# 1 Divided by 3

Staff Emeritus
When you divide 1 by 3, you get .33333... repeating forever of course. My question is whether this operation could ever be considered to end. It looks to me like it's an invalid problem since you could never get a final answer, but simply keeping adding threes to the end of it when you try to solve. Does this make any sense?

Homework Helper
Gold Member
Dearly Missed
When you divide 1 by 3, you get .33333... repeating forever of course. My question is whether this operation could ever be considered to end. It looks to me like it's an invalid problem since you could never get a final answer, but simply keeping adding threes to the end of it when you try to solve. Does this make any sense?

Staff Emeritus

Hrmm. Why is that?

stevenb
When you divide 1 by 3, you get .33333... repeating forever of course. My question is whether this operation could ever be considered to end. It looks to me like it's an invalid problem since you could never get a final answer, but simply keeping adding threes to the end of it when you try to solve. Does this make any sense?

Do the math in a base 3 number system rather than a decimal system.

Here, 1/3 = 0.1, and 2/3 = 0.2

Any number system will be inadaquate for representing some rational numbers with a finite number of digits. This is just an artifact.

Staff Emeritus
Gold Member
When you divide 1 by 3, you get .33333... repeating forever of course. My question is whether this operation could ever be considered to end.
What could it possibly mean for the division to end?

If you instead meant "if I try to compute this quantity by using the long division algorithm, will the long division algorithm ever finish?", then the answer is as arildno said.

Homework Helper
Gold Member
Dearly Missed
Hi, drakkith!
What you are confused about is the difference between a number, and how it is to be represented according to some principle.
This is a subtle difference rarely touched upon in sachool maths, with profound consequences:

What you call "division" is actually "how to represent some number, usually defined as a fraction, by means of powers of ten".

1/3 is, bi its fundamental definition "that number, which multiplied with 3 yields 1".

Now, you CAN of course ask:
"How can this number be represented by powers of ten?".

The school answer to this is "by the process WE call division".

The result is that there is no finite sum of powers of ten that actually equals 1/3, but that, in its infinite limit, equals 1/3

Staff Emeritus
What could it possibly mean for the division to end?

If you instead meant "if I try to compute this quantity by using the long division algorithm, will the long division algorithm ever finish?", then the answer is as arildno said.

If i divide 10 by 5, i get 2, with nothing left over. The operation (the division) has ended, correct? Unless you could say that there are infinite 0's after the 2.0. Thats what I meant.

Staff Emeritus
Hi, drakkith!
What you are confused about is the difference between a number, and how it is to be represented according to some principle.
This is a subtle difference rarely touched upon in sachool maths, with profound consequences:

What you call "division" is actually "how to represent some number, usually defined as a fraction, by means of powers of ten".

1/3 is, bi its fundamental definition "that number, which multiplied with 3 yields 1".

Now, you CAN of course ask:
"How can this number be represented by powers of ten?".

The school answer to this is "by the process WE call division".

The result is that there is no finite sum of powers of ten that actually equals 1/3, but that, in its infinite limit, equals 1/3

I think i see what your saying. If i try to divide by a 0 is this an invalid operation, or is it similar to the above? I've always been decent at math but never actually gotten into some of the details like this before. Thanks for your answer!

slider142
I think i see what your saying. If i try to divide by a 0 is this an invalid operation, or is it similar to the above? I've always been decent at math but never actually gotten into some of the details like this before. Thanks for your answer!

Division by 0 is a different problem, it is not merely a problem of representing the number as powers of some base number; there is no way to define the operation meaningfully.
Division is defined in most cases as being an "undoing" of multiplication. That is a/b = c if and only if b*c = a. In technical terms, we call it an inverse operation to multiplication. However, there are some multiplications that cannot be undone.
When we have 18/9, we are asked to solve the multiplication problem 9*x = 18 for x, which we can see by inspection is 2, and only 2.
What about 1/0 ? We are asked to solve the problem 0*x = 1. But there is no number x for which this statement holds true, thus 1/0 is no number.
0/0 yields the equation 0*x = 0. In this case, x can be any number! We have chosen the convention that this is also not defined, as it does not yield a definite value for x.
There is more formality built around this for purposes of rigor, and there are some algebraic structures where division by 0 is defined, but in the algebra you are used to, that of real numbers, it yields an impossible equation.

