Mixing
Is there a proof of an infinite number of prime numbers using the irrationality of $e$?
$begingroup$
That the set of prime integers is infinite can be proved using the irrationality of $pi$; see this wikipedia link. It analyzes the representation
$tag 1 {displaystyle {frac {pi }{4}}={frac {3}{4}}times {frac {5}{4}}times {frac {7}{8}}times {frac {11}{12}}times {frac {13}{12}}times {frac {17}{16}}times {frac {19}{20}}times {frac {23}{24}}times {frac {29}{28}}times {frac {31}{32}}times cdots }$
It 'seems only fair' that the same thing can be done using $e$:
Question 1: Has a proof that there are an infinite number of primes been
constructed that uses the irrationality of $e$?
If the existence of such a proof is not available, then
Question 1-1: Is a proof available using any properties of $e$?
If again, no answers, then
Question 2: Can someone show that the set of prime integers is
infinite using properties of Euler's constant?
My work
I found this representation of $e$,
$tag 2 {displaystyle e=1+{cfrac {2}{1+{cfrac {1}{6+{cfrac {1}{10+{cfrac {1}{14+{cfrac {1}{18+{cfrac {1}{22+{cfrac {1}{26+ddots ,}}}}}}}}}}}}}}.}$
Note that with $text{(1)}$ there is a 'corresponding' sequence
$tag 3 4,8,12,16,dots$
and that with $text{(2)}$ we can see
$tag 4 6, 10,14,18,dots$
I just find this interesting; perhaps it supplies a 'connection'.
We know that eventually every prime number will divide a term in the $text{(4)}$ sequence.
If the only primes are contained in the finite set $mathcal P = {2,3,5, dots, p}$, then let $beta = 2 times 3 times 5 times dots times p$.
If we analyze $text{(2)}$ using base $beta$, the expansion of $e$ would eventually be repetitive or simply terminate.
(the statement in bold is an unproven bold statement)
soft-question prime-numbers alternative-proof continued-fractions
edited Jan 29 at 16:35
YuiTo Cheng
2,1863937
asked Nov 22 '18 at 14:40
CopyPasteItCopyPasteIt
4,2131728
$endgroup$
|
show 1 more comment
$begingroup$
That the set of prime integers is infinite can be proved using the irrationality of $pi$; see this wikipedia link. It analyzes the representation
$tag 1 {displaystyle {frac {pi }{4}}={frac {3}{4}}times {frac {5}{4}}times {frac {7}{8}}times {frac {11}{12}}times {frac {13}{12}}times {frac {17}{16}}times {frac {19}{20}}times {frac {23}{24}}times {frac {29}{28}}times {frac {31}{32}}times cdots }$
It 'seems only fair' that the same thing can be done using $e$:
Question 1: Has a proof that there are an infinite number of primes been
constructed that uses the irrationality of $e$?
If the existence of such a proof is not available, then
Question 1-1: Is a proof available using any properties of $e$?
If again, no answers, then
Question 2: Can someone show that the set of prime integers is
infinite using properties of Euler's constant?
My work
I found this representation of $e$,
$tag 2 {displaystyle e=1+{cfrac {2}{1+{cfrac {1}{6+{cfrac {1}{10+{cfrac {1}{14+{cfrac {1}{18+{cfrac {1}{22+{cfrac {1}{26+ddots ,}}}}}}}}}}}}}}.}$
Note that with $text{(1)}$ there is a 'corresponding' sequence
$tag 3 4,8,12,16,dots$
and that with $text{(2)}$ we can see
$tag 4 6, 10,14,18,dots$
I just find this interesting; perhaps it supplies a 'connection'.
We know that eventually every prime number will divide a term in the $text{(4)}$ sequence.
If the only primes are contained in the finite set $mathcal P = {2,3,5, dots, p}$, then let $beta = 2 times 3 times 5 times dots times p$.
If we analyze $text{(2)}$ using base $beta$, the expansion of $e$ would eventually be repetitive or simply terminate.
