Mixing
Simple continued fraction for irrational numbers.
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I read it here that: "What you must have read is that a number with an infinite simple continued fraction expansion is irrational. A continued fraction is "simple" if all the partial numerators are ones."
Wolfram has this simple continued fraction for $zeta(5):$ $[1;27,12,1,1,15,...]$. But we don't know if $zeta(5)$ is rational or not, so I understood something wrong.
irrational-numbers continued-fractions
asked Jan 6 at 4:10
PintecoPinteco
731313
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add a comment |
$begingroup$
I read it here that: "What you must have read is that a number with an infinite simple continued fraction expansion is irrational. A continued fraction is "simple" if all the partial numerators are ones."
Wolfram has this simple continued fraction for $zeta(5):$ $[1;27,12,1,1,15,...]$. But we don't know if $zeta(5)$ is rational or not, so I understood something wrong.
irrational-numbers continued-fractions
asked Jan 6 at 4:10
PintecoPinteco
731313
$endgroup$
1
$begingroup$
I think WA is just giving the first few terms of the simple continued fraction.
$endgroup$
– saulspatz
Jan 6 at 4:16
add a comment |
$begingroup$
I read it here that: "What you must have read is that a number with an infinite simple continued fraction expansion is irrational. A continued fraction is "simple" if all the partial numerators are ones."
Wolfram has this simple continued fraction for $zeta(5):$ $[1;27,12,1,1,15,...]$. But we don't know if $zeta(5)$ is rational or not, so I understood something wrong.
irrational-numbers continued-fractions
asked Jan 6 at 4:10
PintecoPinteco
731313
$endgroup$
I read it here that: "What you must have read is that a number with an infinite simple continued fraction expansion is irrational. A continued fraction is "simple" if all the partial numerators are ones."
Wolfram has this simple continued fraction for $zeta(5):$ $[1;27,12,1,1,15,...]$. But we don't know if $zeta(5)$ is rational or not, so I understood something wrong.
irrational-numbers continued-fractions
irrational-numbers continued-fractions
asked Jan 6 at 4:10
PintecoPinteco
731313
asked Jan 6 at 4:10
PintecoPinteco
731313
asked Jan 6 at 4:10
PintecoPinteco
731313
asked Jan 6 at 4:10
PintecoPinteco
731313
asked Jan 6 at 4:10
PintecoPinteco
731313
731313
1
$begingroup$
I think WA is just giving the first few terms of the simple continued fraction.
$endgroup$
– saulspatz
Jan 6 at 4:16
add a comment |
1
$begingroup$
I think WA is just giving the first few terms of the simple continued fraction.
$endgroup$
– saulspatz
Jan 6 at 4:16
1
1
$begingroup$
I think WA is just giving the first few terms of the simple continued fraction.
$endgroup$
– saulspatz
Jan 6 at 4:16
$begingroup$
I think WA is just giving the first few terms of the simple continued fraction.
$endgroup$
– saulspatz
Jan 6 at 4:16
add a comment |
1 Answer
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$begingroup$
The problem is we don't know what the "$ldots$" is. If it terminates after finitely many terms, $zeta(5)$ is rational. If it doesn't, $zeta(5)$ is irrational. We (and WA) can compute as many terms as we want by numerical computation, but there's no sign of a pattern.
answered Jan 6 at 4:16
Robert IsraelRobert Israel
321k23210462
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add a comment |
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$begingroup$
The problem is we don't know what the "$ldots$" is. If it terminates after finitely many terms, $zeta(5)$ is rational. If it doesn't, $zeta(5)$ is irrational. We (and WA) can compute as many terms as we want by numerical computation, but there's no sign of a pattern.
answered Jan 6 at 4:16
Robert IsraelRobert Israel
321k23210462
$endgroup$
add a comment |
$begingroup$
The problem is we don't know what the "$ldots$" is. If it terminates after finitely many terms, $zeta(5)$ is rational. If it doesn't, $zeta(5)$ is irrational. We (and WA) can compute as many terms as we want by numerical computation, but there's no sign of a pattern.
answered Jan 6 at 4:16
Robert IsraelRobert Israel
321k23210462
$endgroup$
add a comment |
$begingroup$
The problem is we don't know what the "$ldots$" is. If it terminates after finitely many terms, $zeta(5)$ is rational. If it doesn't, $zeta(5)$ is irrational. We (and WA) can compute as many terms as we want by numerical computation, but there's no sign of a pattern.
answered Jan 6 at 4:16
Robert IsraelRobert Israel
321k23210462
$endgroup$
The problem is we don't know what the "$ldots$" is. If it terminates after finitely many terms, $zeta(5)$ is rational. If it doesn't, $zeta(5)$ is irrational. We (and WA) can compute as many terms as we want by numerical computation, but there's no sign of a pattern.
answered Jan 6 at 4:16
Robert IsraelRobert Israel
321k23210462
answered Jan 6 at 4:16
Robert IsraelRobert Israel
321k23210462
answered Jan 6 at 4:16
Robert IsraelRobert Israel
321k23210462
answered Jan 6 at 4:16
Robert IsraelRobert Israel
321k23210462
321k23210462
add a comment |
add a comment |
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I think WA is just giving the first few terms of the simple continued fraction.
$endgroup$
– saulspatz
Jan 6 at 4:16