Can you obtain $pi$ using elements of $mathbb{N}$, and finite number of basic arithmetic operations +...












12












$begingroup$


Is it possible to obtain $pi$ from finite amount of operations
${+,-,cdot,div,wedge}$ on $mathbb{N}$ (or $mathbb{Q}$, the answer will still be the same), note that set of all real numbers obtainable this way contains numbers that are not algebraic (for example $2^{2^{1/2}}$ is transcendental)



Bonus: If it happens that the answer is no, is it a solution to some equation generated that way (those $5$ operations performed finitely many times on elements on $mathbb{N}$) ?










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$endgroup$












  • $begingroup$
    Related: math.stackexchange.com/questions/2611084/…
    $endgroup$
    – user202729
    Jul 28 '18 at 13:34






  • 1




    $begingroup$
    (this one is more general. If the answer to this one is "no" then the answer to the other one is "no" as well)
    $endgroup$
    – user202729
    Jul 28 '18 at 13:35










  • $begingroup$
    Of course not, but I can't prove it. There are only countably many finite expressions of this sort, so you can only express countably many real numbers this way. Unless a real has some reason to be expressible it almost certainly can't be.
    $endgroup$
    – Ross Millikan
    Jul 28 '18 at 14:36






  • 1




    $begingroup$
    Yes, there are only countably many of them, so most real numbers do not fall into this category. Still, not much can be deduced about any specific number from that fact. π is arguably somewhat special.
    $endgroup$
    – Mathemagician
    Jul 28 '18 at 14:57
















12












$begingroup$


Is it possible to obtain $pi$ from finite amount of operations
${+,-,cdot,div,wedge}$ on $mathbb{N}$ (or $mathbb{Q}$, the answer will still be the same), note that set of all real numbers obtainable this way contains numbers that are not algebraic (for example $2^{2^{1/2}}$ is transcendental)



Bonus: If it happens that the answer is no, is it a solution to some equation generated that way (those $5$ operations performed finitely many times on elements on $mathbb{N}$) ?










share|cite|improve this question











$endgroup$












  • $begingroup$
    Related: math.stackexchange.com/questions/2611084/…
    $endgroup$
    – user202729
    Jul 28 '18 at 13:34






  • 1




    $begingroup$
    (this one is more general. If the answer to this one is "no" then the answer to the other one is "no" as well)
    $endgroup$
    – user202729
    Jul 28 '18 at 13:35










  • $begingroup$
    Of course not, but I can't prove it. There are only countably many finite expressions of this sort, so you can only express countably many real numbers this way. Unless a real has some reason to be expressible it almost certainly can't be.
    $endgroup$
    – Ross Millikan
    Jul 28 '18 at 14:36






  • 1




    $begingroup$
    Yes, there are only countably many of them, so most real numbers do not fall into this category. Still, not much can be deduced about any specific number from that fact. π is arguably somewhat special.
    $endgroup$
    – Mathemagician
    Jul 28 '18 at 14:57














12












12








12


6



$begingroup$


Is it possible to obtain $pi$ from finite amount of operations
${+,-,cdot,div,wedge}$ on $mathbb{N}$ (or $mathbb{Q}$, the answer will still be the same), note that set of all real numbers obtainable this way contains numbers that are not algebraic (for example $2^{2^{1/2}}$ is transcendental)



Bonus: If it happens that the answer is no, is it a solution to some equation generated that way (those $5$ operations performed finitely many times on elements on $mathbb{N}$) ?










share|cite|improve this question











$endgroup$




Is it possible to obtain $pi$ from finite amount of operations
${+,-,cdot,div,wedge}$ on $mathbb{N}$ (or $mathbb{Q}$, the answer will still be the same), note that set of all real numbers obtainable this way contains numbers that are not algebraic (for example $2^{2^{1/2}}$ is transcendental)



Bonus: If it happens that the answer is no, is it a solution to some equation generated that way (those $5$ operations performed finitely many times on elements on $mathbb{N}$) ?







algebra-precalculus






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share|cite|improve this question













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share|cite|improve this question








edited Jan 1 at 11:32









Eevee Trainer

7,04321337




7,04321337










asked Jul 28 '18 at 13:29









MathemagicianMathemagician

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103116












  • $begingroup$
    Related: math.stackexchange.com/questions/2611084/…
    $endgroup$
    – user202729
    Jul 28 '18 at 13:34






  • 1




    $begingroup$
    (this one is more general. If the answer to this one is "no" then the answer to the other one is "no" as well)
    $endgroup$
    – user202729
    Jul 28 '18 at 13:35










  • $begingroup$
    Of course not, but I can't prove it. There are only countably many finite expressions of this sort, so you can only express countably many real numbers this way. Unless a real has some reason to be expressible it almost certainly can't be.
    $endgroup$
    – Ross Millikan
    Jul 28 '18 at 14:36






  • 1




    $begingroup$
    Yes, there are only countably many of them, so most real numbers do not fall into this category. Still, not much can be deduced about any specific number from that fact. π is arguably somewhat special.
    $endgroup$
    – Mathemagician
    Jul 28 '18 at 14:57


















  • $begingroup$
    Related: math.stackexchange.com/questions/2611084/…
    $endgroup$
    – user202729
    Jul 28 '18 at 13:34






  • 1




    $begingroup$
    (this one is more general. If the answer to this one is "no" then the answer to the other one is "no" as well)
    $endgroup$
    – user202729
    Jul 28 '18 at 13:35










  • $begingroup$
    Of course not, but I can't prove it. There are only countably many finite expressions of this sort, so you can only express countably many real numbers this way. Unless a real has some reason to be expressible it almost certainly can't be.
    $endgroup$
    – Ross Millikan
    Jul 28 '18 at 14:36






  • 1




    $begingroup$
    Yes, there are only countably many of them, so most real numbers do not fall into this category. Still, not much can be deduced about any specific number from that fact. π is arguably somewhat special.
    $endgroup$
    – Mathemagician
    Jul 28 '18 at 14:57
















$begingroup$
Related: math.stackexchange.com/questions/2611084/…
$endgroup$
– user202729
Jul 28 '18 at 13:34




$begingroup$
Related: math.stackexchange.com/questions/2611084/…
$endgroup$
– user202729
Jul 28 '18 at 13:34




1




1




$begingroup$
(this one is more general. If the answer to this one is "no" then the answer to the other one is "no" as well)
$endgroup$
– user202729
Jul 28 '18 at 13:35




$begingroup$
(this one is more general. If the answer to this one is "no" then the answer to the other one is "no" as well)
$endgroup$
– user202729
Jul 28 '18 at 13:35












$begingroup$
Of course not, but I can't prove it. There are only countably many finite expressions of this sort, so you can only express countably many real numbers this way. Unless a real has some reason to be expressible it almost certainly can't be.
$endgroup$
– Ross Millikan
Jul 28 '18 at 14:36




$begingroup$
Of course not, but I can't prove it. There are only countably many finite expressions of this sort, so you can only express countably many real numbers this way. Unless a real has some reason to be expressible it almost certainly can't be.
$endgroup$
– Ross Millikan
Jul 28 '18 at 14:36




1




1




$begingroup$
Yes, there are only countably many of them, so most real numbers do not fall into this category. Still, not much can be deduced about any specific number from that fact. π is arguably somewhat special.
$endgroup$
– Mathemagician
Jul 28 '18 at 14:57




$begingroup$
Yes, there are only countably many of them, so most real numbers do not fall into this category. Still, not much can be deduced about any specific number from that fact. π is arguably somewhat special.
$endgroup$
– Mathemagician
Jul 28 '18 at 14:57










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