First-order logic formula(prime numbers)
$begingroup$
How to write into a first-order logic formula:
1) $m$ is prime number, which consists in $[sqrt{n},n]$
2) $n$ is number of second power of prime number.
$textbf{My work:}$
Prime number can be written like: $prime(x) = 1<x, wedge, forall u, v (x = u cdot v rightarrow u = 1 vee v =1)$.
2) I think that in this case it can be written like $n = 1<n, wedge, forall u, v (n = ucdot u cdot v cdot v rightarrow u = 1 vee v =1)$, but Im not sure
1) Here I have no idea, how to do it...
logic first-order-logic
asked Dec 10 '18 at 20:20
Aleksandra Aleksandra
545
$endgroup$
add a comment |
$begingroup$
How to write into a first-order logic formula:
1) $m$ is prime number, which consists in $[sqrt{n},n]$
2) $n$ is number of second power of prime number.
$textbf{My work:}$
Prime number can be written like: $prime(x) = 1<x, wedge, forall u, v (x = u cdot v rightarrow u = 1 vee v =1)$.
2) I think that in this case it can be written like $n = 1<n, wedge, forall u, v (n = ucdot u cdot v cdot v rightarrow u = 1 vee v =1)$, but Im not sure
1) Here I have no idea, how to do it...
logic first-order-logic
asked Dec 10 '18 at 20:20
Aleksandra Aleksandra
545
$endgroup$
add a comment |
$begingroup$
How to write into a first-order logic formula:
1) $m$ is prime number, which consists in $[sqrt{n},n]$
2) $n$ is number of second power of prime number.
$textbf{My work:}$
Prime number can be written like: $prime(x) = 1<x, wedge, forall u, v (x = u cdot v rightarrow u = 1 vee v =1)$.
2) I think that in this case it can be written like $n = 1<n, wedge, forall u, v (n = ucdot u cdot v cdot v rightarrow u = 1 vee v =1)$, but Im not sure
1) Here I have no idea, how to do it...
logic first-order-logic
asked Dec 10 '18 at 20:20
Aleksandra Aleksandra
545
$endgroup$
How to write into a first-order logic formula:
1) $m$ is prime number, which consists in $[sqrt{n},n]$
2) $n$ is number of second power of prime number.
$textbf{My work:}$
Prime number can be written like: $prime(x) = 1<x, wedge, forall u, v (x = u cdot v rightarrow u = 1 vee v =1)$.
2) I think that in this case it can be written like $n = 1<n, wedge, forall u, v (n = ucdot u cdot v cdot v rightarrow u = 1 vee v =1)$, but Im not sure
1) Here I have no idea, how to do it...
logic first-order-logic
logic first-order-logic
asked Dec 10 '18 at 20:20
Aleksandra Aleksandra
545
asked Dec 10 '18 at 20:20
Aleksandra Aleksandra
545
asked Dec 10 '18 at 20:20
Aleksandra Aleksandra
545
asked Dec 10 '18 at 20:20
Aleksandra Aleksandra
545
asked Dec 10 '18 at 20:20
Aleksandra Aleksandra
545
545
add a comment |
add a comment |
1 Answer
1
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$begingroup$
Hints:
For (1), $m in [sqrt{n}, n]$ iff $n le m^2 le n^2$.
For (2), it is much simpler to use an existential quantifier to express the property that there exists a prime, $p$, such that $n = p^2$. Your suggestion involves a neat idea but it doesn't quite work, e.g., your predicate holds if $n$ is not a square, because then there are no $u$ and $v$ such that $n = u^2v^2$.
answered Dec 10 '18 at 20:39
Rob ArthanRob Arthan
29.3k42966
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add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
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active
oldest
votes
$begingroup$
Hints:
For (1), $m in [sqrt{n}, n]$ iff $n le m^2 le n^2$.
For (2), it is much simpler to use an existential quantifier to express the property that there exists a prime, $p$, such that $n = p^2$. Your suggestion involves a neat idea but it doesn't quite work, e.g., your predicate holds if $n$ is not a square, because then there are no $u$ and $v$ such that $n = u^2v^2$.
answered Dec 10 '18 at 20:39
Rob ArthanRob Arthan
29.3k42966
$endgroup$
add a comment |
$begingroup$
Hints:
For (1), $m in [sqrt{n}, n]$ iff $n le m^2 le n^2$.
For (2), it is much simpler to use an existential quantifier to express the property that there exists a prime, $p$, such that $n = p^2$. Your suggestion involves a neat idea but it doesn't quite work, e.g., your predicate holds if $n$ is not a square, because then there are no $u$ and $v$ such that $n = u^2v^2$.
answered Dec 10 '18 at 20:39
Rob ArthanRob Arthan
29.3k42966
$endgroup$
add a comment |
$begingroup$
Hints:
For (1), $m in [sqrt{n}, n]$ iff $n le m^2 le n^2$.
For (2), it is much simpler to use an existential quantifier to express the property that there exists a prime, $p$, such that $n = p^2$. Your suggestion involves a neat idea but it doesn't quite work, e.g., your predicate holds if $n$ is not a square, because then there are no $u$ and $v$ such that $n = u^2v^2$.
answered Dec 10 '18 at 20:39
Rob ArthanRob Arthan
29.3k42966
$endgroup$
Hints:
For (1), $m in [sqrt{n}, n]$ iff $n le m^2 le n^2$.
For (2), it is much simpler to use an existential quantifier to express the property that there exists a prime, $p$, such that $n = p^2$. Your suggestion involves a neat idea but it doesn't quite work, e.g., your predicate holds if $n$ is not a square, because then there are no $u$ and $v$ such that $n = u^2v^2$.
answered Dec 10 '18 at 20:39
Rob ArthanRob Arthan
29.3k42966
answered Dec 10 '18 at 20:39
Rob ArthanRob Arthan
29.3k42966
answered Dec 10 '18 at 20:39
Rob ArthanRob Arthan
29.3k42966
answered Dec 10 '18 at 20:39
Rob ArthanRob Arthan
29.3k42966
29.3k42966
add a comment |
add a comment |
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