Mixing
How do we mathematically define the meaning of the word “undecidable”?
$begingroup$
I need to understand the meaning of this mathematical concept: "undecided/undecidable".
I know what it means in the English dictionary. But, I don't know what it means mathematically.
If You answer this question with possible mathematical examples, it will be very helpful to understand this issue.
Thank you very much!
soft-question math-history
edited Jan 15 at 15:50
Mauro ALLEGRANZA
66.2k449114
asked Jan 14 at 22:31
ElementaryElementary
354111
$endgroup$
add a comment |
$begingroup$
I need to understand the meaning of this mathematical concept: "undecided/undecidable".
I know what it means in the English dictionary. But, I don't know what it means mathematically.
If You answer this question with possible mathematical examples, it will be very helpful to understand this issue.
Thank you very much!
soft-question math-history
edited Jan 15 at 15:50
Mauro ALLEGRANZA
66.2k449114
asked Jan 14 at 22:31
ElementaryElementary
354111
$endgroup$
$begingroup$
See en.wikipedia.org/wiki/MU_puzzle for an example of a simple system which has an undecidable statement.
$endgroup$
– B. Goddard
Jan 14 at 22:37
$begingroup$
How will you decide between the various answers? Are they undecidable?
$endgroup$
– Michael
Jan 15 at 0:14
$begingroup$
@Michael Although people did not like the question, they gave various answers. I upvoted all anwers. By the way, I'm undecided now.
$endgroup$
– Elementary
Jan 15 at 9:09
$begingroup$
@Beginner : Sorry, I saw your page only to give my silly comment but I didn't think to upvote. I have upvoted now, I think the various upvoted answers means there was interest so you deserve a few upvotes for the question itself.
$endgroup$
– Michael
Jan 15 at 14:59
$begingroup$
@Michael You're being very friendly. And Thank you very much for upvote..
$endgroup$
– Elementary
Jan 15 at 16:00
add a comment |
$begingroup$
I need to understand the meaning of this mathematical concept: "undecided/undecidable".
I know what it means in the English dictionary. But, I don't know what it means mathematically.
If You answer this question with possible mathematical examples, it will be very helpful to understand this issue.
Thank you very much!
soft-question math-history
edited Jan 15 at 15:50
Mauro ALLEGRANZA
66.2k449114
asked Jan 14 at 22:31
ElementaryElementary
354111
$endgroup$
I need to understand the meaning of this mathematical concept: "undecided/undecidable".
I know what it means in the English dictionary. But, I don't know what it means mathematically.
If You answer this question with possible mathematical examples, it will be very helpful to understand this issue.
Thank you very much!
soft-question math-history
soft-question math-history
edited Jan 15 at 15:50
Mauro ALLEGRANZA
66.2k449114
asked Jan 14 at 22:31
ElementaryElementary
354111
edited Jan 15 at 15:50
Mauro ALLEGRANZA
66.2k449114
asked Jan 14 at 22:31
ElementaryElementary
354111
edited Jan 15 at 15:50
Mauro ALLEGRANZA
66.2k449114
edited Jan 15 at 15:50
Mauro ALLEGRANZA
66.2k449114
edited Jan 15 at 15:50
Mauro ALLEGRANZA
66.2k449114
66.2k449114
asked Jan 14 at 22:31
ElementaryElementary
354111
asked Jan 14 at 22:31
ElementaryElementary
354111
asked Jan 14 at 22:31
ElementaryElementary
354111
354111
$begingroup$
See en.wikipedia.org/wiki/MU_puzzle for an example of a simple system which has an undecidable statement.
$endgroup$
– B. Goddard
Jan 14 at 22:37
$begingroup$
How will you decide between the various answers? Are they undecidable?
$endgroup$
– Michael
Jan 15 at 0:14
$begingroup$
@Michael Although people did not like the question, they gave various answers. I upvoted all anwers. By the way, I'm undecided now.
$endgroup$
– Elementary
Jan 15 at 9:09
$begingroup$
@Beginner : Sorry, I saw your page only to give my silly comment but I didn't think to upvote. I have upvoted now, I think the various upvoted answers means there was interest so you deserve a few upvotes for the question itself.
$endgroup$
– Michael
Jan 15 at 14:59
$begingroup$
@Michael You're being very friendly. And Thank you very much for upvote..
$endgroup$
– Elementary
Jan 15 at 16:00
add a comment |
$begingroup$
See en.wikipedia.org/wiki/MU_puzzle for an example of a simple system which has an undecidable statement.
$endgroup$
– B. Goddard
Jan 14 at 22:37
$begingroup$
How will you decide between the various answers? Are they undecidable?
