avewright/proofwiki-math huggingface dataset:
Theorem: Combination Theorem for Sequences/Complex/Sum Rule/Proof 1

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:
{{:Sum Rule for Complex Sequences}}

Proof:
Let $\epsilon > 0$ be given.
Then $\dfrac \epsilon 2 > 0$.
We are given that:
:$\ds \lim_{n \mathop \to \infty} z_n = c$
:$\ds \lim_{n \mathop \to \infty} w_n = d$
By definition of the limit of a complex sequence, we can find $N_1$ such that:
:$\forall n > N_1: \cmod {z_n - c} < \dfrac \epsilon 2$
where $\cmod {z_n - c}$ denotes the complex modulus of $z_n - c$.
Similarly we can find $N_2$ such that:
:$\forall n > N_2: \cmod {w_n - d} < \dfrac \epsilon 2$
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 $\forall n > N$:
{{begin-eqn}}
{{eqn | l = \cmod {\paren {z_n + w_n} - \paren {c + d} }
      | r = \cmod {\paren {z_n - c} + \paren {w_n - d} }
      | c = 
}}
{{eqn | o = \le
      | r = \cmod {z_n - l} + \cmod {w_n - m}
      | c = Triangle Inequality for Complex Numbers
}}
{{eqn | o = <
      | r = \frac \epsilon 2 + \frac \epsilon 2
      | c = 
}}
{{eqn | r = \epsilon
      | c = 
}}
{{end-eqn}}
Hence the result:
:$\ds \lim_{n \mathop \to \infty} \paren {z_n + w_n} = c + d$
{{qed}}
Theorem: Combination Theorem for Sequences/Complex/Sum Rule

:$\ds \lim_{n \mathop \to \infty} \paren {z_n + w_n} = c + d$

Proof:
Let $\epsilon > 0$ be given.
Then $\dfrac \epsilon 2 > 0$.
We are given that:
:$\ds \lim_{n \mathop \to \infty} z_n = c$
:$\ds \lim_{n \mathop \to \infty} w_n = d$
By definition of the limit of a complex sequence, we can find $N_1$ such that:
:$\forall n > N_1: \cmod {z_n - c} < \dfrac \epsilon 2$
where $\cmod {z_n - c}$ denotes the complex modulus of $z_n - c$.
Similarly we can find $N_2$ such that:
:$\forall n > N_2: \cmod {w_n - d} < \dfrac \epsilon 2$
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 $\forall n > N$:
{{begin-eqn}}
{{eqn | l = \cmod {\paren {z_n + w_n} - \paren {c + d} }
      | r = \cmod {\paren {z_n - c} + \paren {w_n - d} }
      | c = 
}}
{{eqn | o = \le
      | r = \cmod {z_n - l} + \cmod {w_n - m}
      | c = Triangle Inequality for Complex Numbers
}}
{{eqn | o = <
      | r = \frac \epsilon 2 + \frac \epsilon 2
      | c = 
}}
{{eqn | r = \epsilon
      | c = 
}}
{{end-eqn}}
Hence the result:
:$\ds \lim_{n \mathop \to \infty} \paren {z_n + w_n} = c + d$
{{qed}}
Theorem: Combination Theorem for Sequences/Complex/Sum Rule

