How can adding an infinite number of rationals yield an irrational number?
For example how come $zeta(2)=sum_{n=1}^{infty}n^{-2}=frac{pi^2}{6}$. It seems counter intuitive that you can add numbers in $mathbb{Q}$ and get an irrational number.
number-theory
|
show 1 more comment
For example how come $zeta(2)=sum_{n=1}^{infty}n^{-2}=frac{pi^2}{6}$. It seems counter intuitive that you can add numbers in $mathbb{Q}$ and get an irrational number.
number-theory
88
Every real number is the sum of countably many rational numbers.
– Did
Mar 4 '12 at 9:47
54
Properties that are preserved under a finite number of operations are not necessarily preserved under an infinite number of them.
– Rahul
Mar 4 '12 at 9:57
6
Real numbers have the property that the bounded infinite sum of positive real numbers is a real number because of completeness. This is not a property of rational numbers, so I would intuitively expect some bounded infinite sums of positive rational numbers not to be a rational number.
– Henry
Mar 4 '12 at 10:15
4
because real number are closure of rational numbers.
– quartz
Mar 4 '12 at 12:12
3
Another example to illustrate Rahul's point: the product of infinite positive reals can yield zero: $prod_{n=1}^infty 1/n = 0$ or more simply $prod_{n=1}^infty 0.5 = 0$
– leonbloy
Apr 30 '12 at 18:36
|
show 1 more comment
For example how come $zeta(2)=sum_{n=1}^{infty}n^{-2}=frac{pi^2}{6}$. It seems counter intuitive that you can add numbers in $mathbb{Q}$ and get an irrational number.
number-theory
For example how come $zeta(2)=sum_{n=1}^{infty}n^{-2}=frac{pi^2}{6}$. It seems counter intuitive that you can add numbers in $mathbb{Q}$ and get an irrational number.
number-theory
number-theory
asked Mar 4 '12 at 9:42
E.O.
5,01452550
5,01452550
88
Every real number is the sum of countably many rational numbers.
– Did
Mar 4 '12 at 9:47
54
Properties that are preserved under a finite number of operations are not necessarily preserved under an infinite number of them.
– Rahul
Mar 4 '12 at 9:57
6
Real numbers have the property that the bounded infinite sum of positive real numbers is a real number because of completeness. This is not a property of rational numbers, so I would intuitively expect some bounded infinite sums of positive rational numbers not to be a rational number.
– Henry
Mar 4 '12 at 10:15
4
because real number are closure of rational numbers.
– quartz
Mar 4 '12 at 12:12
3
Another example to illustrate Rahul's point: the product of infinite positive reals can yield zero: $prod_{n=1}^infty 1/n = 0$ or more simply $prod_{n=1}^infty 0.5 = 0$
– leonbloy
Apr 30 '12 at 18:36
|
show 1 more comment
88
Every real number is the sum of countably many rational numbers.
– Did
Mar 4 '12 at 9:47
54
Properties that are preserved under a finite number of operations are not necessarily preserved under an infinite number of them.
– Rahul
Mar 4 '12 at 9:57
6
Real numbers have the property that the bounded infinite sum of positive real numbers is a real number because of completeness. This is not a property of rational numbers, so I would intuitively expect some bounded infinite sums of positive rational numbers not to be a rational number.
– Henry
Mar 4 '12 at 10:15
4
because real number are closure of rational numbers.
– quartz
Mar 4 '12 at 12:12
3
Another example to illustrate Rahul's point: the product of infinite positive reals can yield zero: $prod_{n=1}^infty 1/n = 0$ or more simply $prod_{n=1}^infty 0.5 = 0$
– leonbloy
Apr 30 '12 at 18:36
88
88
Every real number is the sum of countably many rational numbers.
– Did
Mar 4 '12 at 9:47
Every real number is the sum of countably many rational numbers.
– Did
Mar 4 '12 at 9:47
54
54
Properties that are preserved under a finite number of operations are not necessarily preserved under an infinite number of them.
– Rahul
Mar 4 '12 at 9:57
Properties that are preserved under a finite number of operations are not necessarily preserved under an infinite number of them.
– Rahul
Mar 4 '12 at 9:57
6
6
Real numbers have the property that the bounded infinite sum of positive real numbers is a real number because of completeness. This is not a property of rational numbers, so I would intuitively expect some bounded infinite sums of positive rational numbers not to be a rational number.
– Henry
Mar 4 '12 at 10:15
Real numbers have the property that the bounded infinite sum of positive real numbers is a real number because of completeness. This is not a property of rational numbers, so I would intuitively expect some bounded infinite sums of positive rational numbers not to be a rational number.
– Henry
Mar 4 '12 at 10:15
4
4
because real number are closure of rational numbers.
– quartz
Mar 4 '12 at 12:12
because real number are closure of rational numbers.
