Proof that $e$ is irrational. (Proof verification)
$begingroup$
Here, $e = sum_{k=0}^infty 1/k!$. Define $S_n$ as:
begin{align}
S_n = n!e - n!sum_{k=0}^n frac{1}{k!}
end{align}
where $n!sum_{k=0}^n frac{1}{k!}$ is an integer. Write $e = 1/0!+1/1!+...+1/n!+1/(n+1)!+...$
begin{align}
S_n &= n!left(e - sum_{k=0}^n frac{1}{k!}right) = n!left( frac{1}{0!}+
frac{1}{1!}+...+frac{1}{n!}+frac{1}{(n+1)!} +... - sum_{k=0}^n frac{1}{k!}right)\
&=n!left( frac{1}{0!}+
frac{1}{1!}+...+frac{1}{n!}+frac{1}{(n+1)!} +... - frac{1}{0!} - frac{1}{1!} - ... - frac{1}{n!}
right)\
&=n!left( frac{1}{(n+1)!}+frac{1}{(n+2)!}+... right) \
&= frac{1}{(n+1)}+frac{1}{(n+2)(n+1)}+... < frac{1}{(n+1)}+frac{1}{(n+1)^2}+... = frac{1}{n}
end{align}
So, $0<S_n<1/n$. Now, if $e=p/q$ then we would have an integer between $0$ and $1$ for a large $n$.
$$0 < n!p - qn!sum_{k=0}^n frac{1}{k!}<frac{q}{n}$$
Is this correct?
proof-verification irrational-numbers
$endgroup$
add a comment |
$begingroup$
Here, $e = sum_{k=0}^infty 1/k!$. Define $S_n$ as:
begin{align}
S_n = n!e - n!sum_{k=0}^n frac{1}{k!}
end{align}
where $n!sum_{k=0}^n frac{1}{k!}$ is an integer. Write $e = 1/0!+1/1!+...+1/n!+1/(n+1)!+...$
begin{align}
S_n &= n!left(e - sum_{k=0}^n frac{1}{k!}right) = n!left( frac{1}{0!}+
frac{1}{1!}+...+frac{1}{n!}+frac{1}{(n+1)!} +... - sum_{k=0}^n frac{1}{k!}right)\
&=n!left( frac{1}{0!}+
frac{1}{1!}+...+frac{1}{n!}+frac{1}{(n+1)!} +... - frac{1}{0!} - frac{1}{1!} - ... - frac{1}{n!}
right)\
&=n!left( frac{1}{(n+1)!}+frac{1}{(n+2)!}+... right) \
&= frac{1}{(n+1)}+frac{1}{(n+2)(n+1)}+... < frac{1}{(n+1)}+frac{1}{(n+1)^2}+... = frac{1}{n}
end{align}
So, $0<S_n<1/n$. Now, if $e=p/q$ then we would have an integer between $0$ and $1$ for a large $n$.
$$0 < n!p - qn!sum_{k=0}^n frac{1}{k!}<frac{q}{n}$$
Is this correct?
proof-verification irrational-numbers
$endgroup$
add a comment |
$begingroup$
Here, $e = sum_{k=0}^infty 1/k!$. Define $S_n$ as:
begin{align}
S_n = n!e - n!sum_{k=0}^n frac{1}{k!}
end{align}
where $n!sum_{k=0}^n frac{1}{k!}$ is an integer. Write $e = 1/0!+1/1!+...+1/n!+1/(n+1)!+...$
begin{align}
S_n &= n!left(e - sum_{k=0}^n frac{1}{k!}right) = n!left( frac{1}{0!}+
frac{1}{1!}+...+frac{1}{n!}+frac{1}{(n+1)!} +... - sum_{k=0}^n frac{1}{k!}right)\
&=n!left( frac{1}{0!}+
frac{1}{1!}+...+frac{1}{n!}+frac{1}{(n+1)!} +... - frac{1}{0!} - frac{1}{1!} - ... - frac{1}{n!}
right)\
&=n!left( frac{1}{(n+1)!}+frac{1}{(n+2)!}+... right) \
&= frac{1}{(n+1)}+frac{1}{(n+2)(n+1)}+... < frac{1}{(n+1)}+frac{1}{(n+1)^2}+... = frac{1}{n}
end{align}
So, $0<S_n<1/n$. Now, if $e=p/q$ then we would have an integer between $0$ and $1$ for a large $n$.
$$0 < n!p - qn!sum_{k=0}^n frac{1}{k!}<frac{q}{n}$$
Is this correct?
