Prove the index of the sum of any two computable numbers is computable
$begingroup$
Define a real number $alpha$ to be computable if there is a computable total function $f_alpha$ that, given any rational $epsilon$, yields a rational within $epsilon$-vicinity of $alpha$.
Now, assume some principal universal function $U = U(n, x)$ for the class of computable functions. This $U$ generates a certain numbering of computable real numbers: for each computable $alpha$, its corresponding number is any $n$ such that $U_n = f_alpha$ (note that every $alpha$ has infintely many assigned numbers as per Rice's theorem).
Given that, how does one prove the existence of an algorithm that, given any two numbers $n, m$ assigned to any two computable reals $alpha, beta$, produces some number that's assigned to their sum $alpha + beta$?
So this is a sketch of a proof that came to my mind after writing down this question.
It can be shown that there exists a computable bijection between $mathbb{N}^2$ and $mathbb{N}$, so let's denote $[i, j]$ for the natural that corresponds to the pair $(i, j)$ in that bijection. Now define a binary function $F$ such that $F([i, j], epsilon) = U(i, epsilon/2) + U(j, epsilon/2)$. Note that $F$ is clearly computable, and if $i, j$ are indices of some computable reals, then $F_{[i, j]}$ is the function corresponding to their sum.
Since $U$ is principal and $F$ is computable, there exists a computable total $s_F$ such that $forall n, x : F(n, x) = U(s_F(n), x)$. Combining that with the above, we get that $A(i, j) = s_F([i, j])$ is precisely the algorithm that, given two indices of computable reals, produces an index of their sum.
Does it sound reasonable?
proof-verification logic computability
$endgroup$
add a comment |
$begingroup$
Define a real number $alpha$ to be computable if there is a computable total function $f_alpha$ that, given any rational $epsilon$, yields a rational within $epsilon$-vicinity of $alpha$.
Now, assume some principal universal function $U = U(n, x)$ for the class of computable functions. This $U$ generates a certain numbering of computable real numbers: for each computable $alpha$, its corresponding number is any $n$ such that $U_n = f_alpha$ (note that every $alpha$ has infintely many assigned numbers as per Rice's theorem).
Given that, how does one prove the existence of an algorithm that, given any two numbers $n, m$ assigned to any two computable reals $alpha, beta$, produces some number that's assigned to their sum $alpha + beta$?
So this is a sketch of a proof that came to my mind after writing down this question.
It can be shown that there exists a computable bijection between $mathbb{N}^2$ and $mathbb{N}$, so let's denote $[i, j]$ for the natural that corresponds to the pair $(i, j)$ in that bijection. Now define a binary function $F$ such that $F([i, j], epsilon) = U(i, epsilon/2) + U(j, epsilon/2)$. Note that $F$ is clearly computable, and if $i, j$ are indices of some computable reals, then $F_{[i, j]}$ is the function corresponding to their sum.
Since $U$ is principal and $F$ is computable, there exists a computable total $s_F$ such that $forall n, x : F(n, x) = U(s_F(n), x)$. Combining that with the above, we get that $A(i, j) = s_F([i, j])$ is precisely the algorithm that, given two indices of computable reals, produces an index of their sum.
Does it sound reasonable?
proof-verification logic computability
$endgroup$
$begingroup$
The result follows from examining the usual proof in real analysis that $lim (a_n + b_n) = lim a_n + lim b_n$, using an $epsilon/2$ argument.
$endgroup$
– Carl Mummert
Dec 3 '18 at 18:47
$begingroup$
Indeed, I was able to prove that $alpha + beta$ is computable using that argument. But how do you prove that its index (according to the definitions above) is also computable?
$endgroup$
– 0xd34df00d
Dec 3 '18 at 18:50
$begingroup$
In other words, let $f_alpha$ and $f_beta$ be the corresponding functions that produce the approximations of $alpha$ and $beta$ respectively. The function $f(epsilon) = f_alpha(epsilon/2) + f_beta(epsilon/2)$ is computable and provides the required approximations for their sum, but I'm not sure how to find its $U$-index given $U$-indices of $f_alpha$ and $f_beta$.
$endgroup$
– 0xd34df00d
Dec 3 '18 at 18:54
2
$begingroup$
In programming terms, the problem is asking you how to write a program that represents $alpha + beta$ given programs that represent $alpha$ and $beta$. To do this you need to know what programming language is being used. The choice of $U$ corresponds to the choice of programming language so your answer will depend on how $U$ is defined.
$endgroup$
– Rob Arthan
Dec 3 '18 at 20:26
add a comment |
$begingroup$
Define a real number $alpha$ to be computable if there is a computable total function $f_alpha$ that, given any rational $epsilon$, yields a rational within $epsilon$-vicinity of $alpha$.