– quartz
Mar 4 '12 at 12:12
3
3
Another example to illustrate Rahul's point: the product of infinite positive reals can yield zero: $prod_{n=1}^infty 1/n = 0$ or more simply $prod_{n=1}^infty 0.5 = 0$
– leonbloy
Apr 30 '12 at 18:36
Another example to illustrate Rahul's point: the product of infinite positive reals can yield zero: $prod_{n=1}^infty 1/n = 0$ or more simply $prod_{n=1}^infty 0.5 = 0$
– leonbloy
Apr 30 '12 at 18:36
|
show 1 more comment
7 Answers
7
active
oldest
votes
But for example $$pi=3+0.1+0.04+0.001+0.0005+0.00009+0.000002+cdots$$ and that surely does not seem strange to you...
1
Interesting... as a side note, you can determine the size of an infinite set by seeing if it can fit inside another infinite set. This isn't directly relevant here, but still is something important to understand. en.wikipedia.org/wiki/Aleph_number
– JSWork
Mar 5 '12 at 15:06
45
@JSWork, not only would I say that that isn't directly relevant here: it is quite irrelevant, in fact! :D
– Mariano Suárez-Álvarez
Mar 7 '12 at 4:11
@MarianoSuárez-Álvarez Is infinite sum of rationals always an irrational number ? Because recently I saw an example that the infinite sum of 1/2n for n from 1 to infinity equals 1, and 1 is rational.
– Maths Survivor
Dec 20 '17 at 23:14
@MathsSurvivor No. Your own comment has the proof.
– user3658307
May 11 at 18:37
add a comment |
You can't add an infinite number of rational numbers. What you can do, though, is find a limit of a sequence of partial sums. So, $pi^2/6$ is the limit to infinity of the sequence $1, 1 + 1/4, 1 + 1/4 + 1/9, 1 + 1/4 + 1/9 + 1/16, ldots $. Writing it so that it looks like a sum is really just a shorthand.
In other words, $sum^infty_{i=1} cdots$ is actually kind of an abbreviation for $lim_{ntoinfty} sum^n_{i=1} cdots$.
3
Good point to observe!
– user21436
Mar 4 '12 at 10:25
22
@TonyK, nop, you really cannot. What you can do is to take the limit of sequences of partial sums: there is a difference between that and «summing an infinite sequence of numbers». In particular, the latter simply does not make sense.
– Mariano Suárez-Álvarez
Mar 5 '12 at 2:06
7
@TonyK, Dear TonyK: if you read Rudin with care you will see that it nowhere does that.
– Mariano Suárez-Álvarez
Mar 5 '12 at 20:05
9
Oh well. ${}{}$
– Mariano Suárez-Álvarez
Mar 5 '12 at 22:07
24
@TonyK Quoting from Rudin: "The number $s$ is called the sum of the series, but it should be clearly understood that $s$ is the limit of a sequence of sums and is not obtained simply by addition."
– Pedro Tamaroff♦
Apr 15 '13 at 19:31
|
show 8 more comments
Others have demonstrated some examples that make clear why this can happen, but I wanted to point out the key mathematical concept here is "Completeness" of the metric space. A metric space is any set with "distance" defined between any two elements (in the case of $mathbb{Q}$, we would say $d(x,y) = |x-y|$). A sequence $x_i$ is "Cauchy" if late elements stop moving around very much, a necessary condition for a sequence to have a finite limit. To put it formally, ${x_i}$ is cauchy for $epsilon>0$, there is a sufficiently large $N$ so that for every $m,n>N$ we have $d(x_n,x_m)<epsilon$. A metric space is complete if all Cauchy sequences have a limit in the space. The canonical complete metric space is $mathbb R$, which is in fact the completion of $mathbb{Q}$, or the smallest complete set containing $mathbb Q$.
We think of an infinite sum as the limit of a sequence of partial sums:
$$sum_{n=1}^infty x_n = lim_{Ntoinfty}left( sum_{i=1}^Nx_n right)$$
As others have pointed out with a number of good counter-examples (my favorite of which is the decimal representation of an irrational number), $mathbb{Q}$ is not complete, therefore an infinite sum of elements of $mathbb Q$, for which partial sums are necessarily elements of $mathbb Q$, can converge to a value not in $mathbb Q$.
add a comment |
It is counter-intuitive only if you are adding a "finite" number of rational numbers. Otherwise, as @Mariano implied, any irrational number consists of an infinite number of digits, and thus can be represented as a sum of rational numbers.
2
This is my point. There is nothing magic with irrational numbers and they just use the same digits that rational numbers, so there is no surprise or counter-intuition in having irrational numbers representable as an infinite sum of rational numbers.
– Rafid
Mar 4 '12 at 10:41
2
@RahulNarain: In the decimal expressions of rational numbers, there is always either a terminating or an infinitely repeating string of digits. Irrational numbers have neither of these properties, but can still be expressed as an infinitely long non-repeating sequence digits. Though there may not be an explicit formula for this sequence, it can still be thought of as an infinite sum of rational numbers, as Rafid indicated.
– Paul
Mar 4 '12 at 15:00
1
@Paul, I was making a joke on the phrase "must have digits from 0 to 9" that appeared in the original version of the answer.
