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Can you obtain $pi$ using elements of $mathbbN$, and finite number of basic arithmetic operations + exponentiation?
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Is it possible to obtain $pi$ from finite amount of operations
$+,-,cdot,div,wedge$ on $mathbbN$ (or $mathbbQ$, 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 $mathbbN$) ?
algebra-precalculus
edited Jul 28 at 14:24
Daniel Buck
2,2841623
asked Jul 28 at 13:29
Mathemagician
785
add a comment |Â
up vote
7
down vote
favorite
Is it possible to obtain $pi$ from finite amount of operations
$+,-,cdot,div,wedge$ on $mathbbN$ (or $mathbbQ$, 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 $mathbbN$) ?
algebra-precalculus
edited Jul 28 at 14:24
Daniel Buck
2,2841623
asked Jul 28 at 13:29
Mathemagician
785
Related: math.stackexchange.com/questions/2611084/…
– user202729
Jul 28 at 13:34
1
(this one is more general. If the answer to this one is "no" then the answer to the other one is "no" as well)
– user202729
Jul 28 at 13:35
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.
– Ross Millikan
Jul 28 at 14:36
1
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. À is arguably somewhat special.
– Mathemagician
Jul 28 at 14:57
add a comment |Â
up vote
7
down vote
favorite
up vote
7
down vote
favorite
Is it possible to obtain $pi$ from finite amount of operations
$+,-,cdot,div,wedge$ on $mathbbN$ (or $mathbbQ$, 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 $mathbbN$) ?
algebra-precalculus
edited Jul 28 at 14:24
Daniel Buck
2,2841623
asked Jul 28 at 13:29
Mathemagician
785
Is it possible to obtain $pi$ from finite amount of operations
$+,-,cdot,div,wedge$ on $mathbbN$ (or $mathbbQ$, 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 $mathbbN$) ?
algebra-precalculus
edited Jul 28 at 14:24
Daniel Buck
2,2841623
asked Jul 28 at 13:29
Mathemagician
785
edited Jul 28 at 14:24
Daniel Buck
2,2841623
edited Jul 28 at 14:24
Daniel Buck
2,2841623
edited Jul 28 at 14:24
Daniel Buck
2,2841623
2,2841623
asked Jul 28 at 13:29
Mathemagician
785
asked Jul 28 at 13:29
Mathemagician
785
asked Jul 28 at 13:29
Mathemagician
785
785
Related: math.stackexchange.com/questions/2611084/…
– user202729
Jul 28 at 13:34
1
(this one is more general. If the answer to this one is "no" then the answer to the other one is "no" as well)
– user202729
Jul 28 at 13:35
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.
– Ross Millikan
Jul 28 at 14:36
1
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. À is arguably somewhat special.
– Mathemagician
Jul 28 at 14:57
add a comment |Â
Related: math.stackexchange.com/questions/2611084/…
– user202729
Jul 28 at 13:34
1
(this one is more general. If the answer to this one is "no" then the answer to the other one is "no" as well)
– user202729
Jul 28 at 13:35
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.
– Ross Millikan
Jul 28 at 14:36
1
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. À is arguably somewhat special.
– Mathemagician
Jul 28 at 14:57
Related: math.stackexchange.com/questions/2611084/…
– user202729
Jul 28 at 13:34
Related: math.stackexchange.com/questions/2611084/…
– user202729
Jul 28 at 13:34
1
1
(this one is more general. If the answer to this one is "no" then the answer to the other one is "no" as well)
– user202729
Jul 28 at 13:35
(this one is more general. If the answer to this one is "no" then the answer to the other one is "no" as well)
– user202729
Jul 28 at 13:35
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.
– Ross Millikan
Jul 28 at 14:36
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.
– Ross Millikan
Jul 28 at 14:36
1
1
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. À is arguably somewhat special.
– Mathemagician
Jul 28 at 14:57
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. À is arguably somewhat special.
– Mathemagician
Jul 28 at 14:57
add a comment |Â
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Related: math.stackexchange.com/questions/2611084/…
– user202729
Jul 28 at 13:34
1
(this one is more general. If the answer to this one is "no" then the answer to the other one is "no" as well)
– user202729
Jul 28 at 13:35
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.
– Ross Millikan
Jul 28 at 14:36
1
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. À is arguably somewhat special.
– Mathemagician
Jul 28 at 14:57