:$\ds \lim_{n \mathop \to \infty} \paren {z_n + w_n} = c + d$

Proof:
Let $\epsilon > 0$ be given.
Then $\dfrac \epsilon 2 > 0$.
We are given that:
:$\ds \lim_{n \mathop \to \infty} z_n = c$
:$\ds \lim_{n \mathop \to \infty} w_n = d$
Let:
:$z_n = x_n + i y_n$
:$w_n = r_n + i s_n$
:$c = a + i b$
:$d = l + i m$
where each of $x_n, y_n, r_n, s_n, a, b, l, m \in \R$ are real.
By definition:
$\sequence {z_n}$ '''converges to the limit $c = a + i b$''' {{iff}} both:
:$\forall \epsilon \in \R_{>0}: \exists N \in \R: n > N \implies \size {x_n - a} < \epsilon \text { and } \size {y_n - b} < \epsilon$
$\sequence {w_n}$ '''converges to the limit $d = l + i m$''' {{iff}} both:
:$\forall \epsilon \in \R_{>0}: \exists N \in \R: n > N \implies \size {r_n - l} < \epsilon \text { and } \size {s_n - m} < \epsilon$
where $\size x$ denotes the absolute value of $x \in \R$.
Then:
{{begin-eqn}}
{{eqn | l = \lim_{n \mathop \to \infty} \paren {z_n + w_n}
      | r = \lim_{n \mathop \to \infty} \paren {x_n + i y_n + r_n + i s_n}
      | c = 
}}
{{eqn | r = \lim_{n \mathop \to \infty} \paren {x_n + r_n + i \paren {y_n + s_n} }
      | c = 
}}
{{eqn | r = \lim_{n \mathop \to \infty} \paren {x_n + r_n} + i \lim_{n \mathop \to \infty} \paren {y_n + s_n}
      | c = {{Defof|Convergent Sequence|subdef = Complex Numbers|index = 1}}
}}
{{eqn | r = \paren {a + l} + i \paren {b + m}
      | c = 
}}
{{eqn | r = \paren {a + i b} + \paren {l + i m}
      | c = 
}}
{{eqn | r = c + d
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Combination Theorem for Sequences/Complex/Sum Rule/Proof 2

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:
{{:Combination Theorem for Sequences/Complex/Sum Rule}}

Proof:
Let $\epsilon > 0$ be given.
Then $\dfrac \epsilon 2 > 0$.
We are given that:
:$\ds \lim_{n \mathop \to \infty} z_n = c$
:$\ds \lim_{n \mathop \to \infty} w_n = d$
Let:
:$z_n = x_n + i y_n$
:$w_n = r_n + i s_n$
:$c = a + i b$
:$d = l + i m$
where each of $x_n, y_n, r_n, s_n, a, b, l, m \in \R$ are real.
By definition:
$\sequence {z_n}$ '''converges to the limit $c = a + i b$''' {{iff}} both:
:$\forall \epsilon \in \R_{>0}: \exists N \in \R: n > N \implies \size {x_n - a} < \epsilon \text { and } \size {y_n - b} < \epsilon$
$\sequence {w_n}$ '''converges to the limit $d = l + i m$''' {{iff}} both:
:$\forall \epsilon \in \R_{>0}: \exists N \in \R: n > N \implies \size {r_n - l} < \epsilon \text { and } \size {s_n - m} < \epsilon$
where $\size x$ denotes the absolute value of $x \in \R$.
Then:
{{begin-eqn}}
{{eqn | l = \lim_{n \mathop \to \infty} \paren {z_n + w_n}
      | r = \lim_{n \mathop \to \infty} \paren {x_n + i y_n + r_n + i s_n}
      | c = 
}}
{{eqn | r = \lim_{n \mathop \to \infty} \paren {x_n + r_n + i \paren {y_n + s_n} }
      | c = 
}}
{{eqn | r = \lim_{n \mathop \to \infty} \paren {x_n + r_n} + i \lim_{n \mathop \to \infty} \paren {y_n + s_n}
      | c = {{Defof|Convergent Sequence|subdef = Complex Numbers|index = 1}}
}}
{{eqn | r = \paren {a + l} + i \paren {b + m}
      | c = 
}}
{{eqn | r = \paren {a + i b} + \paren {l + i m}
      | c = 
}}
{{eqn | r = c + d
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Combination Theorem for Sequences/Complex

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:
:$\ds \lim_{n \mathop \to \infty} z_n = c$
:$\ds \lim_{n \mathop \to \infty} w_n = d$
Let $\lambda, \mu \in \C$.
Then the following results hold:
=== Sum Rule ===
{{:Combination Theorem for Sequences/Complex/Sum Rule}}
=== Difference Rule ===
{{:Combination Theorem for Sequences/Complex/Difference Rule}}
=== Multiple Rule ===
{{:Combination Theorem for Sequences/Complex/Multiple Rule}}
=== Combined Sum Rule ===
{{:Combination Theorem for Sequences/Complex/Combined Sum Rule}}
=== Product Rule ===
{{:Combination Theorem for Sequences/Complex/Product Rule}}
=== Quotient Rule ===
{{:Combination Theorem for Sequences/Complex/Quotient Rule}}

Proof:
Because $\sequence {z_n}$ converges, it is bounded by Convergent Sequence is Bounded.
Suppose $\cmod {z_n} \le K$ for $n = 1, 2, 3, \ldots$.
Then:
{{begin-eqn}}
{{eqn | l = \cmod {z_n w_n - c d}
      | r = \cmod {z_n w_n - z_n d + z_n d - c d}
      | c = 
}}
{{eqn | o = \le
      | r = \cmod {z_n w_n - z_n d} + \cmod {z_n d - c d}
      | c = Triangle Inequality for Complex Numbers
}}
{{eqn | r = \cmod {z_n} \cmod {w_n - d} + m \cdot \size {z_n - c}
      | c = Complex Modulus of Product of Complex Numbers
}}
{{eqn | o = \le
      | r = K \cdot \cmod {w_n - d} + \cmod d \cdot \cmod {z_n - c}
      | c = 
}}
{{eqn | o = =:
      | r = \phi_n
      | c = 
}}
{{end-eqn}}
But $z_n \to c$ as $n \to \infty$.
So $\cmod {z_n - c} \to 0$ as $n \to \infty$ from Convergent Sequence Minus Limit.
Similarly $\cmod {w_n - d} \to 0$ as $n \to \infty$.
From the Combined Sum Rule for Real Sequences:
:$\ds \lim_{n \mathop \to \infty} \paren {\lambda z_n + \mu w_n} = \lambda c + \mu d$, $\phi_n \to 0$ as $n \to \infty$
The result follows by the Squeeze Theorem for Complex Sequences.
{{qed}}
Theorem: Combination Theorem for Sequences/Complex

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:
:$\ds \lim_{n \mathop \to \infty} z_n = c$
:$\ds \lim_{n \mathop \to \infty} w_n = d$
Let $\lambda, \mu \in \C$.
Then the following results hold:
=== Sum Rule ===
{{:Combination Theorem for Sequences/Complex/Sum Rule}}
=== Difference Rule ===
{{:Combination Theorem for Sequences/Complex/Difference Rule}}
=== Multiple Rule ===
{{:Combination Theorem for Sequences/Complex/Multiple Rule}}
=== Combined Sum Rule ===
{{:Combination Theorem for Sequences/Complex/Combined Sum Rule}}
=== Product Rule ===
{{:Combination Theorem for Sequences/Complex/Product Rule}}
=== Quotient Rule ===
{{:Combination Theorem for Sequences/Complex/Quotient Rule}}

Proof:
Let $z_n = x_n + i y_n$.
Let $w_n = u_n + i v_n$.
Let $c = a + i b$
Let $d = e + i f$.
By definition of convergent complex sequence:
{{begin-eqn}}
{{eqn | l = \lim_{n \mathop \to \infty} z_n
      | r = c
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \lim_{n \mathop \to \infty} x_n + i \lim_{n \mathop \to \infty} y_n
      | r = a + i b
      | c = {{Defof|Convergent Complex Sequence}}
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \lim_{n \mathop \to \infty} w_n
      | r = d
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \lim_{n \mathop \to \infty} u_n + i \lim_{n \mathop \to \infty} v_n
      | r = e + i f
      | c = {{Defof|Convergent Complex Sequence}}
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \lim_{n \mathop \to \infty} z_n w_n
      | r = \lim_{n \mathop \to \infty} \paren {\paren {x_n u_n - y_n v_n} + i \paren {y_n u_n + x_n v_n} }
      | c = {{Defof|Complex Multiplication}}
}}
{{eqn | r = \lim_{n \mathop \to \infty} \paren {x_n u_n - y_n v_n} + i \lim_{n \mathop \to \infty} \paren {y_n u_n + x_n v_n}
      | c = {{Defof|Convergent Complex Sequence}}
}}
{{eqn | r = \paren {\lim_{n \mathop \to \infty} \paren {x_n u_n} - \lim_{n \mathop \to \infty} \paren {y_n v_n} } + i \paren {\lim_{n \mathop \to \infty} \paren {y_n u_n} + \lim_{n \mathop \to \infty} \paren {x_n v_n} }
      | c = Sum Rule for Real Sequences
}}
{{eqn | r = \paren {\lim_{n \mathop \to \infty} \paren {x_n} \lim_{n \mathop \to \infty} \paren {u_n} - \lim_{n \mathop \to \infty} \paren {y_n} \lim_{n \mathop \to \infty} \paren {v_n} } + i \paren {\lim_{n \mathop \to \infty} \paren {y_n} \lim_{n \mathop \to \infty} \paren {u_n} + \lim_{n \mathop \to \infty} \paren {x_n} \lim_{n \mathop \to \infty} \paren {v_n} }
      | c = Product Rule for Real Sequences
}}
{{eqn | r = \paren {a e - b f} + i \paren {b e + a f}
      | c = 
}}
{{eqn | r = \paren {a + i b} \paren {e + i f}
      | c = {{Defof|Complex Multiplication}}
}}
{{eqn | r = c d
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Combination Theorem for Sequences/Complex