– Rahul
Mar 4 '12 at 22:31
add a comment |
Besides that. Given a sequence of positive rational numbers such that their sum converges and such that $a_n > sum_{k = n + 1}^infty a_k$. Then choosing a $pm$ sign for each term of the sequence gives a new convergent series, each to a different number. By a countable-uncountable argument you get nonnumerable examples of that kind of series :).
add a comment |
For example look at
$$e = sum_{k=0}^{infty}{frac{1}{k!}} $$
add a comment |
This has to see with the rate at which the sum/series converges to its limit and the Roth-Thue-Siegel theorem which allows you to use the rate of convergence to decide if the limit is rational or not.
Maybe Emile (the OP) meant to ask something of this sort (please let me know, Emile): why do some (convergent, of course) infinite sums of rationals are rational and others are irrational?
this is a different question..
– AIB
Mar 6 '12 at 9:41
How so? The conditions/rate of convergence affect whether the limit is rational or not. Don't they?
– AQP
Mar 6 '12 at 16:54
add a comment |
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7 Answers
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But for example $$pi=3+0.1+0.04+0.001+0.0005+0.00009+0.000002+cdots$$ and that surely does not seem strange to you...
1
Interesting... as a side note, you can determine the size of an infinite set by seeing if it can fit inside another infinite set. This isn't directly relevant here, but still is something important to understand. en.wikipedia.org/wiki/Aleph_number
– JSWork
Mar 5 '12 at 15:06
45
@JSWork, not only would I say that that isn't directly relevant here: it is quite irrelevant, in fact! :D
– Mariano Suárez-Álvarez
Mar 7 '12 at 4:11
@MarianoSuárez-Álvarez Is infinite sum of rationals always an irrational number ? Because recently I saw an example that the infinite sum of 1/2n for n from 1 to infinity equals 1, and 1 is rational.
– Maths Survivor
Dec 20 '17 at 23:14
@MathsSurvivor No. Your own comment has the proof.
– user3658307
May 11 at 18:37
add a comment |
But for example $$pi=3+0.1+0.04+0.001+0.0005+0.00009+0.000002+cdots$$ and that surely does not seem strange to you...
1
Interesting... as a side note, you can determine the size of an infinite set by seeing if it can fit inside another infinite set. This isn't directly relevant here, but still is something important to understand. en.wikipedia.org/wiki/Aleph_number
– JSWork
Mar 5 '12 at 15:06
45
@JSWork, not only would I say that that isn't directly relevant here: it is quite irrelevant, in fact! :D
– Mariano Suárez-Álvarez
Mar 7 '12 at 4:11
@MarianoSuárez-Álvarez Is infinite sum of rationals always an irrational number ? Because recently I saw an example that the infinite sum of 1/2n for n from 1 to infinity equals 1, and 1 is rational.
– Maths Survivor
Dec 20 '17 at 23:14
@MathsSurvivor No. Your own comment has the proof.
– user3658307
May 11 at 18:37
add a comment |
But for example $$pi=3+0.1+0.04+0.001+0.0005+0.00009+0.000002+cdots$$ and that surely does not seem strange to you...
But for example $$pi=3+0.1+0.04+0.001+0.0005+0.00009+0.000002+cdots$$ and that surely does not seem strange to you...
answered Mar 4 '12 at 9:45
Mariano Suárez-Álvarez
110k7155280
110k7155280
1
Interesting... as a side note, you can determine the size of an infinite set by seeing if it can fit inside another infinite set. This isn't directly relevant here, but still is something important to understand. en.wikipedia.org/wiki/Aleph_number
– JSWork
Mar 5 '12 at 15:06
45
@JSWork, not only would I say that that isn't directly relevant here: it is quite irrelevant, in fact! :D
– Mariano Suárez-Álvarez
Mar 7 '12 at 4:11
@MarianoSuárez-Álvarez Is infinite sum of rationals always an irrational number ? Because recently I saw an example that the infinite sum of 1/2n for n from 1 to infinity equals 1, and 1 is rational.
– Maths Survivor
Dec 20 '17 at 23:14
@MathsSurvivor No. Your own comment has the proof.
– user3658307
May 11 at 18:37
add a comment |
1
Interesting... as a side note, you can determine the size of an infinite set by seeing if it can fit inside another infinite set. This isn't directly relevant here, but still is something important to understand. en.wikipedia.org/wiki/Aleph_number
– JSWork
Mar 5 '12 at 15:06
45
@JSWork, not only would I say that that isn't directly relevant here: it is quite irrelevant, in fact! :D
– Mariano Suárez-Álvarez
Mar 7 '12 at 4:11
@MarianoSuárez-Álvarez Is infinite sum of rationals always an irrational number ? Because recently I saw an example that the infinite sum of 1/2n for n from 1 to infinity equals 1, and 1 is rational.
– Maths Survivor
Dec 20 '17 at 23:14
@MathsSurvivor No. Your own comment has the proof.
