended to represent is the real number limit of a certain sequence. each decimal corresponds to exactly one real number. none of them correspond to a process that never ends, as some would tend to believe about 0.999...

 

[UrbanXrisis]

1/3 = .333333...
1/3 * 3= 1
.333333... * 3 = .9999999... = 1

 

[okidream]

Here's a proof that I think is sound. Tell me if I'm wrong

x=.9 repeating
10x=9.9 repeating
10x-x=9.9 repeating - .9 repeating
9x=9
x=1

Cosmo16, you're proof actually reinstate the point that 1/3 is 0.333... but not the other way.

If you work backwards, (eg. starting from x = 1), I believe the logic, the first premise will already make it impossible to reach the conclusion that x = 0.333... (or .999...).

But if work forward as above:
After the step 9x = 9, there is one more step , i.e x = 9/9
which brings out a division. (hence, fraction.

Now, if you examine this case: Is 9/9 = 0.999.../0.999... ?
Note first that it looks 'messy' affair to write infinity for repeating 9's as above...but what if the case otherwise, say for the case there is a last unique digit that 'ends' the infinity, will look not, as the case, you cross multiply the above, we get:

8.999...1 = 8.999...1 (no limits/open ends whatsoever)

And then we easily can agree. I am just trying to show how mathematical manipulation can determine the proof to what extent, and that's one of the problem as I said earlier, I guess, with anything point system, other than fractions.

 

[Integral]

...but what if the case otherwise, say for the case there is a last unique digit that 'ends' the infinity, will look not, as the case, you cross multiply the above, we get:

8.999...1 = 8.999...1 (no limits/open ends whatsoever)

...

Didn't you even feel a slight pang of guilt when you speak of the "ends the infinity" Is not this such an obviously self contradicting statement that even someone without a clue about mathematics would have trouble making such a statement.

We MUST stick to real numbers, what you have fabricated is NOT a real number. A basic requirement for the decimal expansion of a real number is that every digit be associated with a integer representing its place value. What is the place value of you 1? As soon as you assign the required place value you have just numbered all of the preceding zeros, therefore there are not, and CANNOT be an infinite number of zeros before the one.

 

[ohwilleke]

Presumes that the arithmetic operations have been defined on infinite strings.

This is definitional, but once you introduce infinite strings, it is necessary to define operations on those infinite strings, and the definition conventionally used has the virtue of being analogous in all respects to those same arithmetic operations when defined on finite strings. Any other definition would not preserve a host of standard algebric properties of real numbers.

For example, this definition preserves the relation of B*(A/B)=A, and it is difficult to imagine any other definition which would preserve this property.

If you choose any other definition division would not have a well defined inverse function for numbers on the real number line, which would be a very undesirable feature for most ordinary mathematics.

Likewise, this definition is necessary to preserve the relation that A/B=C/D

Where A and B are in one base number system and C and D are in another base number system and A=C and B=D in parts of each number system where the mapping from the AB number system to the CD number system are the well defined (for example, where A and B are whole numbers in one base number system and C and D are whole numbers in another base number system and there is a definitional rule that establishes a mapping from whole numbers in the AB system to whole numbers in the CD system).

If the standard definition of the infinite string is not adopted, you have done the equivalent of adopting a preferred reference frame in GR.

Alternately one could define 0.33333. . . as the limit as the number of digits approaches infinity of the series 0.3, 0.33, 0.333, . . . which would uniquely produce the same natural definition of 0.3333. . . . which can be restated:

The limit a the number of digits approaches infinity of 1/3-0.3, 1/3-0.33, 1/3-0.333, . . . which is zero.

 

[StatusX]

My conclusion is: 1/3 is never 0.333..., but 0.3333... is 1/3.
(Just as a square is a rectangle but a rectangle is not a square, kind of arg)

You know, equality is symmetric, by definition, (ie., x=y iff y=x) so such a statement makes no sense. The reason the square/rectangle argument works has to do with the fact that the set of squares is a proper subset of the set of rectangles, and this is not symmetrical.

