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.

:rofl:
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" --- :rofl: :rofl:

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]

:rofl:
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" --- :rofl: :rofl:

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.