– user3658307
May 11 at 18:37
1
1
Interesting... as a side note, you can determine the size of an infinite set by seeing if it can fit inside another infinite set. This isn't directly relevant here, but still is something important to understand. en.wikipedia.org/wiki/Aleph_number
– JSWork
Mar 5 '12 at 15:06
Interesting... as a side note, you can determine the size of an infinite set by seeing if it can fit inside another infinite set. This isn't directly relevant here, but still is something important to understand. en.wikipedia.org/wiki/Aleph_number
– JSWork
Mar 5 '12 at 15:06
45
45
@JSWork, not only would I say that that isn't directly relevant here: it is quite irrelevant, in fact! :D
– Mariano Suárez-Álvarez
Mar 7 '12 at 4:11
@JSWork, not only would I say that that isn't directly relevant here: it is quite irrelevant, in fact! :D
– Mariano Suárez-Álvarez
Mar 7 '12 at 4:11
@MarianoSuárez-Álvarez Is infinite sum of rationals always an irrational number ? Because recently I saw an example that the infinite sum of 1/2n for n from 1 to infinity equals 1, and 1 is rational.
– Maths Survivor
Dec 20 '17 at 23:14
@MarianoSuárez-Álvarez Is infinite sum of rationals always an irrational number ? Because recently I saw an example that the infinite sum of 1/2n for n from 1 to infinity equals 1, and 1 is rational.
– Maths Survivor
Dec 20 '17 at 23:14
@MathsSurvivor No. Your own comment has the proof.
– user3658307
May 11 at 18:37
@MathsSurvivor No. Your own comment has the proof.
– user3658307
May 11 at 18:37
add a comment |
You can't add an infinite number of rational numbers. What you can do, though, is find a limit of a sequence of partial sums. So, $pi^2/6$ is the limit to infinity of the sequence $1, 1 + 1/4, 1 + 1/4 + 1/9, 1 + 1/4 + 1/9 + 1/16, ldots $. Writing it so that it looks like a sum is really just a shorthand.
In other words, $sum^infty_{i=1} cdots$ is actually kind of an abbreviation for $lim_{ntoinfty} sum^n_{i=1} cdots$.
3
Good point to observe!
– user21436
Mar 4 '12 at 10:25
22
@TonyK, nop, you really cannot. What you can do is to take the limit of sequences of partial sums: there is a difference between that and «summing an infinite sequence of numbers». In particular, the latter simply does not make sense.
– Mariano Suárez-Álvarez
Mar 5 '12 at 2:06
7
@TonyK, Dear TonyK: if you read Rudin with care you will see that it nowhere does that.
– Mariano Suárez-Álvarez
Mar 5 '12 at 20:05
9
Oh well. ${}{}$
– Mariano Suárez-Álvarez
Mar 5 '12 at 22:07
24
@TonyK Quoting from Rudin: "The number $s$ is called the sum of the series, but it should be clearly understood that $s$ is the limit of a sequence of sums and is not obtained simply by addition."
– Pedro Tamaroff♦
Apr 15 '13 at 19:31
|
show 8 more comments
You can't add an infinite number of rational numbers. What you can do, though, is find a limit of a sequence of partial sums. So, $pi^2/6$ is the limit to infinity of the sequence $1, 1 + 1/4, 1 + 1/4 + 1/9, 1 + 1/4 + 1/9 + 1/16, ldots $. Writing it so that it looks like a sum is really just a shorthand.
In other words, $sum^infty_{i=1} cdots$ is actually kind of an abbreviation for $lim_{ntoinfty} sum^n_{i=1} cdots$.
3
Good point to observe!
– user21436
Mar 4 '12 at 10:25
22
@TonyK, nop, you really cannot. What you can do is to take the limit of sequences of partial sums: there is a difference between that and «summing an infinite sequence of numbers». In particular, the latter simply does not make sense.
– Mariano Suárez-Álvarez
Mar 5 '12 at 2:06
7
@TonyK, Dear TonyK: if you read Rudin with care you will see that it nowhere does that.
– Mariano Suárez-Álvarez
Mar 5 '12 at 20:05
9
Oh well. ${}{}$
– Mariano Suárez-Álvarez
Mar 5 '12 at 22:07
24
@TonyK Quoting from Rudin: "The number $s$ is called the sum of the series, but it should be clearly understood that $s$ is the limit of a sequence of sums and is not obtained simply by addition."
– Pedro Tamaroff♦
Apr 15 '13 at 19:31
|
show 8 more comments
You can't add an infinite number of rational numbers. What you can do, though, is find a limit of a sequence of partial sums. So, $pi^2/6$ is the limit to infinity of the sequence $1, 1 + 1/4, 1 + 1/4 + 1/9, 1 + 1/4 + 1/9 + 1/16, ldots $. Writing it so that it looks like a sum is really just a shorthand.
In other words, $sum^infty_{i=1} cdots$ is actually kind of an abbreviation for $lim_{ntoinfty} sum^n_{i=1} cdots$.
You can't add an infinite number of rational numbers. What you can do, though, is find a limit of a sequence of partial sums. So, $pi^2/6$ is the limit to infinity of the sequence $1, 1 + 1/4, 1 + 1/4 + 1/9, 1 + 1/4 + 1/9 + 1/16, ldots $. Writing it so that it looks like a sum is really just a shorthand.