 

[okidream]

Didn't you even feel a slight pang of guilt when you speak of the "ends the infinity" Is not this such an obviously self contradicting statement that even someone without a clue about mathematics would have trouble making such a statement.

We MUST stick to real numbers, what you have fabricated is NOT a real number. A basic requirement for the decimal expansion of a real number is that every digit be associated with a integer representing its place value. What is the place value of you 1? As soon as you assign the required place value you have just numbered all of the preceding zeros, therefore there are not, and CANNOT be an infinite number of zeros before the one.

Please don't pour your emotions on ME, dearest sir.

Please don't ask me further more: like what is the place of value of '1' in my above lines. If you wish the answer, you can consult the calculator. If you think this a fabrication, then the calculator had done it.

And they're not zeroes, they're 9's.

"therefore there are not, and CANNOT be an infinite number of zeros before the one" ---

You obviously have a good sense of humour in your argument.

How on earth, in the universe, without the slight pangs of guilt, can you even mention it CANNOT be infinite when you later mention 'before the 1' ?
So what is it? Is it infinite or not?

At least I did have slight guilt -->which is why I wrote the word ends to infinity, with 'ends' encapsulated with ''.

Thanks for making my days with laughter, and I have work/trash to collect this weekend.

 

[chroot]

okidream,

Your construction 8.999...1 is not possible within the boundaries of the definitions of the real numbers and infinity. It is impossible to have an infinite number of nines followed by a one. Integral is correct, and you are not. It's rather callous of you to laugh in the face of someone more educated than you. If you stick around, you'll learn a lot here.

- Warren

 

[okidream]

You know, equality is symmetric, by definition, (ie., x=y iff y=x) so such a statement makes no sense. The reason the square/rectangle argument works has to do with the fact that the set of squares is a proper subset of the set of rectangles, and this is not symmetrical.

There is difference between symmetric and equivalence. In the above, seems equivalence, if talking about sets, and not equations.
there no sets involved (as yet). And neither even is symmetry.

Its meant a statement which goes forward, but not backwards. That's all. (Maybe it could be the case x=y if y=x, I don't know, and I am not investigating it.)

 

[Integral]

Please don't pour your emotions on ME, dearest sir.

Please don't ask me further more: like what is the place of value of '1' in my above lines. If you wish the answer, you can consult the calculator. If you think this a fabrication, then the calculator had done it.

What has a calculator to do with my question?

And they're not zeroes, they're 9's.

"therefore there are not, and CANNOT be an infinite number of zeros before the one" ---

Fine, there is a finite number of 9s before the one. There must then be an infinite number of zeros AFTER the 1.

You obviously have a good sense of humor in your argument.

How on earth, in the universe, without the slight pangs of guilt, can you even mention it CANNOT be infinite when you later mention 'before the 1' ?
So what is it? Is it infinite or not?

At least I did have slight guilt -->which is why I wrote the word ends to infinity, with 'ends' encapsulated with ''.

So if you knew it was wrong why do you think your statement has any meaning? Clearly there is a finite number of 9s in what you have written.

Thanks for making my days with laughter, and I have work/trash to collect this weekend.

I find it hard to laugh at misrepresentations and blatant errors, especially when you seem to be unwilling to learn. We also have some trash collection to do here.

 

[okidream]

okidream,

Your construction 8.999...1 is not possible within the boundaries of the definitions of the real numbers and infinity. It is impossible to have an infinite number of nines followed by a one. Integral is correct, and you are not. It's rather callous of you to laugh in the face of someone more educated than you. If you stick around, you'll learn a lot here.

- Warren

Who says it's impossible? Tell me what do you get when you multiply 9 with 9. You get 81, don't you?. The last, of the last of the last place is the number 1, isn't it? Isn't that different from 9? So isn't the boundary the number 1?

the issue here is, how are you going to write the infinite number 8.999...1, (encapsulated with the last of the last value that ends with '1') in the definition given in all of maths? I can't think of any way, except like I wrote it--- 8.999...1.

 

[okidream]

Fine, there is a finite number of 9s before the one. There must then be an infinite number of zeros AFTER the 1.