$endgroup$
– Michael
Jan 15 at 0:14
$begingroup$
@Michael Although people did not like the question, they gave various answers. I upvoted all anwers. By the way, I'm undecided now.
$endgroup$
– Elementary
Jan 15 at 9:09
$begingroup$
@Beginner : Sorry, I saw your page only to give my silly comment but I didn't think to upvote. I have upvoted now, I think the various upvoted answers means there was interest so you deserve a few upvotes for the question itself.
$endgroup$
– Michael
Jan 15 at 14:59
$begingroup$
@Michael You're being very friendly. And Thank you very much for upvote..
$endgroup$
– Elementary
Jan 15 at 16:00
$begingroup$
See en.wikipedia.org/wiki/MU_puzzle for an example of a simple system which has an undecidable statement.
$endgroup$
– B. Goddard
Jan 14 at 22:37
$begingroup$
See en.wikipedia.org/wiki/MU_puzzle for an example of a simple system which has an undecidable statement.
$endgroup$
– B. Goddard
Jan 14 at 22:37
$begingroup$
How will you decide between the various answers? Are they undecidable?
$endgroup$
– Michael
Jan 15 at 0:14
$begingroup$
How will you decide between the various answers? Are they undecidable?
$endgroup$
– Michael
Jan 15 at 0:14
$begingroup$
@Michael Although people did not like the question, they gave various answers. I upvoted all anwers. By the way, I'm undecided now.
$endgroup$
– Elementary
Jan 15 at 9:09
$begingroup$
@Michael Although people did not like the question, they gave various answers. I upvoted all anwers. By the way, I'm undecided now.
$endgroup$
– Elementary
Jan 15 at 9:09
$begingroup$
@Beginner : Sorry, I saw your page only to give my silly comment but I didn't think to upvote. I have upvoted now, I think the various upvoted answers means there was interest so you deserve a few upvotes for the question itself.
$endgroup$
– Michael
Jan 15 at 14:59
$begingroup$
@Beginner : Sorry, I saw your page only to give my silly comment but I didn't think to upvote. I have upvoted now, I think the various upvoted answers means there was interest so you deserve a few upvotes for the question itself.
$endgroup$
– Michael
Jan 15 at 14:59
$begingroup$
@Michael You're being very friendly. And Thank you very much for upvote..
$endgroup$
– Elementary
Jan 15 at 16:00
$begingroup$
@Michael You're being very friendly. And Thank you very much for upvote..
$endgroup$
– Elementary
Jan 15 at 16:00
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
Given a set of axioms, a statement is undecidable if neither it nor its negation follow from the axioms.
Example:
If your only axiom is:
$$
forall z forall x forall y (y=x) vee(y=z)vee (x=z)
$$
(in English, "for any three things, two of them are equal")
then it is undecidable whether
$$
forall x forall y (x=y)
$$
(English, "there is only one thing.")
By contrast, it is decidable (and false) that
$$
exists x exists y exists z (x neq y) wedge (xneq z) wedge (yneq z).
$$
(English, "there exist three distinct things.")
answered Jan 14 at 22:39
hunterhunter
15k32540
$endgroup$
add a comment |
$begingroup$
Saying that a statement is "undecidable" means that there can be no proof, even theoretically, that the true nor can there be a proof that it is false.
answered Jan 14 at 22:34
user247327user247327
11.1k1515
$endgroup$
add a comment |
$begingroup$
If a proposition $p$ can be stated in the language of a theory $T$, we say $p$ is undecidable in $T$ if $T$ contains neither a proof nor a disproof of $p$.
answered Jan 14 at 22:35
J.G.J.G.
27.1k22843
$endgroup$
add a comment |
$begingroup$
A statement $P$ is undecidable in an theory $T$ if $t cup {P}$ and $t cup {neg P}$ are both consistent. In practice, it's usually used more broadly: $P$ is undecidable in $T$ if $T cup{P}$ and $Tcup{neg P}$ are equiconsistent, becuase Gödel's Second Incompleteness Theorem is annoying that way: specifically, for most systems that we're interested in, we can't prove that the system is consistent (or, more precisely, we can't prove it in that system unless the system is inconsistent). Thus, rather than requiring two systems to be consistent (which we can't prove), we instead require that at least there's no consistency reason to prefer one over the other.
answered Jan 14 at 22:40
user3482749user3482749
4,266919
$endgroup$
add a comment |
$begingroup$
Suppose you have some logical system $S$ equipped with some initial axioms and some process for generating statements from previous statements. Suppose you have a set $T$ of statements such that each statement is either an axiom in $S$, or can be derived from other statements in $T$ using the process given by $S$ for generating statements (note that $T$ is a recursively defined set, in that each statement in $T$ is either an axiom, or a statement that can be derived from axioms, or a statement that can be derived from statements that can be derived from axioms, etc.) A statement is undecidable in $S$ if neither the statement nor its negation is in $T$.
Note that undecidability is a property of a statement and a logical system. A statement might be decidable in one logical system, but undecidable in another. For example, the Axiom of Choice is undecidable in Zermelo–Fraenkel set theory, but is taken as an axiom in ZFC (that is what the C stands for), and thus is decidable in ZFC.
answered Jan 14 at 22:49
AcccumulationAcccumulation
6,9892618
$endgroup$
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Given a set of axioms, a statement is undecidable if neither it nor its negation follow from the axioms.
Example:
If your only axiom is:
$$
forall z forall x forall y (y=x) vee(y=z)vee (x=z)
$$
(in English, "for any three things, two of them are equal")
then it is undecidable whether
$$
forall x forall y (x=y)
$$
(English, "there is only one thing.")
By contrast, it is decidable (and false) that
$$
exists x exists y exists z (x neq y) wedge (xneq z) wedge (yneq z).
$$
(English, "there exist three distinct things.")
answered Jan 14 at 22:39
hunterhunter
15k32540
$endgroup$
add a comment |
$begingroup$
Given a set of axioms, a statement is undecidable if neither it nor its negation follow from the axioms.
Example:
If your only axiom is:
$$
forall z forall x forall y (y=x) vee(y=z)vee (x=z)
$$
(in English, "for any three things, two of them are equal")
then it is undecidable whether
$$
forall x forall y (x=y)
$$
(English, "there is only one thing.")
By contrast, it is decidable (and false) that
$$
exists x exists y exists z (x neq y) wedge (xneq z) wedge (yneq z).
$$
(English, "there exist three distinct things.")
answered Jan 14 at 22:39
hunterhunter
15k32540
$endgroup$
add a comment |
$begingroup$
Given a set of axioms, a statement is undecidable if neither it nor its negation follow from the axioms.
Example:
If your only axiom is:
$$
forall z forall x forall y (y=x) vee(y=z)vee (x=z)
$$
(in English, "for any three things, two of them are equal")
then it is undecidable whether
$$
forall x forall y (x=y)
$$
(English, "there is only one thing.")
By contrast, it is decidable (and false) that
$$
exists x exists y exists z (x neq y) wedge (xneq z) wedge (yneq z).
$$
(English, "there exist three distinct things.")
answered Jan 14 at 22:39
hunterhunter
15k32540
$endgroup$
Given a set of axioms, a statement is undecidable if neither it nor its negation follow from the axioms.
Example:
If your only axiom is:
$$
forall z forall x forall y (y=x) vee(y=z)vee (x=z)
$$
(in English, "for any three things, two of them are equal")
then it is undecidable whether
$$
forall x forall y (x=y)
$$
(English, "there is only one thing.")
By contrast, it is decidable (and false) that
$$
exists x exists y exists z (x neq y) wedge (xneq z) wedge (yneq z).
$$
(English, "there exist three distinct things.")
answered Jan 14 at 22:39
hunterhunter
15k32540
edited Jan 18 at 1:42
answered Jan 14 at 22:39
hunterhunter
15k32540
answered Jan 14 at 22:39
hunterhunter
15k32540
answered Jan 14 at 22:39
hunterhunter
15k32540
15k32540
add a comment |
add a comment |
$begingroup$
Saying that a statement is "undecidable" means that there can be no proof, even theoretically, that the true nor can there be a proof that it is false.
answered Jan 14 at 22:34
user247327user247327
11.1k1515
$endgroup$
add a comment |
$begingroup$
Saying that a statement is "undecidable" means that there can be no proof, even theoretically, that the true nor can there be a proof that it is false.
answered Jan 14 at 22:34
user247327user247327
11.1k1515
$endgroup$
add a comment |
$begingroup$
Saying that a statement is "undecidable" means that there can be no proof, even theoretically, that the true nor can there be a proof that it is false.
answered Jan 14 at 22:34
user247327user247327
11.1k1515
$endgroup$
Saying that a statement is "undecidable" means that there can be no proof, even theoretically, that the true nor can there be a proof that it is false.
answered Jan 14 at 22:34
user247327user247327
11.1k1515
answered Jan 14 at 22:34
user247327user247327
11.1k1515
answered Jan 14 at 22:34
user247327user247327
11.1k1515
answered Jan 14 at 22:34
user247327user247327
11.1k1515
11.1k1515
add a comment |
add a comment |
$begingroup$
If a proposition $p$ can be stated in the language of a theory $T$, we say $p$ is undecidable in $T$ if $T$ contains neither a proof nor a disproof of $p$.
answered Jan 14 at 22:35
J.G.J.G.
27.1k22843
$endgroup$
add a comment |
$begingroup$
If a proposition $p$ can be stated in the language of a theory $T$, we say $p$ is undecidable in $T$ if $T$ contains neither a proof nor a disproof of $p$.
answered Jan 14 at 22:35
J.G.J.G.
27.1k22843
$endgroup$
add a comment |
$begingroup$
If a proposition $p$ can be stated in the language of a theory $T$, we say $p$ is undecidable in $T$ if $T$ contains neither a proof nor a disproof of $p$.
answered Jan 14 at 22:35
J.G.J.G.
27.1k22843
$endgroup$
If a proposition $p$ can be stated in the language of a theory $T$, we say $p$ is undecidable in $T$ if $T$ contains neither a proof nor a disproof of $p$.
answered Jan 14 at 22:35
J.G.J.G.
27.1k22843
answered Jan 14 at 22:35
J.G.J.G.
27.1k22843
answered Jan 14 at 22:35
J.G.J.G.
27.1k22843
answered Jan 14 at 22:35
J.G.J.G.
27.1k22843
27.1k22843
add a comment |
add a comment |
$begingroup$
A statement $P$ is undecidable in an theory $T$ if $t cup {P}$ and $t cup {neg P}$ are both consistent. In practice, it's usually used more broadly: $P$ is undecidable in $T$ if $T cup{P}$ and $Tcup{neg P}$ are equiconsistent, becuase Gödel's Second Incompleteness Theorem is annoying that way: specifically, for most systems that we're interested in, we can't prove that the system is consistent (or, more precisely, we can't prove it in that system unless the system is inconsistent). Thus, rather than requiring two systems to be consistent (which we can't prove), we instead require that at least there's no consistency reason to prefer one over the other.
answered Jan 14 at 22:40
user3482749user3482749
4,266919
$endgroup$
add a comment |
$begingroup$
A statement $P$ is undecidable in an theory $T$ if $t cup {P}$ and $t cup {neg P}$ are both consistent. In practice, it's usually used more broadly: $P$ is undecidable in $T$ if $T cup{P}$ and $Tcup{neg P}$ are equiconsistent, becuase Gödel's Second Incompleteness Theorem is annoying that way: specifically, for most systems that we're interested in, we can't prove that the system is consistent (or, more precisely, we can't prove it in that system unless the system is inconsistent). Thus, rather than requiring two systems to be consistent (which we can't prove), we instead require that at least there's no consistency reason to prefer one over the other.
answered Jan 14 at 22:40
user3482749user3482749
4,266919
$endgroup$
add a comment |
$begingroup$
A statement $P$ is undecidable in an theory $T$ if $t cup {P}$ and $t cup {neg P}$ are both consistent. In practice, it's usually used more broadly: $P$ is undecidable in $T$ if $T cup{P}$ and $Tcup{neg P}$ are equiconsistent, becuase Gödel's Second Incompleteness Theorem is annoying that way: specifically, for most systems that we're interested in, we can't prove that the system is consistent (or, more precisely, we can't prove it in that system unless the system is inconsistent). Thus, rather than requiring two systems to be consistent (which we can't prove), we instead require that at least there's no consistency reason to prefer one over the other.
answered Jan 14 at 22:40
user3482749user3482749
4,266919
$endgroup$
A statement $P$ is undecidable in an theory $T$ if $t cup {P}$ and $t cup {neg P}$ are both consistent. In practice, it's usually used more broadly: $P$ is undecidable in $T$ if $T cup{P}$ and $Tcup{neg P}$ are equiconsistent, becuase Gödel's Second Incompleteness Theorem is annoying that way: specifically, for most systems that we're interested in, we can't prove that the system is consistent (or, more precisely, we can't prove it in that system unless the system is inconsistent). Thus, rather than requiring two systems to be consistent (which we can't prove), we instead require that at least there's no consistency reason to prefer one over the other.
answered Jan 14 at 22:40
user3482749user3482749
4,266919
answered Jan 14 at 22:40
user3482749user3482749
4,266919
answered Jan 14 at 22:40
user3482749user3482749
4,266919
answered Jan 14 at 22:40
user3482749user3482749
4,266919
4,266919
add a comment |
add a comment |
$begingroup$
Suppose you have some logical system $S$ equipped with some initial axioms and some process for generating statements from previous statements. Suppose you have a set $T$ of statements such that each statement is either an axiom in $S$, or can be derived from other statements in $T$ using the process given by $S$ for generating statements (note that $T$ is a recursively defined set, in that each statement in $T$ is either an axiom, or a statement that can be derived from axioms, or a statement that can be derived from statements that can be derived from axioms, etc.) A statement is undecidable in $S$ if neither the statement nor its negation is in $T$.
