nyuuzyou/wikis huggingface dataset:
Combination Theorem for Sequences/Normed Division Ring/Multiple Rule

== Theorem == 
Let $\struct {R, \norm {\, \cdot \,} }$ be a normed division ring.
Let $\sequence {x_n}$ be a sequence in $R$.
Let $\sequence {x_n}$ be convergent in the norm $\norm {\, \cdot \,}$ to the following limit:
:$\ds \lim_{n \mathop \to \infty} x_n = l$
Let $\lambda \in R$.
Then:
:$\sequence {\lambda x_n}$ is convergent and $\ds \lim_{n \mathop \to \infty} \paren {\lambda x_n} = \lambda l$
== Proof ==
Let
:$\sequence {\tilde{x}_n} := \tuple {\lambda, \lambda, \lambda, \ldots}$
and:
:$\sequence {y_n} := \sequence {x_n}$
The claim follows from Product Rule for Sequences in Normed Division Ring, since:
:$\sequence {\lambda x_n} = \sequence {\tilde{x}_n y_n}$
== Also see ==
* Product Rule for Sequences in Normed Division Ring
== Sources ==
Category:Combination Theorem for Sequences in Normed Division Rings
Combination Theorem for Sequences/Normed Division Ring/Product Rule

== Theorem == 
Let $\struct {R, \norm {\,\cdot\,}}$ be a normed division ring.
Let $\sequence {x_n}$, $\sequence {y_n} $ be sequences in $R$.
Let $\sequence {x_n}$ and $\sequence {y_n}$ be convergent in the norm $\norm {\,\cdot\,}$ to the following limits:
:$\ds \lim_{n \mathop \to \infty} x_n = l$
:$\ds \lim_{n \mathop \to \infty} y_n = m$
Then:
:$\sequence {x_n y_n}$ is convergent to the limit $\ds \lim_{n \mathop \to \infty} \paren {x_n y_n} = l m$
== Proof 1 ==
== Proof 2 ==
== Sources ==
Category:Combination Theorem for Sequences in Normed Division Rings
Combination Theorem for Sequences/Normed Division Ring/Product Rule/Proof 1

== Theorem == 
Let $\struct {R, \norm {\,\cdot\,}}$ be a normed division ring.
Let $\sequence {x_n}$, $\sequence {y_n} $ be sequences in $R$.
Let $\sequence {x_n}$ and $\sequence {y_n}$ be convergent in the norm $\norm {\,\cdot\,}$ to the following limits:
:$\ds \lim_{n \mathop \to \infty} x_n = l$
:$\ds \lim_{n \mathop \to \infty} y_n = m$
Then:
== Proof ==
By Convergent Sequence in Normed Division Ring is Bounded, $\sequence {x_n}$ is bounded.
Suppose $\norm {x_n} \le K$ for $n = 1, 2, 3, \ldots$.
Let $M = \max \set {K, \norm m}$.
Then:
:$\norm m \le M$
and:
:$\forall n: \norm{x_n} \le M$
Let $\epsilon > 0$ be given.
Then $\dfrac \epsilon {2 M} > 0$.
As $\sequence {x_n}$ converges to $l$, we can find $N_1$ such that:
:$\forall n > N_1: \norm {x_n - l} < \dfrac \epsilon {2 M}$
Similarly, for $\sequence {y_n}$ we can find $N_2$ such that:
:$\forall n > N_2: \norm {y_n - m} < \dfrac \epsilon {2 M}$
Now let $N = \max \set {N_1, N_2}$.
Then if $n > N$, both the above inequalities will be true.
Thus $\forall n > N$:
Hence:
:$\sequence {x_n y_n}$ is convergent.
It follows that:
:$\ds \lim_{n \mathop \to \infty} \paren {x_n y_n} = l m$
Category:Combination Theorem for Sequences in Normed Division Rings
Combination Theorem for Sequences/Normed Division Ring/Product Rule/Proof 2