No. There aren't. See my reply to choot. 1 is the boundary.

 

[Tide]

Now, if you examine this case: Is 9/9 = 0.999.../0.999... ?

Absolutely! You can easily show it directly with long division. 9 goes into 9.0 0.9 times with a remainder of 0.09 - repeat forever and you get 9/9 = 0.\bar 9.

 

[Integral]

Don't you understand that placing the one TERMINATES the string of 9s? The string of 9 has an end therefore it is not infinite. Saying it is, DOES NOT MAKE IT SO.

You need to learn the definition of a decimal representation. These are not things I am making up as I go. What we are trying to explain to you are the results of well over a centuries worth of work by mathematicians. The real number line is a construction. It has been carefully constructed by Mathematicians. What we are trying to explain to you is a fragment of the results of this construction.

Would you argue with an architect the existence of a beam in the building he designed? That is what you are currently doing. I, and the other mathematicians, on this board are attempting to show you a bit of the blue print. Pay attention and you might even learn something.

 

[master_coda]

Who says it's impossible? Tell me what do you get when you multiply 9 with 9. You get 81, don't you?. The last, of the last of the last place is the number 1, isn't it? Isn't that different from 9? So isn't the boundary the number 1?

the issue here is, how are you going to write the infinite number 8.999...1, (encapsulated with the last of the last value that ends with '1') in the definition given in all of maths? I can't think of any way, except like I wrote it--- 8.999...1.

It's impossible by definition. If there is a last value, then by definition there are only a finite number of values. If you have an infinite number of digits, there is no last one. Saying "1 is the boundary" doesn't magically change this fact.

 

[okidream]

It's impossible by definition. If there is a last value, then by definition there are only a finite number of values. If you have an infinite number of digits, there is no last one. Saying "1 is the boundary" doesn't magically change this fact.

Which is the reason I think the decimal point system, even if it has been worked on for centuries old, as Integral put it, which case I don't really give a toss, is pretty darn useless and only creates fallacies of its own.

Just as how you point it's impossible for me to do so by definition, why can't I point out that it's impossible too by definition that:
0.333... is the sum of the G.P of the decimal point additions which reaches the sum 1/3, is all but plain fallacy.

The same argument I can say 1/3 is the sum of the G.P with a = 1/4 and r = 1/4, (the series looks like 1/4, 1/4^2, 1/4^3, ... ), and then I conclude that my number base was 4 and the number was 0.111...

Think about it, and stop shooting me with what I did wrong, because from the start (which is possibly the definition itself) it seems already wrong.

 

[StatusX]

Which is the reason I think the decimal point system, even if it has been worked on for centuries old, as Integral put it, which case I don't really give a toss, is pretty darn useless and only creates fallacies of its own.

This is just something to consider. Many people claim they have found basic flaws in math that mathematicians either haven't noticed or refuse to acknowledge. You have to realize, mathematicians are very smart people, and they have no motive to ignore problems. In fact, they would be very inclined to publish their findings if they really had found a flaw. But with all the mathematicians working with basic concepts like the decimal system everyday, many of whom know much more than these people about these topics, they never do find these simple flaws. Its much, much more likely that these people are just misunderstanding the problem.

Just as how you point it's impossible for me to do so by definition, why can't I point out that it's impossible too by definition that:
0.333... is the sum of the G.P of the decimal point additions which reaches the sum 1/3, is all but plain fallacy.

The same argument I can say 1/3 is the sum of the G.P with a = 1/4 and r = 1/4, (the series looks like 1/4, 1/4^2, 1/4^3, ... ), and then I conclude that my number base was 4 and the number was 0.111...

What exactly are you saying? Numbers have different representations in different bases. Theres nothing wrong with this, is there?

 

[master_coda]

Which is the reason I think the decimal point system, even if it has been worked on for centuries old, as Integral put it, which case I don't really give a toss, is pretty darn useless and only creates fallacies of its own.

Oh, I understand that you think the current system sucks. But since your posts clearly demonstrate that you don't even understand the current system anyway, it's unlikely that your opinion is correct.