In other words, $sum^infty_{i=1} cdots$ is actually kind of an abbreviation for $lim_{ntoinfty} sum^n_{i=1} cdots$.
answered Mar 4 '12 at 10:20
user22805
3
Good point to observe!
– user21436
Mar 4 '12 at 10:25
22
@TonyK, nop, you really cannot. What you can do is to take the limit of sequences of partial sums: there is a difference between that and «summing an infinite sequence of numbers». In particular, the latter simply does not make sense.
– Mariano Suárez-Álvarez
Mar 5 '12 at 2:06
7
@TonyK, Dear TonyK: if you read Rudin with care you will see that it nowhere does that.
– Mariano Suárez-Álvarez
Mar 5 '12 at 20:05
9
Oh well. ${}{}$
– Mariano Suárez-Álvarez
Mar 5 '12 at 22:07
24
@TonyK Quoting from Rudin: "The number $s$ is called the sum of the series, but it should be clearly understood that $s$ is the limit of a sequence of sums and is not obtained simply by addition."
– Pedro Tamaroff♦
Apr 15 '13 at 19:31
|
show 8 more comments
3
Good point to observe!
– user21436
Mar 4 '12 at 10:25
22
@TonyK, nop, you really cannot. What you can do is to take the limit of sequences of partial sums: there is a difference between that and «summing an infinite sequence of numbers». In particular, the latter simply does not make sense.
– Mariano Suárez-Álvarez
Mar 5 '12 at 2:06
7
@TonyK, Dear TonyK: if you read Rudin with care you will see that it nowhere does that.
– Mariano Suárez-Álvarez
Mar 5 '12 at 20:05
9
Oh well. ${}{}$
– Mariano Suárez-Álvarez
Mar 5 '12 at 22:07
24
@TonyK Quoting from Rudin: "The number $s$ is called the sum of the series, but it should be clearly understood that $s$ is the limit of a sequence of sums and is not obtained simply by addition."
– Pedro Tamaroff♦
Apr 15 '13 at 19:31
3
3
Good point to observe!
– user21436
Mar 4 '12 at 10:25
Good point to observe!
– user21436
Mar 4 '12 at 10:25
22
22
@TonyK, nop, you really cannot. What you can do is to take the limit of sequences of partial sums: there is a difference between that and «summing an infinite sequence of numbers». In particular, the latter simply does not make sense.
– Mariano Suárez-Álvarez
Mar 5 '12 at 2:06
@TonyK, nop, you really cannot. What you can do is to take the limit of sequences of partial sums: there is a difference between that and «summing an infinite sequence of numbers». In particular, the latter simply does not make sense.
– Mariano Suárez-Álvarez
Mar 5 '12 at 2:06
7
7
@TonyK, Dear TonyK: if you read Rudin with care you will see that it nowhere does that.
– Mariano Suárez-Álvarez
Mar 5 '12 at 20:05
@TonyK, Dear TonyK: if you read Rudin with care you will see that it nowhere does that.
– Mariano Suárez-Álvarez
Mar 5 '12 at 20:05
9
9
Oh well. ${}{}$
– Mariano Suárez-Álvarez
Mar 5 '12 at 22:07
Oh well. ${}{}$
– Mariano Suárez-Álvarez
Mar 5 '12 at 22:07
24
24
@TonyK Quoting from Rudin: "The number $s$ is called the sum of the series, but it should be clearly understood that $s$ is the limit of a sequence of sums and is not obtained simply by addition."
– Pedro Tamaroff♦
Apr 15 '13 at 19:31
@TonyK Quoting from Rudin: "The number $s$ is called the sum of the series, but it should be clearly understood that $s$ is the limit of a sequence of sums and is not obtained simply by addition."
– Pedro Tamaroff♦
Apr 15 '13 at 19:31
|
show 8 more comments
Others have demonstrated some examples that make clear why this can happen, but I wanted to point out the key mathematical concept here is "Completeness" of the metric space. A metric space is any set with "distance" defined between any two elements (in the case of $mathbb{Q}$, we would say $d(x,y) = |x-y|$). A sequence $x_i$ is "Cauchy" if late elements stop moving around very much, a necessary condition for a sequence to have a finite limit. To put it formally, ${x_i}$ is cauchy for $epsilon>0$, there is a sufficiently large $N$ so that for every $m,n>N$ we have $d(x_n,x_m)<epsilon$. A metric space is complete if all Cauchy sequences have a limit in the space. The canonical complete metric space is $mathbb R$, which is in fact the completion of $mathbb{Q}$, or the smallest complete set containing $mathbb Q$.