Now, assume some principal universal function $U = U(n, x)$ for the class of computable functions. This $U$ generates a certain numbering of computable real numbers: for each computable $alpha$, its corresponding number is any $n$ such that $U_n = f_alpha$ (note that every $alpha$ has infintely many assigned numbers as per Rice's theorem).
Given that, how does one prove the existence of an algorithm that, given any two numbers $n, m$ assigned to any two computable reals $alpha, beta$, produces some number that's assigned to their sum $alpha + beta$?
So this is a sketch of a proof that came to my mind after writing down this question.
It can be shown that there exists a computable bijection between $mathbb{N}^2$ and $mathbb{N}$, so let's denote $[i, j]$ for the natural that corresponds to the pair $(i, j)$ in that bijection. Now define a binary function $F$ such that $F([i, j], epsilon) = U(i, epsilon/2) + U(j, epsilon/2)$. Note that $F$ is clearly computable, and if $i, j$ are indices of some computable reals, then $F_{[i, j]}$ is the function corresponding to their sum.
Since $U$ is principal and $F$ is computable, there exists a computable total $s_F$ such that $forall n, x : F(n, x) = U(s_F(n), x)$. Combining that with the above, we get that $A(i, j) = s_F([i, j])$ is precisely the algorithm that, given two indices of computable reals, produces an index of their sum.
Does it sound reasonable?
proof-verification logic computability
$endgroup$
Define a real number $alpha$ to be computable if there is a computable total function $f_alpha$ that, given any rational $epsilon$, yields a rational within $epsilon$-vicinity of $alpha$.
Now, assume some principal universal function $U = U(n, x)$ for the class of computable functions. This $U$ generates a certain numbering of computable real numbers: for each computable $alpha$, its corresponding number is any $n$ such that $U_n = f_alpha$ (note that every $alpha$ has infintely many assigned numbers as per Rice's theorem).
Given that, how does one prove the existence of an algorithm that, given any two numbers $n, m$ assigned to any two computable reals $alpha, beta$, produces some number that's assigned to their sum $alpha + beta$?
So this is a sketch of a proof that came to my mind after writing down this question.
It can be shown that there exists a computable bijection between $mathbb{N}^2$ and $mathbb{N}$, so let's denote $[i, j]$ for the natural that corresponds to the pair $(i, j)$ in that bijection. Now define a binary function $F$ such that $F([i, j], epsilon) = U(i, epsilon/2) + U(j, epsilon/2)$. Note that $F$ is clearly computable, and if $i, j$ are indices of some computable reals, then $F_{[i, j]}$ is the function corresponding to their sum.
Since $U$ is principal and $F$ is computable, there exists a computable total $s_F$ such that $forall n, x : F(n, x) = U(s_F(n), x)$. Combining that with the above, we get that $A(i, j) = s_F([i, j])$ is precisely the algorithm that, given two indices of computable reals, produces an index of their sum.
Does it sound reasonable?
proof-verification logic computability
proof-verification logic computability
edited Dec 3 '18 at 19:18
0xd34df00d
asked Dec 3 '18 at 18:36
0xd34df00d0xd34df00d
412212
412212
$begingroup$
The result follows from examining the usual proof in real analysis that $lim (a_n + b_n) = lim a_n + lim b_n$, using an $epsilon/2$ argument.
$endgroup$
– Carl Mummert
Dec 3 '18 at 18:47
$begingroup$
Indeed, I was able to prove that $alpha + beta$ is computable using that argument. But how do you prove that its index (according to the definitions above) is also computable?
$endgroup$
– 0xd34df00d
Dec 3 '18 at 18:50
$begingroup$
In other words, let $f_alpha$ and $f_beta$ be the corresponding functions that produce the approximations of $alpha$ and $beta$ respectively. The function $f(epsilon) = f_alpha(epsilon/2) + f_beta(epsilon/2)$ is computable and provides the required approximations for their sum, but I'm not sure how to find its $U$-index given $U$-indices of $f_alpha$ and $f_beta$.
$endgroup$
– 0xd34df00d
Dec 3 '18 at 18:54
2
$begingroup$
In programming terms, the problem is asking you how to write a program that represents $alpha + beta$ given programs that represent $alpha$ and $beta$. To do this you need to know what programming language is being used. The choice of $U$ corresponds to the choice of programming language so your answer will depend on how $U$ is defined.
$endgroup$
– Rob Arthan
Dec 3 '18 at 20:26
add a comment |
$begingroup$
The result follows from examining the usual proof in real analysis that $lim (a_n + b_n) = lim a_n + lim b_n$, using an $epsilon/2$ argument.
$endgroup$
– Carl Mummert
Dec 3 '18 at 18:47
$begingroup$
Indeed, I was able to prove that $alpha + beta$ is computable using that argument. But how do you prove that its index (according to the definitions above) is also computable?
$endgroup$
– 0xd34df00d
Dec 3 '18 at 18:50
$begingroup$
In other words, let $f_alpha$ and $f_beta$ be the corresponding functions that produce the approximations of $alpha$ and $beta$ respectively. The function $f(epsilon) = f_alpha(epsilon/2) + f_beta(epsilon/2)$ is computable and provides the required approximations for their sum, but I'm not sure how to find its $U$-index given $U$-indices of $f_alpha$ and $f_beta$.
$endgroup$
– 0xd34df00d
Dec 3 '18 at 18:54
2
$begingroup$
In programming terms, the problem is asking you how to write a program that represents $alpha + beta$ given programs that represent $alpha$ and $beta$. To do this you need to know what programming language is being used. The choice of $U$ corresponds to the choice of programming language so your answer will depend on how $U$ is defined.
$endgroup$
– Rob Arthan
Dec 3 '18 at 20:26
$begingroup$
The result follows from examining the usual proof in real analysis that $lim (a_n + b_n) = lim a_n + lim b_n$, using an $epsilon/2$ argument.
$endgroup$
– Carl Mummert
Dec 3 '18 at 18:47
$begingroup$
The result follows from examining the usual proof in real analysis that $lim (a_n + b_n) = lim a_n + lim b_n$, using an $epsilon/2$ argument.
$endgroup$
– Carl Mummert
Dec 3 '18 at 18:47
$begingroup$
Indeed, I was able to prove that $alpha + beta$ is computable using that argument. But how do you prove that its index (according to the definitions above) is also computable?
$endgroup$
– 0xd34df00d
Dec 3 '18 at 18:50
$begingroup$
Indeed, I was able to prove that $alpha + beta$ is computable using that argument. But how do you prove that its index (according to the definitions above) is also computable?
$endgroup$
– 0xd34df00d
Dec 3 '18 at 18:50
$begingroup$
In other words, let $f_alpha$ and $f_beta$ be the corresponding functions that produce the approximations of $alpha$ and $beta$ respectively. The function $f(epsilon) = f_alpha(epsilon/2) + f_beta(epsilon/2)$ is computable and provides the required approximations for their sum, but I'm not sure how to find its $U$-index given $U$-indices of $f_alpha$ and $f_beta$.
$endgroup$
– 0xd34df00d
Dec 3 '18 at 18:54
$begingroup$
In other words, let $f_alpha$ and $f_beta$ be the corresponding functions that produce the approximations of $alpha$ and $beta$ respectively. The function $f(epsilon) = f_alpha(epsilon/2) + f_beta(epsilon/2)$ is computable and provides the required approximations for their sum, but I'm not sure how to find its $U$-index given $U$-indices of $f_alpha$ and $f_beta$.
$endgroup$
– 0xd34df00d
Dec 3 '18 at 18:54
2
2
$begingroup$
In programming terms, the problem is asking you how to write a program that represents $alpha + beta$ given programs that represent $alpha$ and $beta$. To do this you need to know what programming language is being used. The choice of $U$ corresponds to the choice of programming language so your answer will depend on how $U$ is defined.
$endgroup$
– Rob Arthan
Dec 3 '18 at 20:26
$begingroup$
In programming terms, the problem is asking you how to write a program that represents $alpha + beta$ given programs that represent $alpha$ and $beta$. To do this you need to know what programming language is being used. The choice of $U$ corresponds to the choice of programming language so your answer will depend on how $U$ is defined.
$endgroup$
– Rob Arthan
Dec 3 '18 at 20:26
add a comment |
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$begingroup$
The result follows from examining the usual proof in real analysis that $lim (a_n + b_n) = lim a_n + lim b_n$, using an $epsilon/2$ argument.
$endgroup$
– Carl Mummert
Dec 3 '18 at 18:47
$begingroup$
Indeed, I was able to prove that $alpha + beta$ is computable using that argument. But how do you prove that its index (according to the definitions above) is also computable?
$endgroup$
– 0xd34df00d
Dec 3 '18 at 18:50
$begingroup$
In other words, let $f_alpha$ and $f_beta$ be the corresponding functions that produce the approximations of $alpha$ and $beta$ respectively. The function $f(epsilon) = f_alpha(epsilon/2) + f_beta(epsilon/2)$ is computable and provides the required approximations for their sum, but I'm not sure how to find its $U$-index given $U$-indices of $f_alpha$ and $f_beta$.
$endgroup$
– 0xd34df00d
Dec 3 '18 at 18:54
2
$begingroup$
In programming terms, the problem is asking you how to write a program that represents $alpha + beta$ given programs that represent $alpha$ and $beta$. To do this you need to know what programming language is being used. The choice of $U$ corresponds to the choice of programming language so your answer will depend on how $U$ is defined.
$endgroup$
– Rob Arthan
Dec 3 '18 at 20:26