proof-verification irrational-numbers
$endgroup$
Here, $e = sum_{k=0}^infty 1/k!$. Define $S_n$ as:
begin{align}
S_n = n!e - n!sum_{k=0}^n frac{1}{k!}
end{align}
where $n!sum_{k=0}^n frac{1}{k!}$ is an integer. Write $e = 1/0!+1/1!+...+1/n!+1/(n+1)!+...$
begin{align}
S_n &= n!left(e - sum_{k=0}^n frac{1}{k!}right) = n!left( frac{1}{0!}+
frac{1}{1!}+...+frac{1}{n!}+frac{1}{(n+1)!} +... - sum_{k=0}^n frac{1}{k!}right)\
&=n!left( frac{1}{0!}+
frac{1}{1!}+...+frac{1}{n!}+frac{1}{(n+1)!} +... - frac{1}{0!} - frac{1}{1!} - ... - frac{1}{n!}
right)\
&=n!left( frac{1}{(n+1)!}+frac{1}{(n+2)!}+... right) \
&= frac{1}{(n+1)}+frac{1}{(n+2)(n+1)}+... < frac{1}{(n+1)}+frac{1}{(n+1)^2}+... = frac{1}{n}
end{align}
So, $0<S_n<1/n$. Now, if $e=p/q$ then we would have an integer between $0$ and $1$ for a large $n$.
$$0 < n!p - qn!sum_{k=0}^n frac{1}{k!}<frac{q}{n}$$
Is this correct?
proof-verification irrational-numbers
proof-verification irrational-numbers
edited Jan 6 at 7:59
Eevee Trainer
5,4431936
5,4431936
asked Jan 6 at 7:43
PintecoPinteco
731313
731313
add a comment |
add a comment |
2 Answers
2
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oldest
votes
$begingroup$
It's basically fine. To tie up the details, you should specify exactly how large $n$ needs to be.
Also, writing "$0<S_n =$ (something that isn't equal to $S_n$) is bad notation. I'm not even sure exactly what you were trying to say there. If you're not sure how to write what you mean in symbols, use some words to clear things up.
$endgroup$
$begingroup$
I believe he multiplied through by $q$, to touch on that point made in your second question. I agree with the overall point though.
$endgroup$
– Eevee Trainer
Jan 6 at 7:55
$begingroup$
I've fixed that part, thanks.
$endgroup$
– Pinteco
Jan 6 at 7:56
add a comment |
$begingroup$
Your proof seems sufficient enough, as far as I can tell. You did make a slight typo in the final inequality (multiplying through by $q$ would mean you have $qS_n$ there); this was touched on in another answer and you already edited your answer to rectify this.
If I had to nitpick anything of significance...
Personally it could use a lot of elaboration to explain various manipulations and results - where they come from and so on. I found myself having to fill in the holes myself; it can make it a bit difficult for someone to see where you're going.
But this may just be a personal issue. If nothing else, at least everything seems to logically follow. But taking some time to explain everything for the reader will help. (In particular if this is a proof for a class or something: no sense in getting points taken off because you thought it was obvious what was going on.)
$endgroup$
add a comment |
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2 Answers
2
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votes
2 Answers
2
active
oldest
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votes
$begingroup$
It's basically fine. To tie up the details, you should specify exactly how large $n$ needs to be.
Also, writing "$0<S_n =$ (something that isn't equal to $S_n$) is bad notation. I'm not even sure exactly what you were trying to say there. If you're not sure how to write what you mean in symbols, use some words to clear things up.
$endgroup$
$begingroup$
I believe he multiplied through by $q$, to touch on that point made in your second question. I agree with the overall point though.
$endgroup$
– Eevee Trainer
Jan 6 at 7:55
$begingroup$
I've fixed that part, thanks.
$endgroup$
– Pinteco
Jan 6 at 7:56
add a comment |
$begingroup$
It's basically fine. To tie up the details, you should specify exactly how large $n$ needs to be.
Also, writing "$0<S_n =$ (something that isn't equal to $S_n$) is bad notation. I'm not even sure exactly what you were trying to say there. If you're not sure how to write what you mean in symbols, use some words to clear things up.
$endgroup$
$begingroup$
I believe he multiplied through by $q$, to touch on that point made in your second question. I agree with the overall point though.
$endgroup$
– Eevee Trainer
Jan 6 at 7:55
$begingroup$
I've fixed that part, thanks.
$endgroup$
– Pinteco
Jan 6 at 7:56
add a comment |
$begingroup$
It's basically fine. To tie up the details, you should specify exactly how large $n$ needs to be.
