Is the sum of all rational numbers between two integers infinity
If there are infinite numbers between two rational numbers then would that entail that the sum of all numbers, say between 1 and 2, be infinity?
I believe that this cannot be true and has to do something with area under a curve?
calculus integration
asked Nov 26 at 1:18
John Rawls
1,253519
|
show 3 more comments
If there are infinite numbers between two rational numbers then would that entail that the sum of all numbers, say between 1 and 2, be infinity?
I believe that this cannot be true and has to do something with area under a curve?
calculus integration
asked Nov 26 at 1:18
John Rawls
1,253519
2
No, you cannot conclude from the fact that there are infinitely many rationals that the sum must be infinite. This is the error behind Zeno's paradox of the Achilles (e.g., $sum frac{1}{2^n}$ is finite, even though you are adding infinitely many positive rational numbers).
– Arturo Magidin
Nov 26 at 1:21
4
However, the sum is potentially infinite anyway; certainly it is infinite for all rationals between $1$ and $2$, since this includes countably many rationals that are greater than or equal to $1$, so the sum is strictly larger than any positive integer.
– Arturo Magidin
Nov 26 at 1:23
I was just about to type that last comment, @ArturoMagidin. Summing infinitely many numbers greater than $1$ must, almost by definition, be infinite.
– The Count
Nov 26 at 1:24
1
And the rationals between $0$ and $1$ include all rationals of the form $frac{1}{n}$, and since the harmonic series diverges... so, the answer is that the sum of all rationals between two integers is always infinite, but not for the reason you provide; and the sum of all rationals between any two rationals should also be infinite, verifiable by using a linear transformation to take, say, $[1,2]$ to $[q_1,q_2]$ in a way that maps rationals to rationals, and use it to put lower bounds on the sum.
– Arturo Magidin
Nov 26 at 1:27
You want to get into thorny matters? Ask whether the sum of all rational numbers between $-1$ and $1$ is infinite.
– Brian Tung
Nov 26 at 1:48
|
show 3 more comments
If there are infinite numbers between two rational numbers then would that entail that the sum of all numbers, say between 1 and 2, be infinity?
I believe that this cannot be true and has to do something with area under a curve?
calculus integration
asked Nov 26 at 1:18
John Rawls
1,253519
If there are infinite numbers between two rational numbers then would that entail that the sum of all numbers, say between 1 and 2, be infinity?
I believe that this cannot be true and has to do something with area under a curve?
calculus integration
calculus integration
asked Nov 26 at 1:18
John Rawls
1,253519
asked Nov 26 at 1:18
John Rawls
1,253519
asked Nov 26 at 1:18
John Rawls
1,253519
asked Nov 26 at 1:18
John Rawls
1,253519
asked Nov 26 at 1:18
John Rawls
1,253519
1,253519
2
No, you cannot conclude from the fact that there are infinitely many rationals that the sum must be infinite. This is the error behind Zeno's paradox of the Achilles (e.g., $sum frac{1}{2^n}$ is finite, even though you are adding infinitely many positive rational numbers).
– Arturo Magidin
Nov 26 at 1:21
4
However, the sum is potentially infinite anyway; certainly it is infinite for all rationals between $1$ and $2$, since this includes countably many rationals that are greater than or equal to $1$, so the sum is strictly larger than any positive integer.
– Arturo Magidin
Nov 26 at 1:23
I was just about to type that last comment, @ArturoMagidin. Summing infinitely many numbers greater than $1$ must, almost by definition, be infinite.
– The Count
Nov 26 at 1:24
1
And the rationals between $0$ and $1$ include all rationals of the form $frac{1}{n}$, and since the harmonic series diverges... so, the answer is that the sum of all rationals between two integers is always infinite, but not for the reason you provide; and the sum of all rationals between any two rationals should also be infinite, verifiable by using a linear transformation to take, say, $[1,2]$ to $[q_1,q_2]$ in a way that maps rationals to rationals, and use it to put lower bounds on the sum.
– Arturo Magidin
Nov 26 at 1:27
You want to get into thorny matters? Ask whether the sum of all rational numbers between $-1$ and $1$ is infinite.
– Brian Tung
Nov 26 at 1:48
|
show 3 more comments
2
No, you cannot conclude from the fact that there are infinitely many rationals that the sum must be infinite. This is the error behind Zeno's paradox of the Achilles (e.g., $sum frac{1}{2^n}$ is finite, even though you are adding infinitely many positive rational numbers).