== Theorem == 
Let $\struct {R, \norm {\,\cdot\,} }$ be a normed division ring.
Let $\sequence {x_n}$ and $\sequence {y_n}$ be sequences in $R$.
Let $\sequence {x_n}$ and $\sequence {y_n}$ be convergent in the norm $\norm {\,\cdot\,}$ to the following limits:
:$\ds \lim_{n \mathop \to \infty} x_n = l$
:$\ds \lim_{n \mathop \to \infty} y_n = m$
Then:
== Proof ==
By Convergent Sequence in Normed Division Ring is Bounded, $\sequence {x_n}$ is bounded.
Suppose $\norm {x_n} \le K$ for $n = 1, 2, 3, \ldots$.
Then for $n = 1, 2, 3, \ldots$:
We note that $\sequence {z_n}$ is a real sequence.
But $x_n \to l$ as $n \to \infty$.
So from Definition:Convergent Sequence in Normed Division Ring:
:$\norm {x_n - l} \to 0$ as $n \to \infty$
Similarly $\norm {y_n - m} \to 0$ as $n \to \infty$.
From the Combined Sum Rule for Real Sequences:
:$\ds \lim_{n \mathop \to \infty} \paren {\lambda x'_n + \mu y'_n} = \lambda l' + \mu m'$, $z_n \to 0$ as $n \to \infty$
By applying the Squeeze Theorem for Complex Sequences (which applies as well to real as to complex sequences):
:$\sequence {\norm {x_n y_n - l m}}$ converges to $0$ in $\R$.
By definition of a convergent sequence in a normed division ring:
:$\sequence{x_n y_n}$ is convergent in $R$
It follows that:
:$\ds \lim_{n \mathop \to \infty} \paren {x_n y_n} = l m$
Category:Combination Theorem for Sequences in Normed Division Rings
Combination Theorem for Sequences/Normed Division Ring/Quotient Rule

== Theorem == 
Let $\struct {R, \norm {\, \cdot \,} }$ be a normed division ring with zero: $0$.
Let $\sequence {x_n}$ and $\sequence {y_n}$ be sequences in $R$.
Let $\sequence {x_n}$ and $\sequence {y_n}$ be convergent in the norm $\norm {\, \cdot \,}$ to the following limits:
:$\ds \lim_{n \mathop \to \infty} x_n = l$
:$\ds \lim_{n \mathop \to \infty} y_n = m$
Suppose $m \ne 0$.
Then:
:$\exists k \in \N : \forall n \in \N: y_{k + n} \ne 0$
and the sequences:
:$\sequence {x_{k + n} \ {y_{k + n} }^{-1} }$ and $\sequence { {y_{k + n} }^{-1} \ x_{k + n} }$ are well-defined and convergent with:
:$\ds \lim_{n \mathop \to \infty} x_{k + n} \ {y_{k + n} }^{-1} = l m^{-1}$
:$\ds \lim_{n \mathop \to \infty} {y_{k + n} }^{-1} \ x_{k + n} = m^{-1} l$
== Proof ==
From Inverse Rule for Sequences in Normed Division Ring:
:$\exists k \in \N : \forall n \in \N : y_{k + n} \ne 0$
and the sequence:
:$\sequence { {y_{k + n} }^{-1} }$
is well-defined and convergent with:
:$\ds \lim_{n \mathop \to \infty} {y_{k + n} }^{-1} = m^{-1}$
From Limit of Subsequence equals Limit of Sequence, $\sequence {x_{k + n} }$ is convergent with:
:$\ds \lim_{n \mathop \to \infty} x_{k + n} = l$
From Product Rule for Sequences in Normed Division Ring:
:$\ds \lim_{n \mathop \to \infty} x_{k + n} \ {y_{k + n} }^{-1} = l m^{-1}$
:$\ds \lim_{n \mathop \to \infty} {y_{k + n} }^{-1} \ x_{k + n} = m^{-1} l$
Category:Combination Theorem for Sequences in Normed Division Rings
Combination Theorem for Sequences/Normed Division Ring/Sum Rule