Also, writing "$0<S_n =$ (something that isn't equal to $S_n$) is bad notation. I'm not even sure exactly what you were trying to say there. If you're not sure how to write what you mean in symbols, use some words to clear things up.
$endgroup$
It's basically fine. To tie up the details, you should specify exactly how large $n$ needs to be.
Also, writing "$0<S_n =$ (something that isn't equal to $S_n$) is bad notation. I'm not even sure exactly what you were trying to say there. If you're not sure how to write what you mean in symbols, use some words to clear things up.
answered Jan 6 at 7:53
jmerryjmerry
3,727514
3,727514
$begingroup$
I believe he multiplied through by $q$, to touch on that point made in your second question. I agree with the overall point though.
$endgroup$
– Eevee Trainer
Jan 6 at 7:55
$begingroup$
I've fixed that part, thanks.
$endgroup$
– Pinteco
Jan 6 at 7:56
add a comment |
$begingroup$
I believe he multiplied through by $q$, to touch on that point made in your second question. I agree with the overall point though.
$endgroup$
– Eevee Trainer
Jan 6 at 7:55
$begingroup$
I've fixed that part, thanks.
$endgroup$
– Pinteco
Jan 6 at 7:56
$begingroup$
I believe he multiplied through by $q$, to touch on that point made in your second question. I agree with the overall point though.
$endgroup$
– Eevee Trainer
Jan 6 at 7:55
$begingroup$
I believe he multiplied through by $q$, to touch on that point made in your second question. I agree with the overall point though.
$endgroup$
– Eevee Trainer
Jan 6 at 7:55
$begingroup$
I've fixed that part, thanks.
$endgroup$
– Pinteco
Jan 6 at 7:56
$begingroup$
I've fixed that part, thanks.
$endgroup$
– Pinteco
Jan 6 at 7:56
add a comment |
$begingroup$
Your proof seems sufficient enough, as far as I can tell. You did make a slight typo in the final inequality (multiplying through by $q$ would mean you have $qS_n$ there); this was touched on in another answer and you already edited your answer to rectify this.
If I had to nitpick anything of significance...
Personally it could use a lot of elaboration to explain various manipulations and results - where they come from and so on. I found myself having to fill in the holes myself; it can make it a bit difficult for someone to see where you're going.
But this may just be a personal issue. If nothing else, at least everything seems to logically follow. But taking some time to explain everything for the reader will help. (In particular if this is a proof for a class or something: no sense in getting points taken off because you thought it was obvious what was going on.)
$endgroup$
add a comment |
$begingroup$
Your proof seems sufficient enough, as far as I can tell. You did make a slight typo in the final inequality (multiplying through by $q$ would mean you have $qS_n$ there); this was touched on in another answer and you already edited your answer to rectify this.
If I had to nitpick anything of significance...
Personally it could use a lot of elaboration to explain various manipulations and results - where they come from and so on. I found myself having to fill in the holes myself; it can make it a bit difficult for someone to see where you're going.
But this may just be a personal issue. If nothing else, at least everything seems to logically follow. But taking some time to explain everything for the reader will help. (In particular if this is a proof for a class or something: no sense in getting points taken off because you thought it was obvious what was going on.)
$endgroup$
add a comment |
$begingroup$
Your proof seems sufficient enough, as far as I can tell. You did make a slight typo in the final inequality (multiplying through by $q$ would mean you have $qS_n$ there); this was touched on in another answer and you already edited your answer to rectify this.
If I had to nitpick anything of significance...
Personally it could use a lot of elaboration to explain various manipulations and results - where they come from and so on. I found myself having to fill in the holes myself; it can make it a bit difficult for someone to see where you're going.
But this may just be a personal issue. If nothing else, at least everything seems to logically follow. But taking some time to explain everything for the reader will help. (In particular if this is a proof for a class or something: no sense in getting points taken off because you thought it was obvious what was going on.)
$endgroup$
Your proof seems sufficient enough, as far as I can tell. You did make a slight typo in the final inequality (multiplying through by $q$ would mean you have $qS_n$ there); this was touched on in another answer and you already edited your answer to rectify this.
If I had to nitpick anything of significance...
Personally it could use a lot of elaboration to explain various manipulations and results - where they come from and so on. I found myself having to fill in the holes myself; it can make it a bit difficult for someone to see where you're going.
But this may just be a personal issue. If nothing else, at least everything seems to logically follow. But taking some time to explain everything for the reader will help. (In particular if this is a proof for a class or something: no sense in getting points taken off because you thought it was obvious what was going on.)
answered Jan 6 at 7:57
Eevee TrainerEevee Trainer
5,4431936
5,4431936
add a comment |
add a comment |
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