We think of an infinite sum as the limit of a sequence of partial sums:
$$sum_{n=1}^infty x_n = lim_{Ntoinfty}left( sum_{i=1}^Nx_n right)$$
As others have pointed out with a number of good counter-examples (my favorite of which is the decimal representation of an irrational number), $mathbb{Q}$ is not complete, therefore an infinite sum of elements of $mathbb Q$, for which partial sums are necessarily elements of $mathbb Q$, can converge to a value not in $mathbb Q$.
add a comment |
Others have demonstrated some examples that make clear why this can happen, but I wanted to point out the key mathematical concept here is "Completeness" of the metric space. A metric space is any set with "distance" defined between any two elements (in the case of $mathbb{Q}$, we would say $d(x,y) = |x-y|$). A sequence $x_i$ is "Cauchy" if late elements stop moving around very much, a necessary condition for a sequence to have a finite limit. To put it formally, ${x_i}$ is cauchy for $epsilon>0$, there is a sufficiently large $N$ so that for every $m,n>N$ we have $d(x_n,x_m)<epsilon$. A metric space is complete if all Cauchy sequences have a limit in the space. The canonical complete metric space is $mathbb R$, which is in fact the completion of $mathbb{Q}$, or the smallest complete set containing $mathbb Q$.
We think of an infinite sum as the limit of a sequence of partial sums:
$$sum_{n=1}^infty x_n = lim_{Ntoinfty}left( sum_{i=1}^Nx_n right)$$
As others have pointed out with a number of good counter-examples (my favorite of which is the decimal representation of an irrational number), $mathbb{Q}$ is not complete, therefore an infinite sum of elements of $mathbb Q$, for which partial sums are necessarily elements of $mathbb Q$, can converge to a value not in $mathbb Q$.
add a comment |
Others have demonstrated some examples that make clear why this can happen, but I wanted to point out the key mathematical concept here is "Completeness" of the metric space. A metric space is any set with "distance" defined between any two elements (in the case of $mathbb{Q}$, we would say $d(x,y) = |x-y|$). A sequence $x_i$ is "Cauchy" if late elements stop moving around very much, a necessary condition for a sequence to have a finite limit. To put it formally, ${x_i}$ is cauchy for $epsilon>0$, there is a sufficiently large $N$ so that for every $m,n>N$ we have $d(x_n,x_m)<epsilon$. A metric space is complete if all Cauchy sequences have a limit in the space. The canonical complete metric space is $mathbb R$, which is in fact the completion of $mathbb{Q}$, or the smallest complete set containing $mathbb Q$.
We think of an infinite sum as the limit of a sequence of partial sums:
$$sum_{n=1}^infty x_n = lim_{Ntoinfty}left( sum_{i=1}^Nx_n right)$$
As others have pointed out with a number of good counter-examples (my favorite of which is the decimal representation of an irrational number), $mathbb{Q}$ is not complete, therefore an infinite sum of elements of $mathbb Q$, for which partial sums are necessarily elements of $mathbb Q$, can converge to a value not in $mathbb Q$.
Others have demonstrated some examples that make clear why this can happen, but I wanted to point out the key mathematical concept here is "Completeness" of the metric space. A metric space is any set with "distance" defined between any two elements (in the case of $mathbb{Q}$, we would say $d(x,y) = |x-y|$). A sequence $x_i$ is "Cauchy" if late elements stop moving around very much, a necessary condition for a sequence to have a finite limit. To put it formally, ${x_i}$ is cauchy for $epsilon>0$, there is a sufficiently large $N$ so that for every $m,n>N$ we have $d(x_n,x_m)<epsilon$. A metric space is complete if all Cauchy sequences have a limit in the space. The canonical complete metric space is $mathbb R$, which is in fact the completion of $mathbb{Q}$, or the smallest complete set containing $mathbb Q$.
We think of an infinite sum as the limit of a sequence of partial sums:
$$sum_{n=1}^infty x_n = lim_{Ntoinfty}left( sum_{i=1}^Nx_n right)$$
As others have pointed out with a number of good counter-examples (my favorite of which is the decimal representation of an irrational number), $mathbb{Q}$ is not complete, therefore an infinite sum of elements of $mathbb Q$, for which partial sums are necessarily elements of $mathbb Q$, can converge to a value not in $mathbb Q$.
answered Mar 4 '12 at 22:17
Neil Peterman
32114
32114
add a comment |
add a comment |
It is counter-intuitive only if you are adding a "finite" number of rational numbers. Otherwise, as @Mariano implied, any irrational number consists of an infinite number of digits, and thus can be represented as a sum of rational numbers.
2
This is my point. There is nothing magic with irrational numbers and they just use the same digits that rational numbers, so there is no surprise or counter-intuition in having irrational numbers representable as an infinite sum of rational numbers.
– Rafid
Mar 4 '12 at 10:41
2
@RahulNarain: In the decimal expressions of rational numbers, there is always either a terminating or an infinitely repeating string of digits. Irrational numbers have neither of these properties, but can still be expressed as an infinitely long non-repeating sequence digits. Though there may not be an explicit formula for this sequence, it can still be thought of as an infinite sum of rational numbers, as Rafid indicated.
– Paul
Mar 4 '12 at 15:00
1
@Paul, I was making a joke on the phrase "must have digits from 0 to 9" that appeared in the original version of the answer.