(the statement in bold is an unproven bold statement)
soft-question prime-numbers alternative-proof continued-fractions
edited Jan 29 at 16:35
YuiTo Cheng
2,1863937
asked Nov 22 '18 at 14:40
CopyPasteItCopyPasteIt
4,2131728
$endgroup$
$begingroup$
It's hard to see why one would want to prove it that way, as the usual $1+prod_{i=1}^n p_i$ argument does it easily.
$endgroup$
– Richard Martin
Nov 22 '18 at 14:46
$begingroup$
Or, better still, try and figure it out from what I have just said.
$endgroup$
– Richard Martin
Nov 22 '18 at 14:50
$begingroup$
@RichardMartin Ok - Euclid's proof. Just interested in seeing a 'later stage' proof.
$endgroup$
– CopyPasteIt
Nov 22 '18 at 14:52
$begingroup$
@RichardMartin Someone putting up the wiki article thought it was interesting enough to put down the $pi$ proof...
$endgroup$
– CopyPasteIt
Nov 22 '18 at 14:53
5
$begingroup$
@RichardMartin More proofs = more fun. But seriously, new proofs to existing theorems are interesting because of any novel ideas they might contain, not because they achieve the same conclusion.
$endgroup$
– Klangen
Nov 23 '18 at 9:12
|
show 1 more comment
$begingroup$
That the set of prime integers is infinite can be proved using the irrationality of $pi$; see this wikipedia link. It analyzes the representation
$tag 1 {displaystyle {frac {pi }{4}}={frac {3}{4}}times {frac {5}{4}}times {frac {7}{8}}times {frac {11}{12}}times {frac {13}{12}}times {frac {17}{16}}times {frac {19}{20}}times {frac {23}{24}}times {frac {29}{28}}times {frac {31}{32}}times cdots }$
It 'seems only fair' that the same thing can be done using $e$:
Question 1: Has a proof that there are an infinite number of primes been
constructed that uses the irrationality of $e$?
If the existence of such a proof is not available, then
Question 1-1: Is a proof available using any properties of $e$?
If again, no answers, then
Question 2: Can someone show that the set of prime integers is
infinite using properties of Euler's constant?
My work
I found this representation of $e$,
$tag 2 {displaystyle e=1+{cfrac {2}{1+{cfrac {1}{6+{cfrac {1}{10+{cfrac {1}{14+{cfrac {1}{18+{cfrac {1}{22+{cfrac {1}{26+ddots ,}}}}}}}}}}}}}}.}$
Note that with $text{(1)}$ there is a 'corresponding' sequence
$tag 3 4,8,12,16,dots$
and that with $text{(2)}$ we can see
$tag 4 6, 10,14,18,dots$
I just find this interesting; perhaps it supplies a 'connection'.
We know that eventually every prime number will divide a term in the $text{(4)}$ sequence.
If the only primes are contained in the finite set $mathcal P = {2,3,5, dots, p}$, then let $beta = 2 times 3 times 5 times dots times p$.
If we analyze $text{(2)}$ using base $beta$, the expansion of $e$ would eventually be repetitive or simply terminate.
(the statement in bold is an unproven bold statement)
soft-question prime-numbers alternative-proof continued-fractions
edited Jan 29 at 16:35
YuiTo Cheng
2,1863937
asked Nov 22 '18 at 14:40
CopyPasteItCopyPasteIt
4,2131728
$endgroup$
That the set of prime integers is infinite can be proved using the irrationality of $pi$; see this wikipedia link. It analyzes the representation
$tag 1 {displaystyle {frac {pi }{4}}={frac {3}{4}}times {frac {5}{4}}times {frac {7}{8}}times {frac {11}{12}}times {frac {13}{12}}times {frac {17}{16}}times {frac {19}{20}}times {frac {23}{24}}times {frac {29}{28}}times {frac {31}{32}}times cdots }$
It 'seems only fair' that the same thing can be done using $e$:
Question 1: Has a proof that there are an infinite number of primes been
constructed that uses the irrationality of $e$?