Just as how you point it's impossible for me to do so by definition, why can't I point out that it's impossible too by definition that:
0.333... is the sum of the G.P of the decimal point additions which reaches the sum 1/3, is all but plain fallacy.

In order to say that something is impossible by definition, you have to use the actual definitions (which is what I did). You can't just make up your own and complain that stuff breaks when you try to use them.

The same argument I can say 1/3 is the sum of the G.P with a = 1/4 and r = 1/4, (the series looks like 1/4, 1/4^2, 1/4^3, ... ), and then I conclude that my number base was 4 and the number was 0.111...

Think about it, and stop shooting me with what I did wrong, because from the start (which is possibly the definition itself) it seems already wrong.

That argument doesn't even make sense. The current definition is wrong because in two different bases we get two different representations?

 

[matt grime]

This is definitional, but once you introduce infinite strings, it is necessary to define operations on those infinite strings, and the definition conventionally used has the virtue of being analogous in all respects to those same arithmetic operations when defined on finite strings. Any other definition would not preserve a host of standard algebric properties of real numbers.

For example, this definition preserves the relation of B*(A/B)=A, and it is difficult to imagine any other definition which would preserve this property.

If you choose any other definition division would not have a well defined inverse function for numbers on the real number line, which would be a very undesirable feature for most ordinary mathematics.

Likewise, this definition is necessary to preserve the relation that A/B=C/D

Where A and B are in one base number system and C and D are in another base number system and A=C and B=D in parts of each number system where the mapping from the AB number system to the CD number system are the well defined (for example, where A and B are whole numbers in one base number system and C and D are whole numbers in another base number system and there is a definitional rule that establishes a mapping from whole numbers in the AB system to whole numbers in the CD system).

If the standard definition of the infinite string is not adopted, you have done the equivalent of adopting a preferred reference frame in GR.

Alternately one could define 0.33333. . . as the limit as the number of digits approaches infinity of the series 0.3, 0.33, 0.333, . . . which would uniquely produce the same natural definition of 0.3333. . . . which can be restated:

The limit a the number of digits approaches infinity of 1/3-0.3, 1/3-0.33, 1/3-0.333, . . . which is zero.

Why the hell have you introduced this spurious analogy?

I know how to make the decimals a model of the real numbers, thank you.

 

[okidream]

The limit a the number of digits approaches infinity of 1/3-0.3, 1/3-0.33, 1/3-0.333, . . . which is zero.

I agree.

To answer this question from the beginning:
The reason why 1/3 *3 = 1 where 0.333 (repeating) *3 is not:
is because 1/3 is perfectly rational (or simply as an elementary school kid would call it a fraction) while 0.33(repeating) is not.

0.33(repeating) = 1/3 is iff 0.33(repeating) is taken as sum to infinity, regardless of ANY BASE. Get it... ANY BASE.

Further, but at the same time, 0.111(repeating) is also = 1/3 is iff is taken as sum to infinity, becuase I simply had chosen 4 as my base.

Those who will continue to say this doesn't make sense, please don't just shoot those words. Prove it.

 

[Hurkyl]

I prefer a more formal definition of the decimal numbers, so this is how the proof would go.

By definition of multiplicative inverse, 1/3 is any number x such that x * 3 = 1. 3 is, by definition, 1+1+1. Therefore, I wish to show x+x+x=1.

"0.333..." is more precisely written as the (positive finite) sequence {a_n} given by:

a_n = 0 if n >= 0
a_n = 3 if n < 0

Now, when you add, you get no carry, so

{a_n} + {a_n} + {a_n} = {c_n} where

c_n = 0 if n >= 0
c_n = 9 if n < 0

(more commonly written as 0.999...)

Now, by definition of equality for decimal numbers, {c_n} = {d_n} where d_n is

d_n = 0 if n > 0
d_n = 1 if n = 0
d_n = 0 if n < 0

(more commonly written as 1, of course)

because c_0 < 9 and c_n == 9 for all n < 0, and d_0 = c_0 + 1 and d_0 == 0 for all n < 0, and d_n = c_n for all n > 0.