– Arturo Magidin
Nov 26 at 1:21
4
However, the sum is potentially infinite anyway; certainly it is infinite for all rationals between $1$ and $2$, since this includes countably many rationals that are greater than or equal to $1$, so the sum is strictly larger than any positive integer.
– Arturo Magidin
Nov 26 at 1:23
I was just about to type that last comment, @ArturoMagidin. Summing infinitely many numbers greater than $1$ must, almost by definition, be infinite.
– The Count
Nov 26 at 1:24
1
And the rationals between $0$ and $1$ include all rationals of the form $frac{1}{n}$, and since the harmonic series diverges... so, the answer is that the sum of all rationals between two integers is always infinite, but not for the reason you provide; and the sum of all rationals between any two rationals should also be infinite, verifiable by using a linear transformation to take, say, $[1,2]$ to $[q_1,q_2]$ in a way that maps rationals to rationals, and use it to put lower bounds on the sum.
– Arturo Magidin
Nov 26 at 1:27
You want to get into thorny matters? Ask whether the sum of all rational numbers between $-1$ and $1$ is infinite.
– Brian Tung
Nov 26 at 1:48
2
2
No, you cannot conclude from the fact that there are infinitely many rationals that the sum must be infinite. This is the error behind Zeno's paradox of the Achilles (e.g., $sum frac{1}{2^n}$ is finite, even though you are adding infinitely many positive rational numbers).
– Arturo Magidin
Nov 26 at 1:21
No, you cannot conclude from the fact that there are infinitely many rationals that the sum must be infinite. This is the error behind Zeno's paradox of the Achilles (e.g., $sum frac{1}{2^n}$ is finite, even though you are adding infinitely many positive rational numbers).
– Arturo Magidin
Nov 26 at 1:21
4
4
However, the sum is potentially infinite anyway; certainly it is infinite for all rationals between $1$ and $2$, since this includes countably many rationals that are greater than or equal to $1$, so the sum is strictly larger than any positive integer.
– Arturo Magidin
Nov 26 at 1:23
However, the sum is potentially infinite anyway; certainly it is infinite for all rationals between $1$ and $2$, since this includes countably many rationals that are greater than or equal to $1$, so the sum is strictly larger than any positive integer.
– Arturo Magidin
Nov 26 at 1:23
I was just about to type that last comment, @ArturoMagidin. Summing infinitely many numbers greater than $1$ must, almost by definition, be infinite.
– The Count
Nov 26 at 1:24
I was just about to type that last comment, @ArturoMagidin. Summing infinitely many numbers greater than $1$ must, almost by definition, be infinite.
– The Count
Nov 26 at 1:24
1
1
And the rationals between $0$ and $1$ include all rationals of the form $frac{1}{n}$, and since the harmonic series diverges... so, the answer is that the sum of all rationals between two integers is always infinite, but not for the reason you provide; and the sum of all rationals between any two rationals should also be infinite, verifiable by using a linear transformation to take, say, $[1,2]$ to $[q_1,q_2]$ in a way that maps rationals to rationals, and use it to put lower bounds on the sum.
– Arturo Magidin
Nov 26 at 1:27
And the rationals between $0$ and $1$ include all rationals of the form $frac{1}{n}$, and since the harmonic series diverges... so, the answer is that the sum of all rationals between two integers is always infinite, but not for the reason you provide; and the sum of all rationals between any two rationals should also be infinite, verifiable by using a linear transformation to take, say, $[1,2]$ to $[q_1,q_2]$ in a way that maps rationals to rationals, and use it to put lower bounds on the sum.
– Arturo Magidin
Nov 26 at 1:27
You want to get into thorny matters? Ask whether the sum of all rational numbers between $-1$ and $1$ is infinite.
– Brian Tung
Nov 26 at 1:48
You want to get into thorny matters? Ask whether the sum of all rational numbers between $-1$ and $1$ is infinite.
– Brian Tung
Nov 26 at 1:48
|
show 3 more comments
1 Answer
1
active
oldest
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Before you can answer this question you need a definition for what it means to "add up infinitely many numbers".