== Theorem == 
Let $\struct {R, \norm {\, \cdot \,} }$ be a normed division ring.
Let $\sequence {x_n}$, $\sequence {y_n} $ be sequences in $R$.
Let $\sequence {x_n}$ and $\sequence {y_n}$ be convergent in the norm $\norm {\, \cdot \,}$ to the following limits:
:$\ds \lim_{n \mathop \to \infty} x_n = l$
:$\ds \lim_{n \mathop \to \infty} y_n = m$
Then:
:$\sequence {x_n + y_n}$ is convergent and $\ds \lim_{n \mathop \to \infty} \paren {x_n + y_n} = l + m$
== Proof ==
Let $\epsilon > 0$ be given.
Then $\dfrac \epsilon 2 > 0$.
Since $\sequence {x_n}$ is convergent to $l$, we can find $N_1$ such that:
:$\forall n > N_1: \norm {x_n - l} < \dfrac \epsilon 2$
Similarly, for $\sequence {y_n}$ we can find $N_2$ such that:
:$\forall n > N_2: \norm {y_n - m} < \dfrac \epsilon 2$
Now let $N = \max \set {N_1, N_2}$.
Then if $n > N$, both the above inequalities will be true.
Thus $\forall n > N$:
Hence:
:$\sequence {x_n + y_n}$ is convergent to $l + m$.
Category:Combination Theorem for Sequences in Normed Division Rings
Combination Theorem for Sequences/Product Rule

== Theorem ==
=== Product Rule for Real Sequences ===
Let $\sequence {x_n}$ and $\sequence {y_n}$ be sequences in $\R$.
Let $\sequence {x_n}$ and $\sequence {y_n}$ be convergent to the following limits:
:$\ds \lim_{n \mathop \to \infty} x_n = l$
:$\ds \lim_{n \mathop \to \infty} y_n = m$
Then:
=== Product Rule for Complex Sequences ===
Let $\sequence {z_n}$ and $\sequence {w_n}$ be sequences in $\C$.
Let $\sequence {z_n}$ and $\sequence {w_n}$ be convergent to the following limits:
:$\ds \lim_{n \mathop \to \infty} z_n = c$
:$\ds \lim_{n \mathop \to \infty} w_n = d$
Then:
== Also see ==
* Sum Rule for Sequences
* Difference Rule for Sequences
* Multiple Rule for Sequences
* Quotient Rule for Sequences
Category:Combination Theorems for Sequences
Combination Theorem for Sequences/Quotient Rule

== Theorem ==
=== Quotient Rule for Real Sequences ===
Let $\sequence {x_n}$ and $\sequence {y_n}$ be sequences in $\R$.
Let $\sequence {x_n}$ and $\sequence {y_n}$ be convergent to the following limits:
:$\ds \lim_{n \mathop \to \infty} x_n = l$
:$\ds \lim_{n \mathop \to \infty} y_n = m$
Then:
=== Quotient Rule for Complex Sequences ===
Let $\sequence {z_n}$ and $\sequence {w_n}$ be sequences in $\C$.
Let $\sequence {z_n}$ and $\sequence {w_n}$ be convergent to the following limits:
:$\ds \lim_{n \mathop \to \infty} z_n = c$
:$\ds \lim_{n \mathop \to \infty} w_n = d$
Then:
== Also presented as ==
== Also see ==
* Sum Rule for Sequences
* Difference Rule for Sequences
* Multiple Rule for Sequences
* Product Rule for Sequences
Category:Combination Theorems for Sequences
Combination Theorem for Sequences/Quotient Rule/Also presented as

== Quotient Rule for Sequences: Also presented as ==
Some treatments of this subject specifically exclude all sequences where the denominators are zero at any point in their domain.
Thus, for example, this is how it is presented in :
:If $z_n \to c$ and $w_m \to d$, then
::... $\text{(iv)} \ z_n / w_n \to c / d$, where ... $w_n \ne 0$ for all $n$ and $d \ne 0$.
However, it is demonstrated within the proof that past a certain $N \in \R$, which is bound to exist, $w_n$ is guaranteed to be non-zero.
The behaviour of the sequence $S = \sequence {\dfrac {z_n} {w_n} }$ in the limit is not dependent upon the existence or otherwise of its terms for $n < N$.
Thus it is not necessary to state that $w_n \ne 0$ for all $n$, and in fact such a statement would unnecessarily restrict the applicability of the theorem to exclude otherwise well-behaved cases where it is desirable that the theorem does apply.
Hence this restriction is not supported on .
Category:Combination Theorems for Sequences
Combination Theorem for Sequences/Real