Staff Emeritus
Division by 0 is a different problem, it is not merely a problem of representing the number as powers of some base number; there is no way to define the operation meaningfully.
Division is defined in most cases as being an "undoing" of multiplication. That is a/b = c if and only if b*c = a. In technical terms, we call it an inverse operation to multiplication. However, there are some multiplications that cannot be undone.
When we have 18/9, we are asked to solve the multiplication problem 9*x = 18 for x, which we can see by inspection is 2, and only 2.
What about 1/0 ? We are asked to solve the problem 0*x = 1. But there is no number x for which this statement holds true, thus 1/0 is no number.
0/0 yields the equation 0*x = 0. In this case, x can be any number! We have chosen the convention that this is also not defined, as it does not yield a definite value for x.
There is more formality built around this for purposes of rigor, and there are some algebraic structures where division by 0 is defined, but in the algebra you are used to, that of real numbers, it yields an impossible equation.

Thanks!

Homework Helper
When you divide 1 by 3, you get .33333... repeating forever of course. My question is whether this operation could ever be considered to end. It looks to me like it's an invalid problem since you could never get a final answer, but simply keeping adding threes to the end of it when you try to solve. Does this make any sense?

Hrmm. Why is that?
Because when you divide 1 by 3 and get 0.3333...., any "process" you used is finished. It makes no sense to say "you keep adding 3's". "0.333..." has already "added" all the threes- that is what the notation means. "0.333..." means exactly the same thing as 1/3 and is completely "finished".

Another way of looking at it: 0.3333.... means .3+ .03+ .003+ .0003+...= $3\sum_{n=1}^\infty .1^n$ and an "infinite sum" is defined as the limit of the partial sums, NOT the partial sums themselves, which is what you are doing in talking about continuing to "add 3s".

0.3333... is merely a conventional representation of 1/3 which serves to illustrate the fact that the division algorithm will produce digits corresponding to this sequence. It is not however the result of the division algorithm. Not surprisingly the infinite sum (limit of partial sums) corresponding to this sequence will converge to the number being represented, so it's at least superficially a consistent method of representation.

Staff Emeritus
Ahh ok i see now. Thanks alot all!

JDude13
0.33333... Is what we call the decimal aproximation.
Mathematicians often opt for fractions rather than decimals because they are more accurate.

The statement,
$$\frac{1}{3}=3.333333$$
is incorrect.

When writing this equation down, one should always use the "aproximately equal to" sign.

This statement is correct:
$$\frac{1}{3}\approx3.333333$$

Homework Helper
0.33333... Is what we call the decimal aproximation.
No, it isn't an approximation. 0.3333..., with the dots meaning "the 3's keep repeating" is exactly the same as 1/3.

Mathematicians often opt for fractions rather than decimals because they are more accurate.
That's a strange statement. "1/3" and "0.3333..." are both exact. But it's especially peculiar since you just said that 0.3333... was a "decimal approximation".

The statement,
$$\frac{1}{3}=3.333333$$
is incorrect.

When writing this equation down, one should always use the "aproximately equal to" sign.

This statement is correct:
$$\frac{1}{3}\approx3.333333$$
But the statement "1/3= 0.3333..." is correct.

(While 1/3 is not any where near "3.33333"! You have misplaced the decimal.)

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anmolnanda
This is not a problem with mathematics but a problem with representation of the number..probably the representation of decimal system has some loophole which produces this result..
for example
consider that peter is sick
now there are two ways of informing his mom
1)by post
2)by phone(where informing by phone is not a good method due to burred voice due to network but it is far better then post)

*Numbers in decimal makes Algeria operations like addition and other stuffs easier where you do not need to find lcm before addition..probably we have failed to invent better system to represent fraction so we have to go with the one with the loophole.

1MileCrash
How we represent numbers has little to do with their actual value.

1/3 = .333.... repeating.

That's not to say the problem doesn't "end." It just means that within our base 10 number system, it can't be expressed with an "end" (besides of course any notation used to symbolize endless repetition.)

0.333.. is just as exact and rational as 0.5. Merely how we represent them varies.

KEY:
No, it isn't an approximation. 0.3333..., with the dots meaning "the 3's keep repeating" is exactly the same as 1/3.

This is correct.

You can see it, in front of you with long division.

Divide 3 into 1, behind the decimal, so 3 into 10, essentially.
3 goes into 10 3 times, with a remainder of 1, so 3 into 10 again.
3 STILL goes into 10 3 times, with a remainder of 1.

You will be doing this infinitely, therefore a representing an infinite line of 3's is EXACTLY equal to 1/3.

Another confusion people often have with this is that they tend to think of repeating decimals as "growing." In other words, they think that since we can never write an infinite number of 3's, the number never "reaches" 1/3. .3 repeating is a number, it's value doesn't "grow" or "reach" anything, it has a fixed, exact value, which is 1/3.

This ties in with misconception that pi or any irrational number don't have fixed, exact values, which they do. Pi's exact value can be described as the circumference divided by the diameter of any circle, or pi. Just because I can't adequately write its exact value in decimal form doesn't mean it has no exact value.