If the existence of such a proof is not available, then
Question 1-1: Is a proof available using any properties of $e$?
If again, no answers, then
Question 2: Can someone show that the set of prime integers is
infinite using properties of Euler's constant?
My work
I found this representation of $e$,
$tag 2 {displaystyle e=1+{cfrac {2}{1+{cfrac {1}{6+{cfrac {1}{10+{cfrac {1}{14+{cfrac {1}{18+{cfrac {1}{22+{cfrac {1}{26+ddots ,}}}}}}}}}}}}}}.}$
Note that with $text{(1)}$ there is a 'corresponding' sequence
$tag 3 4,8,12,16,dots$
and that with $text{(2)}$ we can see
$tag 4 6, 10,14,18,dots$
I just find this interesting; perhaps it supplies a 'connection'.
We know that eventually every prime number will divide a term in the $text{(4)}$ sequence.
If the only primes are contained in the finite set $mathcal P = {2,3,5, dots, p}$, then let $beta = 2 times 3 times 5 times dots times p$.
If we analyze $text{(2)}$ using base $beta$, the expansion of $e$ would eventually be repetitive or simply terminate.
(the statement in bold is an unproven bold statement)
soft-question prime-numbers alternative-proof continued-fractions
soft-question prime-numbers alternative-proof continued-fractions
edited Jan 29 at 16:35
YuiTo Cheng
2,1863937
asked Nov 22 '18 at 14:40
CopyPasteItCopyPasteIt
4,2131728
edited Jan 29 at 16:35
YuiTo Cheng
2,1863937
asked Nov 22 '18 at 14:40
CopyPasteItCopyPasteIt
4,2131728
edited Jan 29 at 16:35
YuiTo Cheng
2,1863937
edited Jan 29 at 16:35
YuiTo Cheng
2,1863937
edited Jan 29 at 16:35
YuiTo Cheng
2,1863937
2,1863937
asked Nov 22 '18 at 14:40
CopyPasteItCopyPasteIt
4,2131728
asked Nov 22 '18 at 14:40
CopyPasteItCopyPasteIt
4,2131728
asked Nov 22 '18 at 14:40
CopyPasteItCopyPasteIt
4,2131728
4,2131728
$begingroup$
It's hard to see why one would want to prove it that way, as the usual $1+prod_{i=1}^n p_i$ argument does it easily.
$endgroup$
– Richard Martin
Nov 22 '18 at 14:46
$begingroup$
Or, better still, try and figure it out from what I have just said.
$endgroup$
– Richard Martin
Nov 22 '18 at 14:50
$begingroup$
@RichardMartin Ok - Euclid's proof. Just interested in seeing a 'later stage' proof.
$endgroup$
– CopyPasteIt
Nov 22 '18 at 14:52
$begingroup$
@RichardMartin Someone putting up the wiki article thought it was interesting enough to put down the $pi$ proof...
$endgroup$
– CopyPasteIt
Nov 22 '18 at 14:53
5
$begingroup$
@RichardMartin More proofs = more fun. But seriously, new proofs to existing theorems are interesting because of any novel ideas they might contain, not because they achieve the same conclusion.
$endgroup$
– Klangen
Nov 23 '18 at 9:12
|
show 1 more comment
$begingroup$
It's hard to see why one would want to prove it that way, as the usual $1+prod_{i=1}^n p_i$ argument does it easily.
$endgroup$
– Richard Martin
Nov 22 '18 at 14:46
$begingroup$
Or, better still, try and figure it out from what I have just said.
$endgroup$
– Richard Martin
Nov 22 '18 at 14:50
$begingroup$
@RichardMartin Ok - Euclid's proof. Just interested in seeing a 'later stage' proof.
$endgroup$
– CopyPasteIt
Nov 22 '18 at 14:52
$begingroup$
@RichardMartin Someone putting up the wiki article thought it was interesting enough to put down the $pi$ proof...