For the moment, assume those numbers are listed in some order:
$$
a_1, a_2, ldots .
$$
Then mathematicians define the infinite sum to be the limit (if it exists) of the numbers
$$
a_1, quad a_1 + a_2, quad a_1 + a_2 + a_3, ldots .
$$
Then, for example, you could show that the "infinite sum"
$$
1 +
frac{1}{2} +
frac{1}{4} +
frac{1}{8} + cdots
$$
is $2$. So sometimes the sum of infinitely many numbers is finite.
With that definition, you clearly can't sum all the rational numbers between $1$ and $2$ since all of them are greater than $1$, so those partial sums will grow without bound. You can't sum all the rational numbers between $0$ and $0.0001$ since infinitely many of them are greater than $0.00001$. So you are more or less correct - mathematicians prefer to say you can't sum them, not that the sum is infinity.
When you learn calculus (that "something with area under a curve") you will see how to add up more and more and more pieces without growing to infinity because at each stage the pieces are smaller and smaller and smaller.
answered Nov 26 at 1:31
Ethan Bolker
41k546108
"You can't sum all the numbers between [...]" See that OP asks for all the rational numbers. It doesn't seems obvious to me that your proof holds...
– rafa11111
Nov 26 at 1:35
1
@rafa11111 There are infinitely many rational numbers in the interval that are larger than the midpoint. I edited the answer.
– Ethan Bolker
Nov 26 at 1:42
Typo: $1+frac12+frac14+frac18+cdots = 2$, not $1$.
– Brian Tung
Nov 26 at 1:47
@BrianTung I fixed it thanks. You could have.
– Ethan Bolker
Nov 26 at 1:48
1
@EthanBolker: I know. I like to allow the answerer to fix it first in the way they think best. If you hadn't come back to fix it, I'd have done it eventually.
– Brian Tung
Nov 26 at 1:49
|
show 2 more comments
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1 Answer
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votes
Before you can answer this question you need a definition for what it means to "add up infinitely many numbers".
For the moment, assume those numbers are listed in some order:
$$
a_1, a_2, ldots .
$$
Then mathematicians define the infinite sum to be the limit (if it exists) of the numbers
$$
a_1, quad a_1 + a_2, quad a_1 + a_2 + a_3, ldots .
$$
Then, for example, you could show that the "infinite sum"
$$
1 +
frac{1}{2} +
frac{1}{4} +
frac{1}{8} + cdots
$$
is $2$. So sometimes the sum of infinitely many numbers is finite.
With that definition, you clearly can't sum all the rational numbers between $1$ and $2$ since all of them are greater than $1$, so those partial sums will grow without bound. You can't sum all the rational numbers between $0$ and $0.0001$ since infinitely many of them are greater than $0.00001$. So you are more or less correct - mathematicians prefer to say you can't sum them, not that the sum is infinity.
When you learn calculus (that "something with area under a curve") you will see how to add up more and more and more pieces without growing to infinity because at each stage the pieces are smaller and smaller and smaller.
answered Nov 26 at 1:31
Ethan Bolker
41k546108
"You can't sum all the numbers between [...]" See that OP asks for all the rational numbers. It doesn't seems obvious to me that your proof holds...
– rafa11111
Nov 26 at 1:35
1
@rafa11111 There are infinitely many rational numbers in the interval that are larger than the midpoint. I edited the answer.
– Ethan Bolker
Nov 26 at 1:42
Typo: $1+frac12+frac14+frac18+cdots = 2$, not $1$.
– Brian Tung
Nov 26 at 1:47
@BrianTung I fixed it thanks. You could have.
– Ethan Bolker
Nov 26 at 1:48
1
@EthanBolker: I know. I like to allow the answerer to fix it first in the way they think best. If you hadn't come back to fix it, I'd have done it eventually.
– Brian Tung
Nov 26 at 1:49
|
show 2 more comments
Before you can answer this question you need a definition for what it means to "add up infinitely many numbers".
For the moment, assume those numbers are listed in some order:
$$
a_1, a_2, ldots .
$$
Then mathematicians define the infinite sum to be the limit (if it exists) of the numbers
$$
a_1, quad a_1 + a_2, quad a_1 + a_2 + a_3, ldots .
$$
Then, for example, you could show that the "infinite sum"
$$
1 +
frac{1}{2} +
frac{1}{4} +
frac{1}{8} + cdots
$$
is $2$. So sometimes the sum of infinitely many numbers is finite.