== Theorem == 
Let $\sequence {x_n}$ and $\sequence {y_n}$ be sequences in $\R$.
Let $\sequence {x_n}$ and $\sequence {y_n}$ be convergent to the following limits:
:$\ds \lim_{n \mathop \to \infty} x_n = l$
:$\ds \lim_{n \mathop \to \infty} y_n = m$
Let $\lambda, \mu \in \R$.
Then the following results hold:
=== Sum Rule ===
=== Difference Rule ===
=== Multiple Rule ===
=== Combined Sum Rule ===
=== Product Rule ===
=== Quotient Rule ===
Category:Combination Theorems for Sequences
Category:Real Analysis
Combination Theorem for Sequences/Real/Combined Sum Rule

== Theorem == 
Let $\sequence {x_n}$ and $\sequence {y_n}$ be sequences in $\R$.
Let $\sequence {x_n}$ and $\sequence {y_n}$ be convergent to the following limits:
:$\ds \lim_{n \mathop \to \infty} x_n = l$
:$\ds \lim_{n \mathop \to \infty} y_n = m$
Let $\lambda, \mu \in \R$.
Then:
:$\ds \lim_{n \mathop \to \infty} \paren {\lambda x_n + \mu y_n} = \lambda l + \mu m$
== Proof ==
From the Multiple Rule for Real Sequences, we have:
:$\ds \lim_{n \mathop \to \infty} \paren {\lambda x_n} = \lambda l$
:$\ds \lim_{n \mathop \to \infty} \paren {\mu y_n} = \mu m$
The result now follows directly from the Sum Rule for Real Sequences:
:$\ds \lim_{n \mathop \to \infty} \paren {\lambda x_n + \mu y_n} = \lambda l + \mu m$
== Sources ==
* : $\S 4$: Convergent Sequences: $\S 4.8 \ \text {(i)}$: Criteria for convergence
Category:Combination Theorems for Sequences
Combination Theorem for Sequences/Real/Difference Rule

== Theorem == 
Let $\sequence {x_n}$ and $\sequence {y_n}$ be sequences in $\R$.
Let $\sequence {x_n}$ and $\sequence {y_n}$ be convergent to the following limits:
:$\ds \lim_{n \mathop \to \infty} x_n = l$
:$\ds \lim_{n \mathop \to \infty} y_n = m$
Then:
:$\ds \lim_{n \mathop \to \infty} \paren {x_n - y_n} = l - m$
== Proof ==
From Sum Rule for Real Sequences:
:$\ds \lim_{n \mathop \to \infty} \paren {x_n + y_n} = l + m$
From Multiple Rule for Real Sequences:
:$\ds \lim_{n \mathop \to \infty} \paren {-y_n} = -m$
Hence:
:$\ds \lim_{n \mathop \to \infty} \paren {x_n + \paren {-y_n} } = l + \paren {-m}$
The result follows.
== Also see ==
* Sum Rule for Real Sequences
Category:Combination Theorems for Sequences
Combination Theorem for Sequences/Real/Multiple Rule

== Theorem == 
Let $\sequence {x_n}$ be a sequences in $\R$.
Let $\sequence {x_n}$ be convergent to the following limit:
:$\ds \lim_{n \mathop \to \infty} x_n = l$
Let $\lambda \in \R$.
Then:
:$\ds \lim_{n \mathop \to \infty} \paren {\lambda x_n} = \lambda l$
== Proof ==
Let $\epsilon > 0$.
We need to find $N$ such that:
:$\forall n > N: \size {\lambda x_n - \lambda l} < \epsilon$
If $\lambda = 0$ the result is trivial.
So, assume $\lambda \ne 0$.
Then $\size \lambda > 0$ from the definition of the absolute value of $\lambda$.
Hence $\dfrac \epsilon {\size \lambda} > 0$.
We have that $x_n \to l$ as $n \to \infty$.
Thus it follows that:
:$\exists N: \forall n > N: \size {x_n - l} < \dfrac \epsilon {\size \lambda}$
That is:
:$\forall n > N: \size \lambda \size {x_n - l} < \epsilon$
But we have:
Hence:
:$\ds \lim_{n \mathop \to \infty} \paren {\lambda x_n} = \lambda l$
== Sources ==
* : $\S 4$: Convergent Sequences: Exercise $\S 4.6 \ (3)$
* : $3.3b$
Category:Combination Theorems for Sequences
Combination Theorem for Sequences/Real/Product Rule