$endgroup$
– CopyPasteIt
Nov 22 '18 at 14:53
5
$begingroup$
@RichardMartin More proofs = more fun. But seriously, new proofs to existing theorems are interesting because of any novel ideas they might contain, not because they achieve the same conclusion.
$endgroup$
– Klangen
Nov 23 '18 at 9:12
$begingroup$
It's hard to see why one would want to prove it that way, as the usual $1+prod_{i=1}^n p_i$ argument does it easily.
$endgroup$
– Richard Martin
Nov 22 '18 at 14:46
$begingroup$
It's hard to see why one would want to prove it that way, as the usual $1+prod_{i=1}^n p_i$ argument does it easily.
$endgroup$
– Richard Martin
Nov 22 '18 at 14:46
$begingroup$
Or, better still, try and figure it out from what I have just said.
$endgroup$
– Richard Martin
Nov 22 '18 at 14:50
$begingroup$
Or, better still, try and figure it out from what I have just said.
$endgroup$
– Richard Martin
Nov 22 '18 at 14:50
$begingroup$
@RichardMartin Ok - Euclid's proof. Just interested in seeing a 'later stage' proof.
$endgroup$
– CopyPasteIt
Nov 22 '18 at 14:52
$begingroup$
@RichardMartin Ok - Euclid's proof. Just interested in seeing a 'later stage' proof.
$endgroup$
– CopyPasteIt
Nov 22 '18 at 14:52
$begingroup$
@RichardMartin Someone putting up the wiki article thought it was interesting enough to put down the $pi$ proof...
$endgroup$
– CopyPasteIt
Nov 22 '18 at 14:53
$begingroup$
@RichardMartin Someone putting up the wiki article thought it was interesting enough to put down the $pi$ proof...
$endgroup$
– CopyPasteIt
Nov 22 '18 at 14:53
5
5
$begingroup$
@RichardMartin More proofs = more fun. But seriously, new proofs to existing theorems are interesting because of any novel ideas they might contain, not because they achieve the same conclusion.
$endgroup$
– Klangen
Nov 23 '18 at 9:12
$begingroup$
@RichardMartin More proofs = more fun. But seriously, new proofs to existing theorems are interesting because of any novel ideas they might contain, not because they achieve the same conclusion.
$endgroup$
– Klangen
Nov 23 '18 at 9:12
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
We can give a proof based on the transcendentality of $e$:
Since $$lim_{nrightarrowinfty}left(prod_{i=1}^{n} p_iright)^{1/p_n}=e,$$
if the set of prime numbers was finite then $e$ would be algebraic.
answered Jan 29 at 17:02
Jose BroxJose Brox
3,39211129
$endgroup$
$begingroup$
Can't find a link to $lim_{nrightarrowinfty}left(prod_{i=1}^{n} p_iright)^{1/p_n}=e,$ ...(guessing if follows from the prime number theorem)
$endgroup$
– CopyPasteIt
Jan 29 at 17:31
$begingroup$
Did you check all the proof to make sure there is no circular reasoning here?
$endgroup$
– CopyPasteIt
Jan 29 at 17:33
1
$begingroup$
@CopyPasteIt Here you have some info: math.stackexchange.com/questions/231578/… ... The fact is strongly related to the PNT; in fact it comes from the asymptotics of Chebyshev function $theta$. You could say that it is a "cooked" relation with $e$, since we could do just with a lower bound on $theta$, which is implicit. Yet again, we can do with way less in order to prove the infinitude of primes :P Not circular, but we would arrive first to the conclusion due to the asymptotics...
$endgroup$
– Jose Brox
Jan 29 at 17:41
add a comment |
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1 Answer
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1 Answer
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oldest
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$begingroup$
We can give a proof based on the transcendentality of $e$:
Since $$lim_{nrightarrowinfty}left(prod_{i=1}^{n} p_iright)^{1/p_n}=e,$$
if the set of prime numbers was finite then $e$ would be algebraic.
answered Jan 29 at 17:02
Jose BroxJose Brox
3,39211129
$endgroup$
$begingroup$
Can't find a link to $lim_{nrightarrowinfty}left(prod_{i=1}^{n} p_iright)^{1/p_n}=e,$ ...(guessing if follows from the prime number theorem)
$endgroup$
– CopyPasteIt
Jan 29 at 17:31
$begingroup$
Did you check all the proof to make sure there is no circular reasoning here?
$endgroup$
– CopyPasteIt
Jan 29 at 17:33
1
$begingroup$
@CopyPasteIt Here you have some info: math.stackexchange.com/questions/231578/… ... The fact is strongly related to the PNT; in fact it comes from the asymptotics of Chebyshev function $theta$. You could say that it is a "cooked" relation with $e$, since we could do just with a lower bound on $theta$, which is implicit. Yet again, we can do with way less in order to prove the infinitude of primes :P Not circular, but we would arrive first to the conclusion due to the asymptotics...
$endgroup$
– Jose Brox
Jan 29 at 17:41
add a comment |
$begingroup$
We can give a proof based on the transcendentality of $e$:
Since $$lim_{nrightarrowinfty}left(prod_{i=1}^{n} p_iright)^{1/p_n}=e,$$
if the set of prime numbers was finite then $e$ would be algebraic.
answered Jan 29 at 17:02
Jose BroxJose Brox
3,39211129
$endgroup$
$begingroup$
Can't find a link to $lim_{nrightarrowinfty}left(prod_{i=1}^{n} p_iright)^{1/p_n}=e,$ ...(guessing if follows from the prime number theorem)
$endgroup$
– CopyPasteIt
Jan 29 at 17:31
$begingroup$
Did you check all the proof to make sure there is no circular reasoning here?
$endgroup$
– CopyPasteIt
Jan 29 at 17:33
1
$begingroup$
@CopyPasteIt Here you have some info: math.stackexchange.com/questions/231578/… ... The fact is strongly related to the PNT; in fact it comes from the asymptotics of Chebyshev function $theta$. You could say that it is a "cooked" relation with $e$, since we could do just with a lower bound on $theta$, which is implicit. Yet again, we can do with way less in order to prove the infinitude of primes :P Not circular, but we would arrive first to the conclusion due to the asymptotics...
$endgroup$
– Jose Brox
Jan 29 at 17:41
add a comment |
$begingroup$
We can give a proof based on the transcendentality of $e$:
Since $$lim_{nrightarrowinfty}left(prod_{i=1}^{n} p_iright)^{1/p_n}=e,$$
if the set of prime numbers was finite then $e$ would be algebraic.
answered Jan 29 at 17:02
Jose BroxJose Brox
3,39211129
$endgroup$
We can give a proof based on the transcendentality of $e$:
Since $$lim_{nrightarrowinfty}left(prod_{i=1}^{n} p_iright)^{1/p_n}=e,$$
if the set of prime numbers was finite then $e$ would be algebraic.
answered Jan 29 at 17:02
Jose BroxJose Brox
3,39211129
answered Jan 29 at 17:02
Jose BroxJose Brox
3,39211129
answered Jan 29 at 17:02
Jose BroxJose Brox
3,39211129
answered Jan 29 at 17:02
Jose BroxJose Brox
3,39211129
3,39211129
$begingroup$
Can't find a link to $lim_{nrightarrowinfty}left(prod_{i=1}^{n} p_iright)^{1/p_n}=e,$ ...(guessing if follows from the prime number theorem)
$endgroup$
– CopyPasteIt
Jan 29 at 17:31
$begingroup$
Did you check all the proof to make sure there is no circular reasoning here?
$endgroup$
– CopyPasteIt
Jan 29 at 17:33
1
$begingroup$
@CopyPasteIt Here you have some info: math.stackexchange.com/questions/231578/… ... The fact is strongly related to the PNT; in fact it comes from the asymptotics of Chebyshev function $theta$. You could say that it is a "cooked" relation with $e$, since we could do just with a lower bound on $theta$, which is implicit. Yet again, we can do with way less in order to prove the infinitude of primes :P Not circular, but we would arrive first to the conclusion due to the asymptotics...