With that definition, you clearly can't sum all the rational numbers between $1$ and $2$ since all of them are greater than $1$, so those partial sums will grow without bound. You can't sum all the rational numbers between $0$ and $0.0001$ since infinitely many of them are greater than $0.00001$. So you are more or less correct - mathematicians prefer to say you can't sum them, not that the sum is infinity.
When you learn calculus (that "something with area under a curve") you will see how to add up more and more and more pieces without growing to infinity because at each stage the pieces are smaller and smaller and smaller.
answered Nov 26 at 1:31
Ethan Bolker
41k546108
"You can't sum all the numbers between [...]" See that OP asks for all the rational numbers. It doesn't seems obvious to me that your proof holds...
– rafa11111
Nov 26 at 1:35
1
@rafa11111 There are infinitely many rational numbers in the interval that are larger than the midpoint. I edited the answer.
– Ethan Bolker
Nov 26 at 1:42
Typo: $1+frac12+frac14+frac18+cdots = 2$, not $1$.
– Brian Tung
Nov 26 at 1:47
@BrianTung I fixed it thanks. You could have.
– Ethan Bolker
Nov 26 at 1:48
1
@EthanBolker: I know. I like to allow the answerer to fix it first in the way they think best. If you hadn't come back to fix it, I'd have done it eventually.
– Brian Tung
Nov 26 at 1:49
|
show 2 more comments
Before you can answer this question you need a definition for what it means to "add up infinitely many numbers".
For the moment, assume those numbers are listed in some order:
$$
a_1, a_2, ldots .
$$
Then mathematicians define the infinite sum to be the limit (if it exists) of the numbers
$$
a_1, quad a_1 + a_2, quad a_1 + a_2 + a_3, ldots .
$$
Then, for example, you could show that the "infinite sum"
$$
1 +
frac{1}{2} +
frac{1}{4} +
frac{1}{8} + cdots
$$
is $2$. So sometimes the sum of infinitely many numbers is finite.
With that definition, you clearly can't sum all the rational numbers between $1$ and $2$ since all of them are greater than $1$, so those partial sums will grow without bound. You can't sum all the rational numbers between $0$ and $0.0001$ since infinitely many of them are greater than $0.00001$. So you are more or less correct - mathematicians prefer to say you can't sum them, not that the sum is infinity.
When you learn calculus (that "something with area under a curve") you will see how to add up more and more and more pieces without growing to infinity because at each stage the pieces are smaller and smaller and smaller.
answered Nov 26 at 1:31
Ethan Bolker
41k546108
Before you can answer this question you need a definition for what it means to "add up infinitely many numbers".
For the moment, assume those numbers are listed in some order:
$$
a_1, a_2, ldots .
$$
Then mathematicians define the infinite sum to be the limit (if it exists) of the numbers
$$
a_1, quad a_1 + a_2, quad a_1 + a_2 + a_3, ldots .
$$
Then, for example, you could show that the "infinite sum"
$$
1 +
frac{1}{2} +
frac{1}{4} +
frac{1}{8} + cdots
$$
is $2$. So sometimes the sum of infinitely many numbers is finite.
With that definition, you clearly can't sum all the rational numbers between $1$ and $2$ since all of them are greater than $1$, so those partial sums will grow without bound. You can't sum all the rational numbers between $0$ and $0.0001$ since infinitely many of them are greater than $0.00001$. So you are more or less correct - mathematicians prefer to say you can't sum them, not that the sum is infinity.
When you learn calculus (that "something with area under a curve") you will see how to add up more and more and more pieces without growing to infinity because at each stage the pieces are smaller and smaller and smaller.
answered Nov 26 at 1:31
Ethan Bolker
41k546108
edited Nov 26 at 1:48
answered Nov 26 at 1:31
Ethan Bolker
41k546108
answered Nov 26 at 1:31
Ethan Bolker
41k546108
answered Nov 26 at 1:31
Ethan Bolker
41k546108
41k546108
"You can't sum all the numbers between [...]" See that OP asks for all the rational numbers. It doesn't seems obvious to me that your proof holds...
– rafa11111
Nov 26 at 1:35
1
@rafa11111 There are infinitely many rational numbers in the interval that are larger than the midpoint. I edited the answer.