== Theorem == 
Let $\sequence {x_n}$ and $\sequence {y_n}$ be sequences in $\R$.
Let $\sequence {x_n}$ and $\sequence {y_n}$ be convergent to the following limits:
:$\ds \lim_{n \mathop \to \infty} x_n = l$
:$\ds \lim_{n \mathop \to \infty} y_n = m$
Then:
:$\ds \lim_{n \mathop \to \infty} \paren {x_n y_n} = l m$
== Proof ==
Because $\sequence {x_n}$ converges, it is bounded by Convergent Sequence is Bounded.
Suppose $\size {x_n} \le K$ for $n = 1, 2, 3, \ldots$.
Then:
But $x_n \to l$ as $n \to \infty$.
So $\size {x_n - l} \to 0$ as $n \to \infty$ from Convergent Sequence Minus Limit.
Similarly $\size {y_n - m} \to 0$ as $n \to \infty$.
From the Combined Sum Rule for Real Sequences:
:$\ds \lim_{n \mathop \to \infty} \paren {\lambda x_n + \mu y_n} = \lambda l + \mu m$, $z_n \to 0$ as $n \to \infty$
The result follows by the Squeeze Theorem for Real Sequences.
== Sources ==
* : $3.3c$
* : $1$: Review of some real analysis: $\S 1.2$: Real Sequences: Proposition $1.2.11 \ \text {(b)}$
* : $\S 4$: Convergent Sequences: $\S 4.8 \ \text {(ii)}$: Criteria for convergence
* : Appendix: $\S 18.3$: Combination theorem
* : $\S 1.2.3$: Sums and Products: Exercise $5$
Category:Limits of Sequences
Combination Theorem for Sequences/Real/Quotient Rule

== Theorem == 
Let $\sequence {x_n}$ and $\sequence {y_n}$ be sequences in $\R$.
Let $\sequence {x_n}$ and $\sequence {y_n}$ be convergent to the following limits:
:$\ds \lim_{n \mathop \to \infty} x_n = l$
:$\ds \lim_{n \mathop \to \infty} y_n = m$
Then:
:$\ds \lim_{n \mathop \to \infty} \frac {x_n} {y_n} = \frac l m$
provided that $m \ne 0$.
=== Corollary ===
== Proof ==
As $y_n \to m$ as $n \to \infty$, it follows from Modulus of Limit that $\size {y_n} \to \size m$ as $n \to \infty$.
As $m \ne 0$, it follows from the definition of the modulus of $m$ that $\size m > 0$.
As the statement is given, it is possible that $y_n = 0$ for some $n$.
At such $n$, the terms $\dfrac {x_n} {y_n}$ are not defined.
However, from Sequence Converges to Within Half Limit, we have:
:$\exists N: \forall n > N: \size {y_n} > \dfrac {\size m} 2$
Hence for all $n > N$ we have that $y_n \ne 0$.
Thus we may restrict our attention to the domain of $\sequence {y_n}$ such that $n > N$, knowing that $\dfrac {x_n} {y_n}$ will be defined in that domain.
So, for $n > N$, consider:
By the above, $m x_n - y_n l \to m l - m l = 0$ as $n \to \infty$.
The result follows by the Squeeze Theorem for Real Sequences.
== Also presented as ==
== Sources ==
* : $3.3 \ \text{d}$
* : $\S 4$: Convergent Sequences: $\S 4.8 \ \text{(iii)}$: Criteria for convergence
* : Appendix: $\S 18.3$: Combination theorem
Category:Combination Theorems for Sequences
Combination Theorem for Sequences/Real/Sum Rule