$endgroup$
– Jose Brox
Jan 29 at 17:41
add a comment |
$begingroup$
Can't find a link to $lim_{nrightarrowinfty}left(prod_{i=1}^{n} p_iright)^{1/p_n}=e,$ ...(guessing if follows from the prime number theorem)
$endgroup$
– CopyPasteIt
Jan 29 at 17:31
$begingroup$
Did you check all the proof to make sure there is no circular reasoning here?
$endgroup$
– CopyPasteIt
Jan 29 at 17:33
1
$begingroup$
@CopyPasteIt Here you have some info: math.stackexchange.com/questions/231578/… ... The fact is strongly related to the PNT; in fact it comes from the asymptotics of Chebyshev function $theta$. You could say that it is a "cooked" relation with $e$, since we could do just with a lower bound on $theta$, which is implicit. Yet again, we can do with way less in order to prove the infinitude of primes :P Not circular, but we would arrive first to the conclusion due to the asymptotics...
$endgroup$
– Jose Brox
Jan 29 at 17:41
$begingroup$
Can't find a link to $lim_{nrightarrowinfty}left(prod_{i=1}^{n} p_iright)^{1/p_n}=e,$ ...(guessing if follows from the prime number theorem)
$endgroup$
– CopyPasteIt
Jan 29 at 17:31
$begingroup$
Can't find a link to $lim_{nrightarrowinfty}left(prod_{i=1}^{n} p_iright)^{1/p_n}=e,$ ...(guessing if follows from the prime number theorem)
$endgroup$
– CopyPasteIt
Jan 29 at 17:31
$begingroup$
Did you check all the proof to make sure there is no circular reasoning here?
$endgroup$
– CopyPasteIt
Jan 29 at 17:33
$begingroup$
Did you check all the proof to make sure there is no circular reasoning here?
$endgroup$
– CopyPasteIt
Jan 29 at 17:33
1
1
$begingroup$
@CopyPasteIt Here you have some info: math.stackexchange.com/questions/231578/… ... The fact is strongly related to the PNT; in fact it comes from the asymptotics of Chebyshev function $theta$. You could say that it is a "cooked" relation with $e$, since we could do just with a lower bound on $theta$, which is implicit. Yet again, we can do with way less in order to prove the infinitude of primes :P Not circular, but we would arrive first to the conclusion due to the asymptotics...
$endgroup$
– Jose Brox
Jan 29 at 17:41
$begingroup$
@CopyPasteIt Here you have some info: math.stackexchange.com/questions/231578/… ... The fact is strongly related to the PNT; in fact it comes from the asymptotics of Chebyshev function $theta$. You could say that it is a "cooked" relation with $e$, since we could do just with a lower bound on $theta$, which is implicit. Yet again, we can do with way less in order to prove the infinitude of primes :P Not circular, but we would arrive first to the conclusion due to the asymptotics...
$endgroup$
– Jose Brox
Jan 29 at 17:41
add a comment |
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$begingroup$
It's hard to see why one would want to prove it that way, as the usual $1+prod_{i=1}^n p_i$ argument does it easily.
$endgroup$
– Richard Martin
Nov 22 '18 at 14:46
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Or, better still, try and figure it out from what I have just said.
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– Richard Martin
Nov 22 '18 at 14:50
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@RichardMartin Ok - Euclid's proof. Just interested in seeing a 'later stage' proof.
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– CopyPasteIt
Nov 22 '18 at 14:52
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@RichardMartin Someone putting up the wiki article thought it was interesting enough to put down the $pi$ proof...
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– CopyPasteIt
Nov 22 '18 at 14:53
5
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@RichardMartin More proofs = more fun. But seriously, new proofs to existing theorems are interesting because of any novel ideas they might contain, not because they achieve the same conclusion.
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– Klangen
Nov 23 '18 at 9:12