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:
:$\ds \lim_{n \mathop \to \infty} z_n = c$
:$\ds \lim_{n \mathop \to \infty} w_n = d$
Let $\lambda, \mu \in \C$.
Then the following results hold:
=== Sum Rule ===
{{:Combination Theorem for Sequences/Complex/Sum Rule}}
=== Difference Rule ===
{{:Combination Theorem for Sequences/Complex/Difference Rule}}
=== Multiple Rule ===
{{:Combination Theorem for Sequences/Complex/Multiple Rule}}
=== Combined Sum Rule ===
{{:Combination Theorem for Sequences/Complex/Combined Sum Rule}}
=== Product Rule ===
{{:Combination Theorem for Sequences/Complex/Product Rule}}
=== Quotient Rule ===
{{:Combination Theorem for Sequences/Complex/Quotient Rule}}

Proof:
Let $\epsilon > 0$ be given.
Then $\dfrac \epsilon 2 > 0$.
We are given that:
:$\ds \lim_{n \mathop \to \infty} z_n = c$
:$\ds \lim_{n \mathop \to \infty} w_n = d$
By definition of the limit of a complex sequence, we can find $N_1$ such that:
:$\forall n > N_1: \cmod {z_n - c} < \dfrac \epsilon 2$
where $\cmod {z_n - c}$ denotes the complex modulus of $z_n - c$.
Similarly we can find $N_2$ such that:
:$\forall n > N_2: \cmod {w_n - d} < \dfrac \epsilon 2$
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 $\forall n > N$:
{{begin-eqn}}
{{eqn | l = \cmod {\paren {z_n + w_n} - \paren {c + d} }
      | r = \cmod {\paren {z_n - c} + \paren {w_n - d} }
      | c = 
}}
{{eqn | o = \le
      | r = \cmod {z_n - l} + \cmod {w_n - m}
      | c = Triangle Inequality for Complex Numbers
}}
{{eqn | o = <
      | r = \frac \epsilon 2 + \frac \epsilon 2
      | c = 
}}
{{eqn | r = \epsilon
      | c = 
}}
{{end-eqn}}
Hence the result:
:$\ds \lim_{n \mathop \to \infty} \paren {z_n + w_n} = c + d$
{{qed}}
Theorem: Combination Theorem for Sequences/Complex

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:
:$\ds \lim_{n \mathop \to \infty} z_n = c$
:$\ds \lim_{n \mathop \to \infty} w_n = d$
Let $\lambda, \mu \in \C$.
Then the following results hold:
=== Sum Rule ===
{{:Combination Theorem for Sequences/Complex/Sum Rule}}
=== Difference Rule ===
{{:Combination Theorem for Sequences/Complex/Difference Rule}}
=== Multiple Rule ===
{{:Combination Theorem for Sequences/Complex/Multiple Rule}}
=== Combined Sum Rule ===
{{:Combination Theorem for Sequences/Complex/Combined Sum Rule}}
=== Product Rule ===
{{:Combination Theorem for Sequences/Complex/Product Rule}}
=== Quotient Rule ===
{{:Combination Theorem for Sequences/Complex/Quotient Rule}}