== Theorem == 
Let $\sequence {x_n}$ and $\sequence {y_n}$ be sequences in $\R$.
Let $\sequence {x_n}$ and $\sequence {y_n}$ be convergent to the following limits:
:$\ds \lim_{n \mathop \to \infty} x_n = l$
:$\ds \lim_{n \mathop \to \infty} y_n = m$
Then:
:$\ds \lim_{n \mathop \to \infty} \paren {x_n + y_n} = l + m$
== Proof ==
Let $\epsilon > 0$ be given.
Then $\dfrac \epsilon 2 > 0$.
We are given that:
:$\ds \lim_{n \mathop \to \infty} x_n = l$
:$\ds \lim_{n \mathop \to \infty} y_n = m$
By definition of the limit of a real sequence, we can find $N_1$ such that:
:$\forall n > N_1: \size {x_n - l} < \dfrac \epsilon 2$
where $\size {x_n - l}$ denotes the absolute value of $x_n - l$
Similarly we can find $N_2$ such that:
:$\forall n > N_2: \size {y_n - m} < \dfrac \epsilon 2$
Let $N = \max \set {N_1, N_2}$.
Then if $n > N$, both the above inequalities will be true:
:$n > N_1$
:$n > N_2$
Thus $\forall n > N$:
Hence the result:
:$\ds \lim_{n \mathop \to \infty} \paren {x_n + y_n} = l + m$
== Also see ==
* Difference Rule for Real Sequences
== Sources ==
* : $3.3a$
* : $1$: Review of some real analysis: $\S 1.2$: Real Sequences: Proposition $1.2.11 \ \text {(a)}$
Category:Combination Theorems for Sequences
Combination Theorem for Sequences/Sum Rule

== Theorem ==
=== Real Sequences ===
Let $\sequence {x_n}$ and $\sequence {y_n}$ be sequences in $\R$.
Let $\sequence {x_n}$ and $\sequence {y_n}$ be convergent to the following limits:
:$\ds \lim_{n \mathop \to \infty} x_n = l$
:$\ds \lim_{n \mathop \to \infty} y_n = m$
Then:
=== Complex Sequences ===
Let $\sequence {z_n}$ and $\sequence {w_n}$ be sequences in $\C$.
Let $\sequence {z_n}$ and $\sequence {w_n}$ be convergent to the following limits:
:$\ds \lim_{n \mathop \to \infty} z_n = c$
:$\ds \lim_{n \mathop \to \infty} w_n = d$
Then:
== Also see ==
* Difference Rule for Sequences
* Multiple Rule for Sequences
* Product Rule for Sequences
* Quotient Rule for Sequences
== Sources ==
* 
* 
Category:Combination Theorems for Sequences
Combination of Recursive Functions

== Theorem ==
Let $f: \N^k \to \N$ and $g: \N^k \to \N$ be recursive functions (not necessarily total), where $k \ge 1$.
Let $\RR$ be a $k$-ary relation such that:
:if $\map \RR {n_1, n_2, \ldots, n_k}$ holds, then $\map f {n_1, n_2, \ldots, n_k}$ is defined
:if $\map \RR {n_1, n_2, \ldots, n_k}$ does not hold, then $\map g {n_1, n_2, \ldots, n_k}$ is defined.
Let $h: \N^k \to \N$ be the function defined as:
:$\map h {n_1, n_2, \ldots, n_k} = \begin {cases}
\map f {n_1, n_2, \ldots, n_k} & : \text{if } \map \RR {n_1, n_2, \ldots, n_k} \text { holds} \\
\map g {n_1, n_2, \ldots, n_k} & : \text{otherwise}
\end {cases}$
so that $h$ is total.
Then $h$ is recursive.
== Proof ==
Let $P_f, P_g, P_\RR$ be the URM programs computing, respectively, the functions $f$ and $g$ and the characteristic function $\chi_\RR$.
From Recursive Function is URM Computable, these programs are guaranteed to exist.
An informal algorithm for computing $h$ is as follows.
# Input $\tuple {n_1, n_2, \ldots, n_k}$.
# Use $P_\RR$ to determine whether $\map \RR {n_1, n_2, \ldots, n_k}$ holds.
# If it holds, use $P_f$ to compute $\map f {n_1, n_2, \ldots, n_k}$.
# If it does not hold, use $P_g$ to compute $\map g {n_1, n_2, \ldots, n_k}$.
The result follows from URM Computable Function is Recursive.
== Fallacious Proof ==
We might attempt to show that $h$ is recursive by using the equation:
where $\overline \sgn$ is the signum bar function, known to be primitive recursive.
Then the is obtained by substitution from known recursive functions, so is recursive, and hence $h$ is recursive.
However, this fails as follows.
Although $\map h {n_1, n_2, \ldots, n_k}$ is always defined, 
might not be.
$f$ and $g$ might not be total, in which case there may be some $\tuple {n_1, n_2, \ldots, n_k}$ for which at least one of $\map f {n_1, n_2, \ldots, n_k}$ and $\map g {n_1, n_2, \ldots, n_k}$ is not defined.
This means $(1)$ is not defined for those values of $\tuple {n_1, n_2, \ldots, n_k}$.
Note that $(1)$ does define a recursive function, but its domain consists only of those $\tuple {n_1, n_2, \ldots, n_k}$ for which $\map f {n_1, n_2, \ldots, n_k}$ and $\map g {n_1, n_2, \ldots, n_k}$ are both defined.
Thus we establish the result by means of the indirect approach requiring the use of the results concerning the equivalence of recursive functions and URM computable functions.
Category:Recursive Functions
Combination of Solutions to Non-Homogeneous LSOODE with same Homogeneous Part