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slider142
As an example of a representation other than the fractional representation that is finite, consider if our civilization used ternary as the primary representation instead of decimal. Then 1/3 (the number that divides a whole into three equal parts) would be represented as 0.1 and 1/2 would be the repeating string 0.111... . An analogue of this conversation would then be taking place about how 1/2 "never ends" and is not "an exact number" , not from any intrinsic property, but simply because of a historical choice of representation.

JDude13
No, it isn't an approximation. 0.3333..., with the dots meaning "the 3's keep repeating" is exactly the same as 1/3.

You got me... that was an error.

That's a strange statement. "1/3" and "0.3333..." are both exact. But it's especially peculiar since you just said that 0.3333... was a "decimal approximation".

Type 3.333333... into a calculator and tell me what you get.

But the statement "1/3= 0.3333..." is correct.

Please note that I did not include an ellipse at the end of that statement.

Mentor
You got me... that was an error.

Type 3.333333... into a calculator and tell me what you get.
I dunno. What do you get? You certainly don't get 1/3, since 3.333333... is off by a factor of 10.

Assuming that you really meant .333333 + (as many more digits as will fit in the calculator display>, some calculators have logic built in to recognize the first few digits of selected fractions. That in no way means that .333333333333 equals 1/3. It's not even all that close to 1/3.
Please note that I did not include an ellipse at the end of that statement.
Or an ellipsis, either. Homework Helper
You got me... that was an error.

Type 3.333333... into a calculator and tell me what you get.

Please note that I did not include an ellipse at the end of that statement.
Yes, I did. That was why I used the word "but".

Tarantula
if you cut mathematical cake(weight 1kg) into 3 ,every piece is 1/3 of originial cake and you calculate that every piece is almost ~0.3kg.You eat the cake.How much weight did you gain - 1kg or 0.9kg?

Mentor
if you cut mathematical cake(weight 1kg) into 3 ,every piece is 1/3 of originial cake and you calculate that every piece is almost ~0.3kg.You eat the cake.How much weight did you gain - 1kg or 0.9kg?
Each piece is NOT "almost ~0.3kg" How do you figure that?

Staff Emeritus
2021 Award
This statement is correct:
$$\frac{1}{3}\approx3.333333$$

Ummm...No, it is not. The left side is smaller than one and the right side is larger.

JDude13
Ummm...No, it is not. The left side is smaller than one and the right side is larger.

*blush*
a careless error...

Homework Helper
if you cut mathematical cake(weight 1kg) into 3 ,every piece is 1/3 of originial cake and you calculate that every piece is almost ~0.3kg.You eat the cake.How much weight did you gain - 1kg or 0.9kg?

This statement doesn't make any sense. If you want to round off by the tenth digit, then ok, 1/3 ~ 0.3. But all you're doing is rounding off (If I'm understanding what you said) Howeverm if you eat three pieces of cake, you gain 3*(1/3 kg) = 1kg

mburt
I think a lot of confusion is started simply because people hate working with fractions!

But remember that fractions are exact values, and in any situation where you think you have to approximate (like the cake example), that is a misconception.

Any algebra or formulae can utilize fractions, approximations are often used in real-life situations where a fractional answer isn't practical.

tedima
1 divided by 3 = 0.3r
0.3r x 3 = 0.9r
missing value of 0.0r1 1 at the end of an infinate amount of 0's
0.0r1 divided by 3 = 0.0r0.3r

so

1 divided by 3 = 0.[3r0r]r
0.[3r0r]r x 3 = 1

Which adds the missing value of 0.0r1

If you understand O.O

(^Dont know if its true just been thinking about it) Tell me if im wrong
OR add your own correction Last edited:
Mentor
1 divided by 3 = 0.3r
0.3r x 3 = 0.9r
missing value of 0.0r1 1 at the end of an infinate amount of 0's
Where exactly is the end of an infinite number of 0's?
0.0r1 divided by 3 = 0.0r0.3r

so

1 divided by 3 = 0.[3r0r]r
0.[3r0r]r x 3 = 1

Which adds the missing value of 0.0r1

If you understand O.O

(^Dont know if its true just been thinking about it) Tell me if im wrong

Staff Emeritus
Tedima, there is no end to an infinitely long number.
The number 0.999... REPRESENTS a value that is equal to the number 1.

tedima
Tedima, there is no end to an infinitely long number.
The number 0.999... REPRESENTS a value that is equal to the number 1.

0.9999 is smaller than 1 if i put 99p into my bank account it wouldnt show as £1 on the machine

tedima
Where exactly is the end of an infinite number of 0's?

There is no end, 1 can not actually be divided by 3 in theory. Infinity is not a number... but yet u divide 1 by 3 and you get an infinite amount of 3's.

Staff Emeritus