Proof:
Let $\epsilon > 0$ be given.
Then $\dfrac \epsilon 2 > 0$.
We are given that:
:$\ds \lim_{n \mathop \to \infty} z_n = c$
:$\ds \lim_{n \mathop \to \infty} w_n = d$
Let:
:$z_n = x_n + i y_n$
:$w_n = r_n + i s_n$
:$c = a + i b$
:$d = l + i m$
where each of $x_n, y_n, r_n, s_n, a, b, l, m \in \R$ are real.
By definition:
$\sequence {z_n}$ '''converges to the limit $c = a + i b$''' {{iff}} both:
:$\forall \epsilon \in \R_{>0}: \exists N \in \R: n > N \implies \size {x_n - a} < \epsilon \text { and } \size {y_n - b} < \epsilon$
$\sequence {w_n}$ '''converges to the limit $d = l + i m$''' {{iff}} both:
:$\forall \epsilon \in \R_{>0}: \exists N \in \R: n > N \implies \size {r_n - l} < \epsilon \text { and } \size {s_n - m} < \epsilon$
where $\size x$ denotes the absolute value of $x \in \R$.
Then:
{{begin-eqn}}
{{eqn | l = \lim_{n \mathop \to \infty} \paren {z_n + w_n}
      | r = \lim_{n \mathop \to \infty} \paren {x_n + i y_n + r_n + i s_n}
      | c = 
}}
{{eqn | r = \lim_{n \mathop \to \infty} \paren {x_n + r_n + i \paren {y_n + s_n} }
      | c = 
}}
{{eqn | r = \lim_{n \mathop \to \infty} \paren {x_n + r_n} + i \lim_{n \mathop \to \infty} \paren {y_n + s_n}
      | c = {{Defof|Convergent Sequence|subdef = Complex Numbers|index = 1}}
}}
{{eqn | r = \paren {a + l} + i \paren {b + m}
      | c = 
}}
{{eqn | r = \paren {a + i b} + \paren {l + i m}
      | c = 
}}
{{eqn | r = c + d
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Combination Theorem for Sequences/Complex/Difference Rule

:$\ds \lim_{n \mathop \to \infty} \paren {z_n - w_n} = c - d$

Proof:
From Sum Rule for Complex Sequences:
:$\ds \lim_{n \mathop \to \infty} \paren {z_n + w_n} = c + d$
From Multiple Rule for Complex Sequences:
:$\ds \lim_{n \mathop \to \infty} \paren {-w_n} = -d$
Hence:
:$\ds \lim_{n \mathop \to \infty} \paren {z_n + \paren {-w_n} } = c + \paren {-d}$
The result follows.
{{qed}}
Theorem: Combination Theorem for Sequences/Complex/Multiple Rule

:$\ds \lim_{n \mathop \to \infty} \paren {\lambda z_n} = \lambda c$

Proof:
Let $\epsilon > 0$.
We need to find $N$ such that:
:$\forall n > N: \cmod {\lambda z_n - \lambda c} < \epsilon$
If $\lambda = 0$ the result is trivial.
So, assume $\lambda \ne 0$.
Then $\cmod \lambda > 0$ from the definition of the modulus of $\lambda$.
Hence $\dfrac \epsilon {\cmod \lambda} > 0$.
We have that $z_n \to c$ as $n \to \infty$.
Thus it follows that:
:$\exists N: \forall n > N: \cmod {z_n - c} < \dfrac \epsilon {\cmod \lambda}$
That is:
:$\forall n > N: \cmod \lambda \cmod {z_n - c} < \epsilon$
But we have:
{{begin-eqn}}
{{eqn | l = \cmod \lambda \cmod {z_n - c}
      | r = \cmod {\lambda \paren {z_n - c} }
      | c = Complex Modulus of Product of Complex Numbers
}}
{{eqn | r = \cmod {\lambda x_n - \lambda l}
      | c = 
}}
{{end-eqn}}
Hence:
:$\ds \lim_{n \mathop \to \infty} \paren {\lambda x_n} = \lambda c$
{{qed}}
Category:Combination Theorems for Sequences
ojhj47pxweiqj1kyydx1et87ekdix6i
Theorem: Combination Theorem for Sequences/Complex/Combined Sum Rule

:$\ds \lim_{n \mathop \to \infty} \paren {\lambda z_n + \mu w_n} = \lambda c + \mu d$

Proof:
From the Multiple Rule for Complex Sequences, we have:
:$\ds \lim_{n \mathop \to \infty} \paren {\lambda z_n} = \lambda c$
:$\ds \lim_{n \mathop \to \infty} \paren {\mu w_n} = \mu d$
The result now follows directly from the Sum Rule for Complex Sequences:
:$\ds \lim_{n \mathop \to \infty} \paren {\lambda z_n + \mu w_n} = \lambda c + \mu d$
{{qed}}
Category:Combination Theorems for Sequences
j577sjk30nn37zao5gysh3p3q0y4fe4
Theorem: Combination Theorem for Sequences/Normed Division Ring/Sum Rule

:$\sequence {x_n + y_n}$ 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 $\forall n > N$:
{{begin-eqn}}
{{eqn | l = \norm {\paren {x_n + y_n} - \paren {l + m} }
      | r = \norm {\paren {x_n - l} + \paren {y_n - m} }
      | c = 
}}
{{eqn | o = \le
      | r = \norm {x_n - l} + \norm {y_n - m}
      | c = {{Norm-axiom-mult|3}}
}}
{{eqn | o = <
      | r = \frac \epsilon 2 + \frac \epsilon 2 = \epsilon
      | c = 
}}
{{end-eqn}}
Hence:
:$\sequence {x_n + y_n}$ is convergent to $l + m$.
{{qed}}
Category:Combination Theorem for Sequences in Normed Division Rings
p85s5t87j343aaeujic6igs98jh75d2
Theorem: Absolute Value Function is Completely Multiplicative

Let $x, y \in \R$ be real numbers.
Then:
:$\size {x y} = \size x \size y$
where $\size x$ denotes the absolute value of $x$.
Thus the absolute value function is completely multiplicative.

Proof:
Let either $x = 0$ or $y = 0$, or both.
We have that $\size 0 = 0$ by definition of absolute value.
Hence:
:$\size x \size y = 0 = x y = \size {x y}$
Let $x > 0$ and $y > 0$.
Then:
{{begin-eqn}}
{{eqn | l = x y
      | o = >
      | r = 0
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \size {x y}
      | r = x y
      | c = {{Defof|Absolute Value}}
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | l = x
      | r = \size x
      | c = {{Defof|Absolute Value}}
}}
{{eqn | l = y
      | r = \size y
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \size x \size y
      | r = x y
      | c = 
}}
{{eqn | r = \size {x y}
      | c = 
}}
{{end-eqn}}
Let $x < 0$ and $y < 0$.
Then:
{{begin-eqn}}
{{eqn | l = x y
      | o = >
      | r = 0
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \size {x y}
      | r = x y
      | c = {{Defof|Absolute Value}}
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | l = -x
      | r = \size x
      | c = {{Defof|Absolute Value}}
}}
{{eqn | l = -y
      | r = \size y
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \size x \size y
      | r = \paren {-x} \paren {-y}
      | c = 
}}
{{eqn | r = xy
      | c = 
}}
{{eqn | r = \size {x y}
      | c = 
}}
{{end-eqn}}
The final case is where one of $x$ and $y$ is positive, and one is negative.
{{WLOG}}, let $x < 0$ and $y > 0$.
Then:
{{begin-eqn}}
{{eqn | l = x y
      | o = <
      | r = 0
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \size {x y}
      | r = -\paren {x y}
      | c = {{Defof|Absolute Value}}
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | l = -x
      | r = \size x
      | c = {{Defof|Absolute Value}}
}}
{{eqn | l = y
      | r = \size y
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \size x \size y
      | r = \paren {-x} y
      | c = 
}}
{{eqn | r = -\paren {x y}
      | c = 
}}
{{eqn | r = \size {x y}
      | c = 
}}
{{end-eqn}}
The case where $x > 0$ and $y < 0$ is the same.
{{qed}}
Theorem: Absolute Value Function is Completely Multiplicative

Let $x, y \in \R$ be real numbers.
Then:
:$\size {x y} = \size x \size y$
where $\size x$ denotes the absolute value of $x$.
Thus the absolute value function is completely multiplicative.