== Theorem ==
Let $\map {y_1} x$ be a particular solution of the linear second order ODE:
:$(1): \quad y + \map P x y' + \map Q x y = \map {R_1} x$
Let $\map {y_2} x$ be a particular solution of the linear second order ODE:
:$(2): \quad y + \map P x y' + \map Q x y = \map {R_2} x$
Then $\map y x = \map {y_1} x + \map {y_2} x$ is a particular solution of the linear second order ODE:
:$(3): \quad y + \map P x y' + \map Q x y = \map {R_1} x + \map {R_2} x$
== Proof ==
Hence the result.
== Sources ==
* : $\S 3.18$: Problem $3$
Category:Linear Second Order ODEs
Commensurability is Transitive

== Theorem ==
Let $a$, $b$, $c$ be three real numbers.
Let $a$ and $b$ be commensurable, and $b$ and $c$ be commensurable.
Then $a$ and $c$ are commensurable.
== Proof ==
From the definition of commensurablility:
:$\dfrac a b, \dfrac b c \in \Q$
where $\Q$ denotes the set of all rational numbers.
From Rational Multiplication is Closed:
:$\dfrac a b \times \dfrac b c \in \Q$
Cancelling $b$, we have:
:$\dfrac a c \in \Q$
Hence the result.
Category:Number Theory
Commensurability is Transitive Relation

== Theorem ==
== Proof ==
Let $A$ and $B$ be magnitudes which are both commensurable with another magnitude $C$.
As $A$ is commensurable with $C$, from Ratio of Commensurable Magnitudes:
: $A$ has to $C$ the ratio which a number has to a number.
Let it have the ratio which $D$ has to $E$, where $D$ and $E$ are numbers.
As $C$ is commensurable with $B$, from Ratio of Commensurable Magnitudes:
: $C$ has to $B$ the ratio which a number has to a number.
Let it have the ratio which $F$ has to $G$, where $F$ and $G$ are numbers.
From Construction of Sequence of Numbers with Given Ratios, let the numbers $H$, $K$, $L$ be assigned the ratios:
:$D : E = H : K$
:$F : G = K : L$
So:
and:
Then:
That is, $A$ has to $B$ the ratio which a number has to a number.
From Magnitudes with Rational Ratio are Commensurable:
:$A$ and $B$ are commensurable with each other.
== Sources ==
* : Book $\text X$. Propositions
Category:Euclidean Number Theory
Commensurability of Elements of Proportional Magnitudes

== Theorem ==
== Proof ==
Let $A$, $B$, $C$ and $D$ be four magnitudes in proportion:
:$A : B = C : D$
Let $A$ be commensurable with $B$.
Then from Ratio of Commensurable Magnitudes:
: $A$ has to $B$ the ratio which a number has to a number.
As $A : B = C : D$ it follows that:
: $C$ has to $D$ the ratio which a number has to a number.
From Magnitudes with Rational Ratio are Commensurable it follows that:
:$C$ is commensurable with $D$.
Let $A$ be incommensurable with $B$.
Then from Incommensurable Magnitudes have Irrational Ratio:
: $A$ does not have to $B$ the ratio which a number has to a number.
As $A : B = C : D$ it follows that:
: $C$ does not have to $D$ the ratio which a number has to a number.
From Magnitudes with Irrational Ratio are Incommensurable it follows that:
:$C$ is incommensurable with $D$.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory