Mixing
Show that the greatest common divisor $gcd(x,y)$ is primitive recursive.
$begingroup$
Let $x,y geq 2 $. Actually I have to show that $CD(x,y)$ is primitive recursive, where
$ C(x,y) = 1$ if $gcd(x,y)=1$, otherwise $CD(x,y) = 0$. But I have showed that it is enough to show that $gcd(x,y)$ is primitive recursive.
Which functions are allowed?
1) constant function
2)projection
3) $s(n) = n+1$
4) $mbox{add}(x,y)$
5) $mbox{mult}(x,y)$
6) $mbox{sub}(x,y)$
7) $x^y$
8) $mbox{sg}(x)$
9) absolute value
10) $mbox{fac}(x) := x!$
11) $ mod(N,m) := N mod m$
12) $ max(x_1,...,x_m) := max${$x_1,...,x_m$} for an arbitrary $m$, but fixed.
My idea:
It has something has to do with $11)$ and $12)$, I think. But remember $m$ in $12)$ has to be fixed. My first attempt: $ gcd(x,y) = max$ { $d in$ {$0,...,x$}$ | mod(x,d) = 0 , mod(y,d) = 0$ }
computability
asked Jan 14 at 19:12
RukiaKuchikiRukiaKuchiki
337211
$endgroup$
add a comment |
$begingroup$
Let $x,y geq 2 $. Actually I have to show that $CD(x,y)$ is primitive recursive, where
$ C(x,y) = 1$ if $gcd(x,y)=1$, otherwise $CD(x,y) = 0$. But I have showed that it is enough to show that $gcd(x,y)$ is primitive recursive.
Which functions are allowed?
1) constant function
2)projection
3) $s(n) = n+1$
4) $mbox{add}(x,y)$
5) $mbox{mult}(x,y)$
6) $mbox{sub}(x,y)$
7) $x^y$
8) $mbox{sg}(x)$
9) absolute value
10) $mbox{fac}(x) := x!$
11) $ mod(N,m) := N mod m$
12) $ max(x_1,...,x_m) := max${$x_1,...,x_m$} for an arbitrary $m$, but fixed.
My idea:
It has something has to do with $11)$ and $12)$, I think. But remember $m$ in $12)$ has to be fixed. My first attempt: $ gcd(x,y) = max$ { $d in$ {$0,...,x$}$ | mod(x,d) = 0 , mod(y,d) = 0$ }
computability
asked Jan 14 at 19:12
RukiaKuchikiRukiaKuchiki
337211
$endgroup$
add a comment |
$begingroup$
Let $x,y geq 2 $. Actually I have to show that $CD(x,y)$ is primitive recursive, where
$ C(x,y) = 1$ if $gcd(x,y)=1$, otherwise $CD(x,y) = 0$. But I have showed that it is enough to show that $gcd(x,y)$ is primitive recursive.
Which functions are allowed?
1) constant function
2)projection
3) $s(n) = n+1$
4) $mbox{add}(x,y)$
5) $mbox{mult}(x,y)$
6) $mbox{sub}(x,y)$
7) $x^y$
8) $mbox{sg}(x)$
9) absolute value
10) $mbox{fac}(x) := x!$
11) $ mod(N,m) := N mod m$
12) $ max(x_1,...,x_m) := max${$x_1,...,x_m$} for an arbitrary $m$, but fixed.
My idea:
It has something has to do with $11)$ and $12)$, I think. But remember $m$ in $12)$ has to be fixed. My first attempt: $ gcd(x,y) = max$ { $d in$ {$0,...,x$}$ | mod(x,d) = 0 , mod(y,d) = 0$ }
computability
asked Jan 14 at 19:12
RukiaKuchikiRukiaKuchiki
337211
$endgroup$
Let $x,y geq 2 $. Actually I have to show that $CD(x,y)$ is primitive recursive, where
$ C(x,y) = 1$ if $gcd(x,y)=1$, otherwise $CD(x,y) = 0$. But I have showed that it is enough to show that $gcd(x,y)$ is primitive recursive.
Which functions are allowed?
1) constant function
2)projection
3) $s(n) = n+1$
4) $mbox{add}(x,y)$
5) $mbox{mult}(x,y)$
6) $mbox{sub}(x,y)$
7) $x^y$
8) $mbox{sg}(x)$
9) absolute value
10) $mbox{fac}(x) := x!$
11) $ mod(N,m) := N mod m$
12) $ max(x_1,...,x_m) := max${$x_1,...,x_m$} for an arbitrary $m$, but fixed.
My idea:
It has something has to do with $11)$ and $12)$, I think. But remember $m$ in $12)$ has to be fixed. My first attempt: $ gcd(x,y) = max$ { $d in$ {$0,...,x$}$ | mod(x,d) = 0 , mod(y,d) = 0$ }
computability
computability
asked Jan 14 at 19:12
RukiaKuchikiRukiaKuchiki
337211
asked Jan 14 at 19:12
RukiaKuchikiRukiaKuchiki
337211
asked Jan 14 at 19:12
RukiaKuchikiRukiaKuchiki
337211
asked Jan 14 at 19:12
RukiaKuchikiRukiaKuchiki
337211
asked Jan 14 at 19:12
RukiaKuchikiRukiaKuchiki
337211
337211
add a comment |
add a comment |
1 Answer
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$begingroup$
Hint: The basic relation you need is divisibility:
$$D = {(x,y)in{Bbb N}_0^2mid x;text{divides}; y}.$$
$x$ divides $y$ if and only if $x·i = y$ for some $1 ≤ i ≤ y$. So the characteristic function of $D$ can be written as
$$chi_D(x, y) = text{sgn}[chi_=(x · 1, y) + chi_=(x · 2, y) + . . . + chi_=(x · y, y)],$$
where $chi_=$ is the (primitive recursive) characteristic function of the equality relation ${(x,x)mid xin{Bbb N}_0}$.
Since the characteristic function $chi_D$ is primitive recursive, the relation D is primitive.
answered Jan 16 at 14:17
WuestenfuxWuestenfux
4,7001413
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1 Answer
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1 Answer
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$begingroup$
Hint: The basic relation you need is divisibility:
$$D = {(x,y)in{Bbb N}_0^2mid x;text{divides}; y}.$$
$x$ divides $y$ if and only if $x·i = y$ for some $1 ≤ i ≤ y$. So the characteristic function of $D$ can be written as
$$chi_D(x, y) = text{sgn}[chi_=(x · 1, y) + chi_=(x · 2, y) + . . . + chi_=(x · y, y)],$$
where $chi_=$ is the (primitive recursive) characteristic function of the equality relation ${(x,x)mid xin{Bbb N}_0}$.
Since the characteristic function $chi_D$ is primitive recursive, the relation D is primitive.
answered Jan 16 at 14:17
WuestenfuxWuestenfux
4,7001413
$endgroup$
add a comment |
$begingroup$
Hint: The basic relation you need is divisibility:
$$D = {(x,y)in{Bbb N}_0^2mid x;text{divides}; y}.$$
$x$ divides $y$ if and only if $x·i = y$ for some $1 ≤ i ≤ y$. So the characteristic function of $D$ can be written as
$$chi_D(x, y) = text{sgn}[chi_=(x · 1, y) + chi_=(x · 2, y) + . . . + chi_=(x · y, y)],$$
where $chi_=$ is the (primitive recursive) characteristic function of the equality relation ${(x,x)mid xin{Bbb N}_0}$.
Since the characteristic function $chi_D$ is primitive recursive, the relation D is primitive.
answered Jan 16 at 14:17
WuestenfuxWuestenfux
4,7001413
$endgroup$
add a comment |
$begingroup$
Hint: The basic relation you need is divisibility:
$$D = {(x,y)in{Bbb N}_0^2mid x;text{divides}; y}.$$
$x$ divides $y$ if and only if $x·i = y$ for some $1 ≤ i ≤ y$. So the characteristic function of $D$ can be written as
$$chi_D(x, y) = text{sgn}[chi_=(x · 1, y) + chi_=(x · 2, y) + . . . + chi_=(x · y, y)],$$
where $chi_=$ is the (primitive recursive) characteristic function of the equality relation ${(x,x)mid xin{Bbb N}_0}$.
Since the characteristic function $chi_D$ is primitive recursive, the relation D is primitive.
answered Jan 16 at 14:17
WuestenfuxWuestenfux
4,7001413
$endgroup$
Hint: The basic relation you need is divisibility:
$$D = {(x,y)in{Bbb N}_0^2mid x;text{divides}; y}.$$
$x$ divides $y$ if and only if $x·i = y$ for some $1 ≤ i ≤ y$. So the characteristic function of $D$ can be written as
$$chi_D(x, y) = text{sgn}[chi_=(x · 1, y) + chi_=(x · 2, y) + . . . + chi_=(x · y, y)],$$
where $chi_=$ is the (primitive recursive) characteristic function of the equality relation ${(x,x)mid xin{Bbb N}_0}$.
Since the characteristic function $chi_D$ is primitive recursive, the relation D is primitive.
answered Jan 16 at 14:17
WuestenfuxWuestenfux
4,7001413
edited Jan 17 at 6:35
answered Jan 16 at 14:17
WuestenfuxWuestenfux
4,7001413
answered Jan 16 at 14:17
WuestenfuxWuestenfux
4,7001413
answered Jan 16 at 14:17
WuestenfuxWuestenfux
4,7001413
4,7001413
add a comment |
add a comment |
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