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

[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?

 

[Rogerio]

...
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.
...

First: there is no 'perfect' rational number - a number is rational or irrational.

And second: 0.33... is rational, since it can be represented as 2/6 , for instance.

:-)

 

[ohwilleke]

Why the hell have you introduced this spurious analogy?

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

Why so hostile? Nothing spurious here. The best way to explain something, when one approach isn't working, it to explain it in another correct manner in that hope that useing a different approach makes the matter more clear.

Your statement to which I replied says that a matter is definitional. But, that really understates the situation. Why did somebody use a particular definition? Because no other definition makes sense to give what we expect the thing we call "real numbers" to do.

The point of my discussion was to illustrate explicitly, as okidream doesn't seem to follow (as indicated his response which seems to erroneously state that the value of a number should be independent of the base of the number system in which it is represented), why the relationship between the base ten number system and what it means to represent numbers in more than one number system, makes the definitions the imply that 0.333 repeating=1/3 the uniquely sensible choice, rather than being an arbitrary definition.

 

[matt grime]

Your analogy was completely bizarre (frames of reference in GR? In what sense can this be 'equivalent'?) and I always get hostile when people use patronizing language to explain to *me* something that is bleeding obvious, and that *I* don't need to be told, especially if they don't explain what it means to produce a model of a complete ordered field.

 

[Hurkyl]

Ok, nobody seems to be convincing anyone of anything (per usual, *sigh*), and temperatures are rising, so I'm going to close this.