Thus, 0.333... is the multiplicative inverse of 3.


Others might prefer a more analytic definition of real numbers, using infinite summations to define the value of a decimal string. In that case, 0.333... is again more precisely represented by the sequence a_n where

a_n = 0 if n >= 0
a_n = 3 if n < 0

and its value is defined as

<br /> \sum_{i=-\infty}^{+\infty} a_i 10^i = \sum_{i=-\infty}^{-1} 3 * 10^i<br />

Which is a familiar geometric series whose sum is (3/10) / (1 - 1/10) = 3 / (10 - 1) = 3/9 = 1/3.

 

[okidream]

...Which is a familiar geometric series whose sum is (3/10) / (1 - 1/10) = 3 / (10 - 1) = 3/9 = 1/3.

well, I can divide 3 by any integers, (although by the defintion of base,
not < 3, since we all love definitions, don't we?) and that number sums to 1/3. The fraction 1/3. For any base - the fraction 1/3 of that base. Period.

I'm already getting tired explaining on this.

 

[Zurtex]

well, I can divide 3 by any integers, (although by the defintion of base,
not < 3, since we all love definitions, don't we?) and that number sums to 1/3. The fraction 1/3. For any base - the fraction 1/3 of that base. Period.

I'm already getting tired explaining on this.

You're tired of explaining this because what you are saying makes no sense (seriously that post means nothing to me).

Have you ever considered what a real number is and how it is defined in maths? It is very interesting.

Consider x to exist in the interval: a₁ < x < b₁, where x is a real number and a and b are rational numbers. The interval between a_n and b_n grows smaller as n grows larger, bit it is always true that:

a_n < x < b_n

The interval always grows smaller so it is always true that where y is another real number there exists some n that |x-y| > |a_n - b_n| for |a_n - b_n| not equal to 0 and therefore x is unique. By this definition it is obviously apparent that there exists no number in between 2 real numbers then they are the same number. So let's look at this example, it is said that:

0.999 \ldots =1

You try to show that these exists a real number between them:

0.999 \ldots a

Where a is some integer in the range [0,8]. However by definition the 9s go on forever and therefore it must be true they do not stop so it must be true that the number is in fact:

0.999 \ldots a \ldots

And by definition of the number a=9. Therefore there exists no number in between 0.999... and 1 and therefore |1 - 0.999...| = 0 and the two numbers must be the same.

 

[arildno]

well, I can divide 3 by any integers, and that number sums to 1/3.

1) Do you know what a number is?
2) Do you know what a sum is?
3) Do you know what a logically coherent statement is?
Evidently not.

 

[Canute]

Not sure this is quite relevant but it seems to me that it is possible to show that there are is an infinite series of fractional numbers between zero and 1, and that this series cannot sum to 1. Is this completely obvious or mathematically heretical, and does it have any bearing on the 0.33.. x 3 =1 question?

 

[master_coda]

I agree.

To answer this question from the beginning:
The reason why 1/3 *3 = 1 where 0.333 (repeating) *3 is not:
is because 1/3 is perfectly rational (or simply as an elementary school kid would call it a fraction) while 0.33(repeating) is not.

0.33(repeating) = 1/3 is iff 0.33(repeating) is taken as sum to infinity, regardless of ANY BASE. Get it... ANY BASE.

Further, but at the same time, 0.111(repeating) is also = 1/3 is iff is taken as sum to infinity, becuase I simply had chosen 4 as my base.

Those who will continue to say this doesn't make sense, please don't just shoot those words. Prove it.

Irregardless of base? How can you use decimal notation regardless of base? That's like trying to use words regardless of language. 0.333... has different meanings in different bases; but when talking about the decimal 0.333... it doesn't matter, because a decimal number is always base 10. And the decimal 0.333...=1/3.

Using a different base, you get a different representation. In base 4, you represent 1/3 using 0.111... instead of 0.333..., so what? When you speak in French instead of English, you have to use different words to mean the same thing. When you use base four instead of base ten, you use different symbols to represent the same numbers. Where exactly is the contradiction?