– Ethan Bolker
Nov 26 at 1:42
Typo: $1+frac12+frac14+frac18+cdots = 2$, not $1$.
– Brian Tung
Nov 26 at 1:47
@BrianTung I fixed it thanks. You could have.
– Ethan Bolker
Nov 26 at 1:48
1
@EthanBolker: I know. I like to allow the answerer to fix it first in the way they think best. If you hadn't come back to fix it, I'd have done it eventually.
– Brian Tung
Nov 26 at 1:49
|
show 2 more comments
"You can't sum all the numbers between [...]" See that OP asks for all the rational numbers. It doesn't seems obvious to me that your proof holds...
– rafa11111
Nov 26 at 1:35
1
@rafa11111 There are infinitely many rational numbers in the interval that are larger than the midpoint. I edited the answer.
– Ethan Bolker
Nov 26 at 1:42
Typo: $1+frac12+frac14+frac18+cdots = 2$, not $1$.
– Brian Tung
Nov 26 at 1:47
@BrianTung I fixed it thanks. You could have.
– Ethan Bolker
Nov 26 at 1:48
1
@EthanBolker: I know. I like to allow the answerer to fix it first in the way they think best. If you hadn't come back to fix it, I'd have done it eventually.
– Brian Tung
Nov 26 at 1:49
"You can't sum all the numbers between [...]" See that OP asks for all the rational numbers. It doesn't seems obvious to me that your proof holds...
– rafa11111
Nov 26 at 1:35
"You can't sum all the numbers between [...]" See that OP asks for all the rational numbers. It doesn't seems obvious to me that your proof holds...
– rafa11111
Nov 26 at 1:35
1
1
@rafa11111 There are infinitely many rational numbers in the interval that are larger than the midpoint. I edited the answer.
– Ethan Bolker
Nov 26 at 1:42
@rafa11111 There are infinitely many rational numbers in the interval that are larger than the midpoint. I edited the answer.
– Ethan Bolker
Nov 26 at 1:42
Typo: $1+frac12+frac14+frac18+cdots = 2$, not $1$.
– Brian Tung
Nov 26 at 1:47
Typo: $1+frac12+frac14+frac18+cdots = 2$, not $1$.
– Brian Tung
Nov 26 at 1:47
@BrianTung I fixed it thanks. You could have.
– Ethan Bolker
Nov 26 at 1:48
@BrianTung I fixed it thanks. You could have.
– Ethan Bolker
Nov 26 at 1:48
1
1
@EthanBolker: I know. I like to allow the answerer to fix it first in the way they think best. If you hadn't come back to fix it, I'd have done it eventually.
– Brian Tung
Nov 26 at 1:49
@EthanBolker: I know. I like to allow the answerer to fix it first in the way they think best. If you hadn't come back to fix it, I'd have done it eventually.
– Brian Tung
Nov 26 at 1:49
|
show 2 more comments
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2
No, you cannot conclude from the fact that there are infinitely many rationals that the sum must be infinite. This is the error behind Zeno's paradox of the Achilles (e.g., $sum frac{1}{2^n}$ is finite, even though you are adding infinitely many positive rational numbers).
– Arturo Magidin
Nov 26 at 1:21
4
However, the sum is potentially infinite anyway; certainly it is infinite for all rationals between $1$ and $2$, since this includes countably many rationals that are greater than or equal to $1$, so the sum is strictly larger than any positive integer.
– Arturo Magidin
Nov 26 at 1:23
I was just about to type that last comment, @ArturoMagidin. Summing infinitely many numbers greater than $1$ must, almost by definition, be infinite.
– The Count
Nov 26 at 1:24
1
And the rationals between $0$ and $1$ include all rationals of the form $frac{1}{n}$, and since the harmonic series diverges... so, the answer is that the sum of all rationals between two integers is always infinite, but not for the reason you provide; and the sum of all rationals between any two rationals should also be infinite, verifiable by using a linear transformation to take, say, $[1,2]$ to $[q_1,q_2]$ in a way that maps rationals to rationals, and use it to put lower bounds on the sum.
– Arturo Magidin
Nov 26 at 1:27
You want to get into thorny matters? Ask whether the sum of all rational numbers between $-1$ and $1$ is infinite.
– Brian Tung
Nov 26 at 1:48