Note that undecidability is a property of a statement and a logical system. A statement might be decidable in one logical system, but undecidable in another. For example, the Axiom of Choice is undecidable in Zermelo–Fraenkel set theory, but is taken as an axiom in ZFC (that is what the C stands for), and thus is decidable in ZFC.
answered Jan 14 at 22:49
AcccumulationAcccumulation
6,9892618
$endgroup$
add a comment |
$begingroup$
Suppose you have some logical system $S$ equipped with some initial axioms and some process for generating statements from previous statements. Suppose you have a set $T$ of statements such that each statement is either an axiom in $S$, or can be derived from other statements in $T$ using the process given by $S$ for generating statements (note that $T$ is a recursively defined set, in that each statement in $T$ is either an axiom, or a statement that can be derived from axioms, or a statement that can be derived from statements that can be derived from axioms, etc.) A statement is undecidable in $S$ if neither the statement nor its negation is in $T$.
Note that undecidability is a property of a statement and a logical system. A statement might be decidable in one logical system, but undecidable in another. For example, the Axiom of Choice is undecidable in Zermelo–Fraenkel set theory, but is taken as an axiom in ZFC (that is what the C stands for), and thus is decidable in ZFC.
answered Jan 14 at 22:49
AcccumulationAcccumulation
6,9892618
$endgroup$
add a comment |
$begingroup$
Suppose you have some logical system $S$ equipped with some initial axioms and some process for generating statements from previous statements. Suppose you have a set $T$ of statements such that each statement is either an axiom in $S$, or can be derived from other statements in $T$ using the process given by $S$ for generating statements (note that $T$ is a recursively defined set, in that each statement in $T$ is either an axiom, or a statement that can be derived from axioms, or a statement that can be derived from statements that can be derived from axioms, etc.) A statement is undecidable in $S$ if neither the statement nor its negation is in $T$.
Note that undecidability is a property of a statement and a logical system. A statement might be decidable in one logical system, but undecidable in another. For example, the Axiom of Choice is undecidable in Zermelo–Fraenkel set theory, but is taken as an axiom in ZFC (that is what the C stands for), and thus is decidable in ZFC.
answered Jan 14 at 22:49
AcccumulationAcccumulation
6,9892618
$endgroup$
Suppose you have some logical system $S$ equipped with some initial axioms and some process for generating statements from previous statements. Suppose you have a set $T$ of statements such that each statement is either an axiom in $S$, or can be derived from other statements in $T$ using the process given by $S$ for generating statements (note that $T$ is a recursively defined set, in that each statement in $T$ is either an axiom, or a statement that can be derived from axioms, or a statement that can be derived from statements that can be derived from axioms, etc.) A statement is undecidable in $S$ if neither the statement nor its negation is in $T$.
Note that undecidability is a property of a statement and a logical system. A statement might be decidable in one logical system, but undecidable in another. For example, the Axiom of Choice is undecidable in Zermelo–Fraenkel set theory, but is taken as an axiom in ZFC (that is what the C stands for), and thus is decidable in ZFC.
answered Jan 14 at 22:49
AcccumulationAcccumulation
6,9892618
answered Jan 14 at 22:49
AcccumulationAcccumulation
6,9892618
answered Jan 14 at 22:49
AcccumulationAcccumulation
6,9892618
answered Jan 14 at 22:49
AcccumulationAcccumulation
6,9892618
6,9892618
add a comment |
add a comment |
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See en.wikipedia.org/wiki/MU_puzzle for an example of a simple system which has an undecidable statement.
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– B. Goddard
Jan 14 at 22:37
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How will you decide between the various answers? Are they undecidable?
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– Michael
Jan 15 at 0:14
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@Michael Although people did not like the question, they gave various answers. I upvoted all anwers. By the way, I'm undecided now.
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– Elementary
Jan 15 at 9:09
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@Beginner : Sorry, I saw your page only to give my silly comment but I didn't think to upvote. I have upvoted now, I think the various upvoted answers means there was interest so you deserve a few upvotes for the question itself.
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– Michael
Jan 15 at 14:59
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@Michael You're being very friendly. And Thank you very much for upvote..
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– Elementary
Jan 15 at 16:00