– Rahul
Mar 4 '12 at 22:31
add a comment |
It is counter-intuitive only if you are adding a "finite" number of rational numbers. Otherwise, as @Mariano implied, any irrational number consists of an infinite number of digits, and thus can be represented as a sum of rational numbers.
2
This is my point. There is nothing magic with irrational numbers and they just use the same digits that rational numbers, so there is no surprise or counter-intuition in having irrational numbers representable as an infinite sum of rational numbers.
– Rafid
Mar 4 '12 at 10:41
2
@RahulNarain: In the decimal expressions of rational numbers, there is always either a terminating or an infinitely repeating string of digits. Irrational numbers have neither of these properties, but can still be expressed as an infinitely long non-repeating sequence digits. Though there may not be an explicit formula for this sequence, it can still be thought of as an infinite sum of rational numbers, as Rafid indicated.
– Paul
Mar 4 '12 at 15:00
1
@Paul, I was making a joke on the phrase "must have digits from 0 to 9" that appeared in the original version of the answer.
– Rahul
Mar 4 '12 at 22:31
add a comment |
It is counter-intuitive only if you are adding a "finite" number of rational numbers. Otherwise, as @Mariano implied, any irrational number consists of an infinite number of digits, and thus can be represented as a sum of rational numbers.
It is counter-intuitive only if you are adding a "finite" number of rational numbers. Otherwise, as @Mariano implied, any irrational number consists of an infinite number of digits, and thus can be represented as a sum of rational numbers.
edited Mar 4 '12 at 10:44
answered Mar 4 '12 at 10:24
Rafid
6121710
6121710
2
This is my point. There is nothing magic with irrational numbers and they just use the same digits that rational numbers, so there is no surprise or counter-intuition in having irrational numbers representable as an infinite sum of rational numbers.
– Rafid
Mar 4 '12 at 10:41
2
@RahulNarain: In the decimal expressions of rational numbers, there is always either a terminating or an infinitely repeating string of digits. Irrational numbers have neither of these properties, but can still be expressed as an infinitely long non-repeating sequence digits. Though there may not be an explicit formula for this sequence, it can still be thought of as an infinite sum of rational numbers, as Rafid indicated.
– Paul
Mar 4 '12 at 15:00
1
@Paul, I was making a joke on the phrase "must have digits from 0 to 9" that appeared in the original version of the answer.
– Rahul
Mar 4 '12 at 22:31
add a comment |
2
This is my point. There is nothing magic with irrational numbers and they just use the same digits that rational numbers, so there is no surprise or counter-intuition in having irrational numbers representable as an infinite sum of rational numbers.
– Rafid
Mar 4 '12 at 10:41
2
@RahulNarain: In the decimal expressions of rational numbers, there is always either a terminating or an infinitely repeating string of digits. Irrational numbers have neither of these properties, but can still be expressed as an infinitely long non-repeating sequence digits. Though there may not be an explicit formula for this sequence, it can still be thought of as an infinite sum of rational numbers, as Rafid indicated.
– Paul
Mar 4 '12 at 15:00
1
@Paul, I was making a joke on the phrase "must have digits from 0 to 9" that appeared in the original version of the answer.
– Rahul
Mar 4 '12 at 22:31
2
2
This is my point. There is nothing magic with irrational numbers and they just use the same digits that rational numbers, so there is no surprise or counter-intuition in having irrational numbers representable as an infinite sum of rational numbers.
– Rafid
Mar 4 '12 at 10:41
This is my point. There is nothing magic with irrational numbers and they just use the same digits that rational numbers, so there is no surprise or counter-intuition in having irrational numbers representable as an infinite sum of rational numbers.
– Rafid
Mar 4 '12 at 10:41
2
2
@RahulNarain: In the decimal expressions of rational numbers, there is always either a terminating or an infinitely repeating string of digits. Irrational numbers have neither of these properties, but can still be expressed as an infinitely long non-repeating sequence digits. Though there may not be an explicit formula for this sequence, it can still be thought of as an infinite sum of rational numbers, as Rafid indicated.
– Paul
Mar 4 '12 at 15:00
@RahulNarain: In the decimal expressions of rational numbers, there is always either a terminating or an infinitely repeating string of digits. Irrational numbers have neither of these properties, but can still be expressed as an infinitely long non-repeating sequence digits. Though there may not be an explicit formula for this sequence, it can still be thought of as an infinite sum of rational numbers, as Rafid indicated.
– Paul
Mar 4 '12 at 15:00
1
1
@Paul, I was making a joke on the phrase "must have digits from 0 to 9" that appeared in the original version of the answer.
– Rahul
Mar 4 '12 at 22:31
@Paul, I was making a joke on the phrase "must have digits from 0 to 9" that appeared in the original version of the answer.
– Rahul
Mar 4 '12 at 22:31
add a comment |
Besides that. Given a sequence of positive rational numbers such that their sum converges and such that $a_n > sum_{k = n + 1}^infty a_k$. Then choosing a $pm$ sign for each term of the sequence gives a new convergent series, each to a different number. By a countable-uncountable argument you get nonnumerable examples of that kind of series :).
add a comment |
Besides that. Given a sequence of positive rational numbers such that their sum converges and such that $a_n > sum_{k = n + 1}^infty a_k$. Then choosing a $pm$ sign for each term of the sequence gives a new convergent series, each to a different number. By a countable-uncountable argument you get nonnumerable examples of that kind of series :).
add a comment |
Besides that. Given a sequence of positive rational numbers such that their sum converges and such that $a_n > sum_{k = n + 1}^infty a_k$. Then choosing a $pm$ sign for each term of the sequence gives a new convergent series, each to a different number. By a countable-uncountable argument you get nonnumerable examples of that kind of series :).
Besides that. Given a sequence of positive rational numbers such that their sum converges and such that $a_n > sum_{k = n + 1}^infty a_k$. Then choosing a $pm$ sign for each term of the sequence gives a new convergent series, each to a different number. By a countable-uncountable argument you get nonnumerable examples of that kind of series :).
answered Mar 4 '12 at 12:45
student
411
411
add a comment |
add a comment |
For example look at
$$e = sum_{k=0}^{infty}{frac{1}{k!}} $$
add a comment |
For example look at
$$e = sum_{k=0}^{infty}{frac{1}{k!}} $$
add a comment |
For example look at
$$e = sum_{k=0}^{infty}{frac{1}{k!}} $$
For example look at
$$e = sum_{k=0}^{infty}{frac{1}{k!}} $$
answered Aug 2 '15 at 7:43
user210387
add a comment |
add a comment |
This has to see with the rate at which the sum/series converges to its limit and the Roth-Thue-Siegel theorem which allows you to use the rate of convergence to decide if the limit is rational or not.
Maybe Emile (the OP) meant to ask something of this sort (please let me know, Emile): why do some (convergent, of course) infinite sums of rationals are rational and others are irrational?
this is a different question..
– AIB
Mar 6 '12 at 9:41
How so? The conditions/rate of convergence affect whether the limit is rational or not. Don't they?
– AQP
Mar 6 '12 at 16:54
add a comment |
This has to see with the rate at which the sum/series converges to its limit and the Roth-Thue-Siegel theorem which allows you to use the rate of convergence to decide if the limit is rational or not.
Maybe Emile (the OP) meant to ask something of this sort (please let me know, Emile): why do some (convergent, of course) infinite sums of rationals are rational and others are irrational?
this is a different question..
– AIB
Mar 6 '12 at 9:41
How so? The conditions/rate of convergence affect whether the limit is rational or not. Don't they?
– AQP
Mar 6 '12 at 16:54
add a comment |
This has to see with the rate at which the sum/series converges to its limit and the Roth-Thue-Siegel theorem which allows you to use the rate of convergence to decide if the limit is rational or not.
Maybe Emile (the OP) meant to ask something of this sort (please let me know, Emile): why do some (convergent, of course) infinite sums of rationals are rational and others are irrational?
This has to see with the rate at which the sum/series converges to its limit and the Roth-Thue-Siegel theorem which allows you to use the rate of convergence to decide if the limit is rational or not.
Maybe Emile (the OP) meant to ask something of this sort (please let me know, Emile): why do some (convergent, of course) infinite sums of rationals are rational and others are irrational?
edited Mar 4 '12 at 10:25
g.castro
37218
37218
answered Mar 4 '12 at 10:09
AQP
12816
12816
this is a different question..
– AIB
Mar 6 '12 at 9:41
How so? The conditions/rate of convergence affect whether the limit is rational or not. Don't they?
– AQP
Mar 6 '12 at 16:54
add a comment |
this is a different question..
– AIB
Mar 6 '12 at 9:41
How so? The conditions/rate of convergence affect whether the limit is rational or not. Don't they?
– AQP
Mar 6 '12 at 16:54
this is a different question..
– AIB
Mar 6 '12 at 9:41
this is a different question..
– AIB
Mar 6 '12 at 9:41
How so? The conditions/rate of convergence affect whether the limit is rational or not. Don't they?
– AQP
Mar 6 '12 at 16:54
How so? The conditions/rate of convergence affect whether the limit is rational or not. Don't they?
– AQP
Mar 6 '12 at 16:54
add a comment |
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88
Every real number is the sum of countably many rational numbers.
– Did
Mar 4 '12 at 9:47
54
Properties that are preserved under a finite number of operations are not necessarily preserved under an infinite number of them.
– Rahul
Mar 4 '12 at 9:57
6
Real numbers have the property that the bounded infinite sum of positive real numbers is a real number because of completeness. This is not a property of rational numbers, so I would intuitively expect some bounded infinite sums of positive rational numbers not to be a rational number.
– Henry
Mar 4 '12 at 10:15
4
because real number are closure of rational numbers.
– quartz
Mar 4 '12 at 12:12
3
Another example to illustrate Rahul's point: the product of infinite positive reals can yield zero: $prod_{n=1}^infty 1/n = 0$ or more simply $prod_{n=1}^infty 0.5 = 0$
– leonbloy
Apr 30 '12 at 18:36