Proof:
Let $x$ and $y$ be considered as complex numbers which are wholly real.
That is:
:$x = x + 0 i, y = y + 0 i$
From Complex Modulus of Real Number equals Absolute Value, the absolute value of $x$ and $y$ equal the complex moduli of $x + 0 i$ and $y + 0 i$.
Thus $\cmod x \cmod y$ can be interpreted as the complex modulus of $x$ multiplied by the complex modulus of $y$.
By Complex Modulus of Product of Complex Numbers:
:$\cmod x \cmod y = \cmod {x y}$
As $x$ and $y$ are both real, so is $x y$.
Thus by Complex Modulus of Real Number equals Absolute Value, $\cmod {x y}$ can be interpreted as the absolute value of $x y$ as well as its complex modulus.
{{qed}}
Theorem: Absolute Value Function is Completely Multiplicative

Let $x, y \in \R$ be real numbers.
Then:
:$\size {x y} = \size x \size y$
where $\size x$ denotes the absolute value of $x$.
Thus the absolute value function is completely multiplicative.

Proof:
{{begin-eqn}}
{{eqn | l = \size {x y}
      | r = \sqrt {\paren {x y}^2}
      | c = {{Defof|Absolute Value|index = 2}}
}}
{{eqn | r = \sqrt {x^2 y^2}
      | c = 
}}
{{eqn | r = \sqrt {x^2} \sqrt{y^2}
      | c = 
}}
{{eqn | r = \size x \cdot \size y
      | c = {{Defof|Absolute Value|index = 2}}
}}
{{end-eqn}}
{{qed}}
Theorem: Absolute Value Function is Completely Multiplicative

Let $x, y \in \R$ be real numbers.
Then:
:$\size {x y} = \size x \size y$
where $\size x$ denotes the absolute value of $x$.
Thus the absolute value function is completely multiplicative.

Proof:
Follows directly from:
:Real Numbers form Ordered Integral Domain
:Product of Absolute Values on Ordered Integral Domain.
{{qed}}
Theorem: Combination Theorem for Sequences/Complex/Product Rule

:$\ds \lim_{n \mathop \to \infty} \paren {z_n w_n} = c d$

Proof:
Because $\sequence {z_n}$ converges, it is bounded by Convergent Sequence is Bounded.
Suppose $\cmod {z_n} \le K$ for $n = 1, 2, 3, \ldots$.
Then:
{{begin-eqn}}
{{eqn | l = \cmod {z_n w_n - c d}
      | r = \cmod {z_n w_n - z_n d + z_n d - c d}
      | c = 
}}
{{eqn | o = \le
      | r = \cmod {z_n w_n - z_n d} + \cmod {z_n d - c d}
      | c = Triangle Inequality for Complex Numbers
}}
{{eqn | r = \cmod {z_n} \cmod {w_n - d} + m \cdot \size {z_n - c}
      | c = Complex Modulus of Product of Complex Numbers
}}
{{eqn | o = \le
      | r = K \cdot \cmod {w_n - d} + \cmod d \cdot \cmod {z_n - c}
      | c = 
}}
{{eqn | o = =:
      | r = \phi_n
      | c = 
}}
{{end-eqn}}
But $z_n \to c$ as $n \to \infty$.
So $\cmod {z_n - c} \to 0$ as $n \to \infty$ from Convergent Sequence Minus Limit.
Similarly $\cmod {w_n - d} \to 0$ as $n \to \infty$.
From the Combined Sum Rule for Real Sequences:
:$\ds \lim_{n \mathop \to \infty} \paren {\lambda z_n + \mu w_n} = \lambda c + \mu d$, $\phi_n \to 0$ as $n \to \infty$
The result follows by the Squeeze Theorem for Complex Sequences.
{{qed}}