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How does this series yield an irrational function [duplicate]
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This question already has an answer here:
How can adding an infinite number of rationals yield an irrational number?
7 answers
It is well known that the euler number $e$ is irrational.
It is also well known that the Taylor expansion of $e$ can be represented as
$$e=sum_k=0^inftyfrac1k!$$
Now, when we look at the term $$T(k)=frac1k!quadforall kinmathbbZ$$
We realize that $T(k)$ has to be a rational number for all $k$. How can a sum of rational numbers yield an irrational number $e$? I am not a mathematician, hence all and any help is appreciated.
number-theory elementary-number-theory taylor-expansion
asked Jul 26 at 20:38
ubuntu_noob
30718
marked as duplicate by M. Winter, Arthur, Clement C., Clayton, Math Lover Jul 26 at 22:30
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |Â
up vote
1
down vote
favorite
This question already has an answer here:
How can adding an infinite number of rationals yield an irrational number?
7 answers
It is well known that the euler number $e$ is irrational.
It is also well known that the Taylor expansion of $e$ can be represented as
$$e=sum_k=0^inftyfrac1k!$$
Now, when we look at the term $$T(k)=frac1k!quadforall kinmathbbZ$$
We realize that $T(k)$ has to be a rational number for all $k$. How can a sum of rational numbers yield an irrational number $e$? I am not a mathematician, hence all and any help is appreciated.
number-theory elementary-number-theory taylor-expansion
asked Jul 26 at 20:38
ubuntu_noob
30718
marked as duplicate by M. Winter, Arthur, Clement C., Clayton, Math Lover Jul 26 at 22:30
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
3
Any number can be written as a (possibly non terminating) decimal. An irrational number is a number.
– copper.hat
Jul 26 at 20:39
1
"How can a sum of rational numbers yield an irrational number $e$?" Because you add together infinitely many of these rational numbers, and infinities tend to ruin intuition. In fact, one common way to formally define the real numbers is (vaguely speaking) as any possible limit of sequences of rational numbers.
– Arthur
Jul 26 at 20:42
Every number, rational or irrational, has a decimal expansion. A decimal expansion is an infinite series of rational terms. It's the infinite number of terms in the series that is a necessary (but not sufficient) condition for the result to be irrational. It is not sufficient e.g because there are rationals with infinite (repeating) decimal expansions.
– NickD
Jul 26 at 20:42
Spivak's calculus book (not the manifolds book) has a great elementary proof.
– Michael Burr
Jul 26 at 20:46
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
This question already has an answer here:
How can adding an infinite number of rationals yield an irrational number?
7 answers
It is well known that the euler number $e$ is irrational.
It is also well known that the Taylor expansion of $e$ can be represented as
$$e=sum_k=0^inftyfrac1k!$$
Now, when we look at the term $$T(k)=frac1k!quadforall kinmathbbZ$$
We realize that $T(k)$ has to be a rational number for all $k$. How can a sum of rational numbers yield an irrational number $e$? I am not a mathematician, hence all and any help is appreciated.
number-theory elementary-number-theory taylor-expansion
asked Jul 26 at 20:38
ubuntu_noob
30718
This question already has an answer here:
How can adding an infinite number of rationals yield an irrational number?
7 answers
It is well known that the euler number $e$ is irrational.
It is also well known that the Taylor expansion of $e$ can be represented as
$$e=sum_k=0^inftyfrac1k!$$
Now, when we look at the term $$T(k)=frac1k!quadforall kinmathbbZ$$
We realize that $T(k)$ has to be a rational number for all $k$. How can a sum of rational numbers yield an irrational number $e$? I am not a mathematician, hence all and any help is appreciated.
This question already has an answer here:
How can adding an infinite number of rationals yield an irrational number?
7 answers
number-theory elementary-number-theory taylor-expansion
asked Jul 26 at 20:38
ubuntu_noob
30718
asked Jul 26 at 20:38
ubuntu_noob
30718
asked Jul 26 at 20:38
ubuntu_noob
30718
asked Jul 26 at 20:38
ubuntu_noob
30718
30718
marked as duplicate by M. Winter, Arthur, Clement C., Clayton, Math Lover Jul 26 at 22:30
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by M. Winter, Arthur, Clement C., Clayton, Math Lover Jul 26 at 22:30
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
3
Any number can be written as a (possibly non terminating) decimal. An irrational number is a number.
– copper.hat
Jul 26 at 20:39
1
"How can a sum of rational numbers yield an irrational number $e$?" Because you add together infinitely many of these rational numbers, and infinities tend to ruin intuition. In fact, one common way to formally define the real numbers is (vaguely speaking) as any possible limit of sequences of rational numbers.
– Arthur
Jul 26 at 20:42
Every number, rational or irrational, has a decimal expansion. A decimal expansion is an infinite series of rational terms. It's the infinite number of terms in the series that is a necessary (but not sufficient) condition for the result to be irrational. It is not sufficient e.g because there are rationals with infinite (repeating) decimal expansions.
– NickD
Jul 26 at 20:42
Spivak's calculus book (not the manifolds book) has a great elementary proof.
– Michael Burr
Jul 26 at 20:46
add a comment |Â
3
Any number can be written as a (possibly non terminating) decimal. An irrational number is a number.
– copper.hat
Jul 26 at 20:39
1
"How can a sum of rational numbers yield an irrational number $e$?" Because you add together infinitely many of these rational numbers, and infinities tend to ruin intuition. In fact, one common way to formally define the real numbers is (vaguely speaking) as any possible limit of sequences of rational numbers.
– Arthur
Jul 26 at 20:42
Every number, rational or irrational, has a decimal expansion. A decimal expansion is an infinite series of rational terms. It's the infinite number of terms in the series that is a necessary (but not sufficient) condition for the result to be irrational. It is not sufficient e.g because there are rationals with infinite (repeating) decimal expansions.
– NickD
Jul 26 at 20:42
Spivak's calculus book (not the manifolds book) has a great elementary proof.
– Michael Burr
Jul 26 at 20:46
3
3
Any number can be written as a (possibly non terminating) decimal. An irrational number is a number.
– copper.hat
Jul 26 at 20:39
Any number can be written as a (possibly non terminating) decimal. An irrational number is a number.
– copper.hat
Jul 26 at 20:39
1
1
"How can a sum of rational numbers yield an irrational number $e$?" Because you add together infinitely many of these rational numbers, and infinities tend to ruin intuition. In fact, one common way to formally define the real numbers is (vaguely speaking) as any possible limit of sequences of rational numbers.
– Arthur
Jul 26 at 20:42
"How can a sum of rational numbers yield an irrational number $e$?" Because you add together infinitely many of these rational numbers, and infinities tend to ruin intuition. In fact, one common way to formally define the real numbers is (vaguely speaking) as any possible limit of sequences of rational numbers.
– Arthur
Jul 26 at 20:42
Every number, rational or irrational, has a decimal expansion. A decimal expansion is an infinite series of rational terms. It's the infinite number of terms in the series that is a necessary (but not sufficient) condition for the result to be irrational. It is not sufficient e.g because there are rationals with infinite (repeating) decimal expansions.
– NickD
Jul 26 at 20:42
Every number, rational or irrational, has a decimal expansion. A decimal expansion is an infinite series of rational terms. It's the infinite number of terms in the series that is a necessary (but not sufficient) condition for the result to be irrational. It is not sufficient e.g because there are rationals with infinite (repeating) decimal expansions.
– NickD
Jul 26 at 20:42
Spivak's calculus book (not the manifolds book) has a great elementary proof.
– Michael Burr
Jul 26 at 20:46
Spivak's calculus book (not the manifolds book) has a great elementary proof.
– Michael Burr
Jul 26 at 20:46
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
2
down vote
accepted
Every irrational number is an infinite sum of rational numbers
For example $$sqrt 2 =1+.4+.01+.004+.....$$
The magic word is infinite sum.
answered Jul 26 at 20:45
Mohammad Riazi-Kermani
27.3k41851
A finite sum of rational numbers is always rational, but an infinite sum of rational numbers can be rational or irrational.
– James Arathoon
Jul 26 at 21:26
@JamesArathoon Sure, every real number is an infinite sum of rationals and an infinite sum of irrationals.
– Mohammad Riazi-Kermani
Jul 26 at 21:34
add a comment |Â
up vote
2
down vote
Take the square root of $2$, which is known to be irrational; it's roughly
$1.41428ldots$. I can write that as
$$
1 + frac410 + frac1100 + frac41000 + frac210000 + frac8100000 + ldots
$$
When I do so, you can see that each individual term is a rational number.
Now you might be thinking "but I know that when I add two rationals, I get a rational:"
$$
fracab + fraccd = fracad + bcbd
$$
... so why isn't the larger sum still rational?
The answer is that any finite part of it is rational. The first three terms, for instance, give the rational
$$
frac141100.
$$
But that doesn't mean that an infinite sum must be rational as well, and indeed, that's not true.
Let me work by analogy with another notion: any finite sum of numbers is finite. But an infinite sum of numbers is not necessarily finite, as
$$
1 + 1 + 1 + ldots
$$
shows.
So just because you've proven some property works for pairs of things or finite collections, you don't necessarily know that it works for infinite things.
That subtle fact is a large part of what the third portion of most serious calculus courses is all about -- the "sequences and series" part. So to really understand it, you've got some work to do, alas.
answered Jul 26 at 20:46
John Hughes
59.4k23785
Nice explained at the targeted recipient(s). (+1)
– Mark Viola
Jul 26 at 21:29
Thanks. I guess all those years of being a teacher have actually taught me something. :)
– John Hughes
Jul 26 at 21:30
The seven years I spent as a professor taught me that most students didn't really care about learning, rather they just wanted a "rubber stamp" in order to get a job. I left academics 22 years ago, but have friends, who are still professors, who have told me that there has been a markedly sharp decline in their student's skills/abilities/attentiveness/motivation etc. Have you seen this phenomenon ("The customer's mentality") at Brown yet?
– Mark Viola
Jul 26 at 21:39
Sure, we all claim to be seeing that, and maybe we really are. Then I look at something like this, and realize that I could be wrong: lunarbaboon.com/comics/youth.html
– John Hughes
Jul 26 at 22:59
add a comment |Â
up vote
1
down vote
If the sum was finite, then of course we could add all the rational numbers into a fraction with the result being rational. But since the sum has infinitely many terms, there is a possibility that the sum does not approach something we can represent as the ratio of two integers. In that case, it is irrational.
If I remember correctly, one of the classic proofs of the irrationality of e does not need more than a bit of calculus to follow.
answered Jul 26 at 20:47
El Gallo Negro
233
add a comment |Â
up vote
0
down vote
That's quite a normal situation. When you “sum a seriesâ€Â, you don't actually do a sum, but take a limit of finite sums:
$$
sum_n=0^infty a_n=lim_ntoinfty(a_0+a_1+dots+a_n)
$$
Even if each $a_n$ is rational, the limit is not a (finite) sum of rational numbers, so there's no compelling reason for it to be rational.
The Taylor-Lagrange theorem applied to the exponential function says that, for any integer $k$, there exists $c_kin(0,x)$ such that
$$
e=sum_n=0^kfrac1^nn!+frace^c_k(k+1)!
$$
because the $n$-th derivative of $e^x$ is $e^x$.
Suppose, to the contrary, that $e=a/b$, for some positive integers $a$ and $b$. Take $k>maxb,3$; then
$$
e=sum_n=0^kfrac1^nn!+frace^c_k(k+1)!<
sum_n=0^kfrac1^nn!+frac3(k+1)!
$$
because $c_k<1$ and $e<3$. Thus
$$
0<left(e-left(1+1+frac12!+dots+frac1k!right)k!right)<
frac3(k+1)!k!=frac3k+1
$$
However, the term
$$
e-left(1+1+frac12!+dots+frac1k!right)k!
$$
is integer by assumption, as $k!$ is a multiple of $b$. On the other hand
$$
frac3k+1<frac33+1=frac34
$$
which is a contradiction.
Thus $e$ is irrational and it's the very structure of
$$
e=sum_n=0^inftyfrac1n!
$$
that makes it so.
answered Jul 26 at 21:02
egreg
164k1180187
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Every irrational number is an infinite sum of rational numbers
For example $$sqrt 2 =1+.4+.01+.004+.....$$
The magic word is infinite sum.
answered Jul 26 at 20:45
Mohammad Riazi-Kermani
27.3k41851
A finite sum of rational numbers is always rational, but an infinite sum of rational numbers can be rational or irrational.
– James Arathoon
Jul 26 at 21:26
@JamesArathoon Sure, every real number is an infinite sum of rationals and an infinite sum of irrationals.
– Mohammad Riazi-Kermani
Jul 26 at 21:34
add a comment |Â
up vote
2
down vote
accepted
Every irrational number is an infinite sum of rational numbers
For example $$sqrt 2 =1+.4+.01+.004+.....$$
The magic word is infinite sum.
answered Jul 26 at 20:45
Mohammad Riazi-Kermani
27.3k41851
A finite sum of rational numbers is always rational, but an infinite sum of rational numbers can be rational or irrational.
– James Arathoon
Jul 26 at 21:26
@JamesArathoon Sure, every real number is an infinite sum of rationals and an infinite sum of irrationals.
– Mohammad Riazi-Kermani
Jul 26 at 21:34
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Every irrational number is an infinite sum of rational numbers
For example $$sqrt 2 =1+.4+.01+.004+.....$$
The magic word is infinite sum.
answered Jul 26 at 20:45
Mohammad Riazi-Kermani
27.3k41851
Every irrational number is an infinite sum of rational numbers
For example $$sqrt 2 =1+.4+.01+.004+.....$$
The magic word is infinite sum.
answered Jul 26 at 20:45
Mohammad Riazi-Kermani
27.3k41851
answered Jul 26 at 20:45
Mohammad Riazi-Kermani
27.3k41851
answered Jul 26 at 20:45
Mohammad Riazi-Kermani
27.3k41851
answered Jul 26 at 20:45
Mohammad Riazi-Kermani
27.3k41851
27.3k41851
A finite sum of rational numbers is always rational, but an infinite sum of rational numbers can be rational or irrational.
– James Arathoon
Jul 26 at 21:26
@JamesArathoon Sure, every real number is an infinite sum of rationals and an infinite sum of irrationals.
– Mohammad Riazi-Kermani
Jul 26 at 21:34
add a comment |Â
A finite sum of rational numbers is always rational, but an infinite sum of rational numbers can be rational or irrational.
– James Arathoon
Jul 26 at 21:26
@JamesArathoon Sure, every real number is an infinite sum of rationals and an infinite sum of irrationals.
– Mohammad Riazi-Kermani
Jul 26 at 21:34
A finite sum of rational numbers is always rational, but an infinite sum of rational numbers can be rational or irrational.
– James Arathoon
Jul 26 at 21:26
A finite sum of rational numbers is always rational, but an infinite sum of rational numbers can be rational or irrational.
– James Arathoon
Jul 26 at 21:26
@JamesArathoon Sure, every real number is an infinite sum of rationals and an infinite sum of irrationals.
– Mohammad Riazi-Kermani
Jul 26 at 21:34
@JamesArathoon Sure, every real number is an infinite sum of rationals and an infinite sum of irrationals.
– Mohammad Riazi-Kermani
Jul 26 at 21:34
add a comment |Â
up vote
2
down vote
Take the square root of $2$, which is known to be irrational; it's roughly
$1.41428ldots$. I can write that as
$$
1 + frac410 + frac1100 + frac41000 + frac210000 + frac8100000 + ldots
$$
When I do so, you can see that each individual term is a rational number.
Now you might be thinking "but I know that when I add two rationals, I get a rational:"
$$
fracab + fraccd = fracad + bcbd
$$
... so why isn't the larger sum still rational?
The answer is that any finite part of it is rational. The first three terms, for instance, give the rational
$$
frac141100.
$$
But that doesn't mean that an infinite sum must be rational as well, and indeed, that's not true.
Let me work by analogy with another notion: any finite sum of numbers is finite. But an infinite sum of numbers is not necessarily finite, as
$$
1 + 1 + 1 + ldots
$$
shows.
So just because you've proven some property works for pairs of things or finite collections, you don't necessarily know that it works for infinite things.
That subtle fact is a large part of what the third portion of most serious calculus courses is all about -- the "sequences and series" part. So to really understand it, you've got some work to do, alas.
answered Jul 26 at 20:46
John Hughes
59.4k23785
Nice explained at the targeted recipient(s). (+1)
– Mark Viola
Jul 26 at 21:29
Thanks. I guess all those years of being a teacher have actually taught me something. :)
– John Hughes
Jul 26 at 21:30
The seven years I spent as a professor taught me that most students didn't really care about learning, rather they just wanted a "rubber stamp" in order to get a job. I left academics 22 years ago, but have friends, who are still professors, who have told me that there has been a markedly sharp decline in their student's skills/abilities/attentiveness/motivation etc. Have you seen this phenomenon ("The customer's mentality") at Brown yet?
– Mark Viola
Jul 26 at 21:39
Sure, we all claim to be seeing that, and maybe we really are. Then I look at something like this, and realize that I could be wrong: lunarbaboon.com/comics/youth.html
– John Hughes
Jul 26 at 22:59
add a comment |Â
up vote
2
down vote
Take the square root of $2$, which is known to be irrational; it's roughly
$1.41428ldots$. I can write that as
$$
1 + frac410 + frac1100 + frac41000 + frac210000 + frac8100000 + ldots
$$
When I do so, you can see that each individual term is a rational number.
Now you might be thinking "but I know that when I add two rationals, I get a rational:"
$$
fracab + fraccd = fracad + bcbd
$$
... so why isn't the larger sum still rational?
The answer is that any finite part of it is rational. The first three terms, for instance, give the rational
$$
frac141100.
$$
But that doesn't mean that an infinite sum must be rational as well, and indeed, that's not true.
Let me work by analogy with another notion: any finite sum of numbers is finite. But an infinite sum of numbers is not necessarily finite, as
$$
1 + 1 + 1 + ldots
$$
shows.
So just because you've proven some property works for pairs of things or finite collections, you don't necessarily know that it works for infinite things.
That subtle fact is a large part of what the third portion of most serious calculus courses is all about -- the "sequences and series" part. So to really understand it, you've got some work to do, alas.
answered Jul 26 at 20:46
John Hughes
59.4k23785
Nice explained at the targeted recipient(s). (+1)
– Mark Viola
Jul 26 at 21:29
Thanks. I guess all those years of being a teacher have actually taught me something. :)
– John Hughes
Jul 26 at 21:30
The seven years I spent as a professor taught me that most students didn't really care about learning, rather they just wanted a "rubber stamp" in order to get a job. I left academics 22 years ago, but have friends, who are still professors, who have told me that there has been a markedly sharp decline in their student's skills/abilities/attentiveness/motivation etc. Have you seen this phenomenon ("The customer's mentality") at Brown yet?
– Mark Viola
Jul 26 at 21:39
Sure, we all claim to be seeing that, and maybe we really are. Then I look at something like this, and realize that I could be wrong: lunarbaboon.com/comics/youth.html
– John Hughes
Jul 26 at 22:59
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Take the square root of $2$, which is known to be irrational; it's roughly
$1.41428ldots$. I can write that as
$$
1 + frac410 + frac1100 + frac41000 + frac210000 + frac8100000 + ldots
$$
When I do so, you can see that each individual term is a rational number.
Now you might be thinking "but I know that when I add two rationals, I get a rational:"
$$
fracab + fraccd = fracad + bcbd
$$
... so why isn't the larger sum still rational?
The answer is that any finite part of it is rational. The first three terms, for instance, give the rational
$$
frac141100.
$$
But that doesn't mean that an infinite sum must be rational as well, and indeed, that's not true.
Let me work by analogy with another notion: any finite sum of numbers is finite. But an infinite sum of numbers is not necessarily finite, as
$$
1 + 1 + 1 + ldots
$$
shows.
So just because you've proven some property works for pairs of things or finite collections, you don't necessarily know that it works for infinite things.
That subtle fact is a large part of what the third portion of most serious calculus courses is all about -- the "sequences and series" part. So to really understand it, you've got some work to do, alas.
answered Jul 26 at 20:46
John Hughes
59.4k23785
Take the square root of $2$, which is known to be irrational; it's roughly
$1.41428ldots$. I can write that as
$$
1 + frac410 + frac1100 + frac41000 + frac210000 + frac8100000 + ldots
$$
When I do so, you can see that each individual term is a rational number.
Now you might be thinking "but I know that when I add two rationals, I get a rational:"
$$
fracab + fraccd = fracad + bcbd
$$
... so why isn't the larger sum still rational?
The answer is that any finite part of it is rational. The first three terms, for instance, give the rational
$$
frac141100.
$$
But that doesn't mean that an infinite sum must be rational as well, and indeed, that's not true.
Let me work by analogy with another notion: any finite sum of numbers is finite. But an infinite sum of numbers is not necessarily finite, as
$$
1 + 1 + 1 + ldots
$$
shows.
So just because you've proven some property works for pairs of things or finite collections, you don't necessarily know that it works for infinite things.
That subtle fact is a large part of what the third portion of most serious calculus courses is all about -- the "sequences and series" part. So to really understand it, you've got some work to do, alas.
answered Jul 26 at 20:46
John Hughes
59.4k23785
answered Jul 26 at 20:46
John Hughes
59.4k23785
answered Jul 26 at 20:46
John Hughes
59.4k23785
answered Jul 26 at 20:46
John Hughes
59.4k23785
59.4k23785
Nice explained at the targeted recipient(s). (+1)
– Mark Viola
Jul 26 at 21:29
Thanks. I guess all those years of being a teacher have actually taught me something. :)
– John Hughes
Jul 26 at 21:30
The seven years I spent as a professor taught me that most students didn't really care about learning, rather they just wanted a "rubber stamp" in order to get a job. I left academics 22 years ago, but have friends, who are still professors, who have told me that there has been a markedly sharp decline in their student's skills/abilities/attentiveness/motivation etc. Have you seen this phenomenon ("The customer's mentality") at Brown yet?
– Mark Viola
Jul 26 at 21:39
Sure, we all claim to be seeing that, and maybe we really are. Then I look at something like this, and realize that I could be wrong: lunarbaboon.com/comics/youth.html
– John Hughes
Jul 26 at 22:59
add a comment |Â
Nice explained at the targeted recipient(s). (+1)
– Mark Viola
Jul 26 at 21:29
Thanks. I guess all those years of being a teacher have actually taught me something. :)
– John Hughes
Jul 26 at 21:30
The seven years I spent as a professor taught me that most students didn't really care about learning, rather they just wanted a "rubber stamp" in order to get a job. I left academics 22 years ago, but have friends, who are still professors, who have told me that there has been a markedly sharp decline in their student's skills/abilities/attentiveness/motivation etc. Have you seen this phenomenon ("The customer's mentality") at Brown yet?
– Mark Viola
Jul 26 at 21:39
Sure, we all claim to be seeing that, and maybe we really are. Then I look at something like this, and realize that I could be wrong: lunarbaboon.com/comics/youth.html
– John Hughes
Jul 26 at 22:59
Nice explained at the targeted recipient(s). (+1)
– Mark Viola
Jul 26 at 21:29
Nice explained at the targeted recipient(s). (+1)
– Mark Viola
Jul 26 at 21:29
Thanks. I guess all those years of being a teacher have actually taught me something. :)
– John Hughes
Jul 26 at 21:30
Thanks. I guess all those years of being a teacher have actually taught me something. :)
– John Hughes
Jul 26 at 21:30
The seven years I spent as a professor taught me that most students didn't really care about learning, rather they just wanted a "rubber stamp" in order to get a job. I left academics 22 years ago, but have friends, who are still professors, who have told me that there has been a markedly sharp decline in their student's skills/abilities/attentiveness/motivation etc. Have you seen this phenomenon ("The customer's mentality") at Brown yet?
– Mark Viola
Jul 26 at 21:39
The seven years I spent as a professor taught me that most students didn't really care about learning, rather they just wanted a "rubber stamp" in order to get a job. I left academics 22 years ago, but have friends, who are still professors, who have told me that there has been a markedly sharp decline in their student's skills/abilities/attentiveness/motivation etc. Have you seen this phenomenon ("The customer's mentality") at Brown yet?
– Mark Viola
Jul 26 at 21:39
Sure, we all claim to be seeing that, and maybe we really are. Then I look at something like this, and realize that I could be wrong: lunarbaboon.com/comics/youth.html
– John Hughes
Jul 26 at 22:59
Sure, we all claim to be seeing that, and maybe we really are. Then I look at something like this, and realize that I could be wrong: lunarbaboon.com/comics/youth.html
– John Hughes
Jul 26 at 22:59
add a comment |Â
up vote
1
down vote
If the sum was finite, then of course we could add all the rational numbers into a fraction with the result being rational. But since the sum has infinitely many terms, there is a possibility that the sum does not approach something we can represent as the ratio of two integers. In that case, it is irrational.
If I remember correctly, one of the classic proofs of the irrationality of e does not need more than a bit of calculus to follow.
answered Jul 26 at 20:47
El Gallo Negro
233
add a comment |Â
up vote
1
down vote
If the sum was finite, then of course we could add all the rational numbers into a fraction with the result being rational. But since the sum has infinitely many terms, there is a possibility that the sum does not approach something we can represent as the ratio of two integers. In that case, it is irrational.
If I remember correctly, one of the classic proofs of the irrationality of e does not need more than a bit of calculus to follow.
answered Jul 26 at 20:47
El Gallo Negro
233
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up vote
1
down vote
up vote
1
down vote
If the sum was finite, then of course we could add all the rational numbers into a fraction with the result being rational. But since the sum has infinitely many terms, there is a possibility that the sum does not approach something we can represent as the ratio of two integers. In that case, it is irrational.
If I remember correctly, one of the classic proofs of the irrationality of e does not need more than a bit of calculus to follow.
answered Jul 26 at 20:47
El Gallo Negro
233
If the sum was finite, then of course we could add all the rational numbers into a fraction with the result being rational. But since the sum has infinitely many terms, there is a possibility that the sum does not approach something we can represent as the ratio of two integers. In that case, it is irrational.
If I remember correctly, one of the classic proofs of the irrationality of e does not need more than a bit of calculus to follow.
answered Jul 26 at 20:47
El Gallo Negro
233
answered Jul 26 at 20:47
El Gallo Negro
233
answered Jul 26 at 20:47
El Gallo Negro
233
answered Jul 26 at 20:47
El Gallo Negro
233
233
add a comment |Â
add a comment |Â
up vote
0
down vote
That's quite a normal situation. When you “sum a seriesâ€Â, you don't actually do a sum, but take a limit of finite sums:
$$
sum_n=0^infty a_n=lim_ntoinfty(a_0+a_1+dots+a_n)
$$
Even if each $a_n$ is rational, the limit is not a (finite) sum of rational numbers, so there's no compelling reason for it to be rational.
The Taylor-Lagrange theorem applied to the exponential function says that, for any integer $k$, there exists $c_kin(0,x)$ such that
$$
e=sum_n=0^kfrac1^nn!+frace^c_k(k+1)!
$$
because the $n$-th derivative of $e^x$ is $e^x$.
Suppose, to the contrary, that $e=a/b$, for some positive integers $a$ and $b$. Take $k>maxb,3$; then
$$
e=sum_n=0^kfrac1^nn!+frace^c_k(k+1)!<
sum_n=0^kfrac1^nn!+frac3(k+1)!
$$
because $c_k<1$ and $e<3$. Thus
$$
0<left(e-left(1+1+frac12!+dots+frac1k!right)k!right)<
frac3(k+1)!k!=frac3k+1
$$
However, the term
$$
e-left(1+1+frac12!+dots+frac1k!right)k!
$$
is integer by assumption, as $k!$ is a multiple of $b$. On the other hand
$$
frac3k+1<frac33+1=frac34
$$
which is a contradiction.
Thus $e$ is irrational and it's the very structure of
$$
e=sum_n=0^inftyfrac1n!
$$
that makes it so.
answered Jul 26 at 21:02
egreg
164k1180187
add a comment |Â
up vote
0
down vote
That's quite a normal situation. When you “sum a seriesâ€Â, you don't actually do a sum, but take a limit of finite sums:
$$
sum_n=0^infty a_n=lim_ntoinfty(a_0+a_1+dots+a_n)
$$
Even if each $a_n$ is rational, the limit is not a (finite) sum of rational numbers, so there's no compelling reason for it to be rational.
The Taylor-Lagrange theorem applied to the exponential function says that, for any integer $k$, there exists $c_kin(0,x)$ such that
$$
e=sum_n=0^kfrac1^nn!+frace^c_k(k+1)!
$$
because the $n$-th derivative of $e^x$ is $e^x$.
Suppose, to the contrary, that $e=a/b$, for some positive integers $a$ and $b$. Take $k>maxb,3$; then
$$
e=sum_n=0^kfrac1^nn!+frace^c_k(k+1)!<
sum_n=0^kfrac1^nn!+frac3(k+1)!
$$
because $c_k<1$ and $e<3$. Thus
$$
0<left(e-left(1+1+frac12!+dots+frac1k!right)k!right)<
frac3(k+1)!k!=frac3k+1
$$
However, the term
$$
e-left(1+1+frac12!+dots+frac1k!right)k!
$$
is integer by assumption, as $k!$ is a multiple of $b$. On the other hand
$$
frac3k+1<frac33+1=frac34
$$
which is a contradiction.
Thus $e$ is irrational and it's the very structure of
$$
e=sum_n=0^inftyfrac1n!
$$
that makes it so.
answered Jul 26 at 21:02
egreg
164k1180187
add a comment |Â
up vote
0
down vote
up vote
0
down vote
That's quite a normal situation. When you “sum a seriesâ€Â, you don't actually do a sum, but take a limit of finite sums:
$$
sum_n=0^infty a_n=lim_ntoinfty(a_0+a_1+dots+a_n)
$$
Even if each $a_n$ is rational, the limit is not a (finite) sum of rational numbers, so there's no compelling reason for it to be rational.
The Taylor-Lagrange theorem applied to the exponential function says that, for any integer $k$, there exists $c_kin(0,x)$ such that
$$
e=sum_n=0^kfrac1^nn!+frace^c_k(k+1)!
$$
because the $n$-th derivative of $e^x$ is $e^x$.
Suppose, to the contrary, that $e=a/b$, for some positive integers $a$ and $b$. Take $k>maxb,3$; then
$$
e=sum_n=0^kfrac1^nn!+frace^c_k(k+1)!<
sum_n=0^kfrac1^nn!+frac3(k+1)!
$$
because $c_k<1$ and $e<3$. Thus
$$
0<left(e-left(1+1+frac12!+dots+frac1k!right)k!right)<
frac3(k+1)!k!=frac3k+1
$$
However, the term
$$
e-left(1+1+frac12!+dots+frac1k!right)k!
$$
is integer by assumption, as $k!$ is a multiple of $b$. On the other hand
$$
frac3k+1<frac33+1=frac34
$$
which is a contradiction.
Thus $e$ is irrational and it's the very structure of
$$
e=sum_n=0^inftyfrac1n!
$$
that makes it so.
answered Jul 26 at 21:02
egreg
164k1180187
That's quite a normal situation. When you “sum a seriesâ€Â, you don't actually do a sum, but take a limit of finite sums:
$$
sum_n=0^infty a_n=lim_ntoinfty(a_0+a_1+dots+a_n)
$$
Even if each $a_n$ is rational, the limit is not a (finite) sum of rational numbers, so there's no compelling reason for it to be rational.
The Taylor-Lagrange theorem applied to the exponential function says that, for any integer $k$, there exists $c_kin(0,x)$ such that
$$
e=sum_n=0^kfrac1^nn!+frace^c_k(k+1)!
$$
because the $n$-th derivative of $e^x$ is $e^x$.
Suppose, to the contrary, that $e=a/b$, for some positive integers $a$ and $b$. Take $k>maxb,3$; then
$$
e=sum_n=0^kfrac1^nn!+frace^c_k(k+1)!<
sum_n=0^kfrac1^nn!+frac3(k+1)!
$$
because $c_k<1$ and $e<3$. Thus
$$
0<left(e-left(1+1+frac12!+dots+frac1k!right)k!right)<
frac3(k+1)!k!=frac3k+1
$$
However, the term
$$
e-left(1+1+frac12!+dots+frac1k!right)k!
$$
is integer by assumption, as $k!$ is a multiple of $b$. On the other hand
$$
frac3k+1<frac33+1=frac34
$$
which is a contradiction.
Thus $e$ is irrational and it's the very structure of
$$
e=sum_n=0^inftyfrac1n!
$$
that makes it so.
answered Jul 26 at 21:02
egreg
164k1180187
answered Jul 26 at 21:02
egreg
164k1180187
answered Jul 26 at 21:02
egreg
164k1180187
answered Jul 26 at 21:02
egreg
164k1180187
164k1180187
add a comment |Â
add a comment |Â
3
Any number can be written as a (possibly non terminating) decimal. An irrational number is a number.
– copper.hat
Jul 26 at 20:39
1
"How can a sum of rational numbers yield an irrational number $e$?" Because you add together infinitely many of these rational numbers, and infinities tend to ruin intuition. In fact, one common way to formally define the real numbers is (vaguely speaking) as any possible limit of sequences of rational numbers.
– Arthur
Jul 26 at 20:42
Every number, rational or irrational, has a decimal expansion. A decimal expansion is an infinite series of rational terms. It's the infinite number of terms in the series that is a necessary (but not sufficient) condition for the result to be irrational. It is not sufficient e.g because there are rationals with infinite (repeating) decimal expansions.
– NickD
Jul 26 at 20:42
Spivak's calculus book (not the manifolds book) has a great elementary proof.
– Michael Burr
Jul 26 at 20:46