avewright/proofwiki-math huggingface dataset:
Theorem: Gaussian Integers are Closed under Multiplication

The set of Gaussian integers $\Z \sqbrk i$ is closed under multiplication:
:$\forall x, y \in \Z \sqbrk i: x \times y \in \Z \sqbrk i$

Proof:
Let $x$ and $y$ be Gaussian integers.
Then:
{{begin-eqn}}
{{eqn | q = \exists a, b \in \Z
      | l = x
      | r = a + b i
      | c = {{Defof|Gaussian Integer}}
}}
{{eqn | q = \exists c, d \in \Z
      | l = y
      | r = c + d i
      | c = {{Defof|Gaussian Integer}}
}}
{{eqn | ll= \leadsto
      | l = x \times y
      | r = \paren {a c - b d} + i \paren {a d + b c}
      | c = {{Defof|Complex Multiplication}}
}}
{{eqn | ll= \leadsto
      | l = x \times y
      | o = \in
      | r = \Z \sqbrk i
      | c = Integer Addition is Closed, Integer Subtraction is Closed, Integer Multiplication is Closed
}}
{{end-eqn}}
{{qed}}
Theorem: Gaussian Integers are not Closed under Division

The set of Gaussian integers $\Z \sqbrk i$ is not closed under division.

Proof:
Proof by Counterexample:
Let:
{{begin-eqn}}
{{eqn | l = x
      | r = 1 + 2 i
}}
{{eqn | l = y
      | r = 3 + 4 i
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | ll= \leadsto
      | l = x \div y
      | r = \dfrac {1 + 2 i} {3 + 4 i}
      | c = {{Defof|Complex Division}}
}}
{{eqn | r = \dfrac {\paren {1 + 2 i} \paren {3 - 4 i} } {\paren {3 + 4 i} \paren {3 - 4 i} }
      | c = 
}}
{{eqn | r = \dfrac {11 + 2 i} {\sqrt {3^2 + 4^2} }
      | c = 
}}
{{eqn | r = \dfrac {11} 5 + \dfrac {2 i} 5
      | c = which is not a Gaussian integer
}}
{{end-eqn}}
{{qed}}
Theorem: Gauss-Markov Theorem

The least-squares estimator gives the unbiased (linear) estimator of a parameter having minimum variance.

Proof:
{{ProofWanted}}
{{Namedfor|Carl Friedrich Gauss|name2 = Andrey Andreyevich Markov|cat = Gauss|cat2 = Markov A A}}
Theorem: Generalized Eigenvalues as Roots of Equation

Let $\mathbf A$ be a square matrix of order $n$.
Let $\lambda$ be a generalized eigenvalue of $\mathbf A$.
Then:
:$\map \det {\mathbf A - \lambda \mathbf B} = 0$
where:
:$\mathbf B$ is another square matrix of order $n$
:$\det$ denotes the determinant.

Proof:
By definition of generalized eigenvalue:
{{begin-eqn}}
{{eqn | l = \mathbf A \mathbf x
      | r = \lambda \mathbf B \mathbf x
      | c = for some non-zero vector $\mathbf x$
}}
{{eqn | ll= \leadsto
      | l = \mathbf A \mathbf x - \lambda \mathbf B \mathbf x
      | r = 0
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \paren {\mathbf A - \lambda \mathbf B} \mathbf x
      | r = 0
      | c = 
}}
{{end-eqn}}
We have that $\mathbf x$ is non-zero {{hypothesis}}.
Hence from Matrix is Singular iff Product with Non-Zero Vector is Zero, the matrix $\mathbf A - \lambda \mathbf B$ is singular.
Hence $\map \det {\mathbf A - \lambda \mathbf B} = 0$ by definition of singular matrix.
{{qed}}
Theorem: Eigenvalue is Instance of Generalized Eigenvalue

Let $\mathbf A$ be a square matrix of order $n$. 
Let $\lambda$ be an eigenvalue of $\mathbf A$.
Then:
:$\lambda$ is a generalized eigenvalue of $\mathbf A$
:the corresponding eigenvector $\mathbf x$ corresponding to $\lambda$ is the generalized eigenvector of $\mathbf A$ corresponding to $\lambda$.

Proof:
By the definition of eigenvalue of $\mathbf A$:
:$\map \det {\mathbf I_n \mathbf x - \mathbf A} = 0$
for some non-zero vector $\mathbf x$.
Recall the definition of generalized eigenvalue of $\mathbf A$:
:$\mathbf A \mathbf x = \lambda \mathbf B \mathbf x$
for:
:some non-zero vector $\mathbf x$
:some square matrix $\mathbf B$ of order $n$.
Then from Generalized Eigenvalues as Roots of Equation:
:$\map \det {\mathbf A - \lambda \mathbf B} = 0$
where:
:$\mathbf B$ is another square matrix of order $n$
:$\det$ denotes the determinant.
Hence setting:
:$\mathbf B = \mathbf I_n$
it is seen that $\lambda$ is a generalized eigenvalue of $\mathbf A$
Hence, by definition, the eigenvector $\mathbf x$ corresponding to $\lambda$ is the generalized eigenvector of $\mathbf A$ corresponding to $\lambda$.
{{qed}}
Theorem: Special Linear Group is Normal Subgroup of General Linear Group

Let $K$ be a field whose zero is $0_K$ and unity is $1_K$.
Let $\SL {n, K}$ be the special linear group of order $n$ over $K$.
Then $\SL {n, K}$ is a normal subgroup of the general linear group $\GL {n, K}$.

Proof:
From Special Linear Group is Subgroup of General Linear Group, we have that $\SL {n, K}$ is a subgroup of $\GL {n, K}$.
{{Proofread}}
Let $\mathbf A \in \SL {n, K}$.
Let $\mathbf B \in \GL {n, K}$.
Then:
{{begin-eqn}}
{{eqn | l = \map \det {\mathbf B \mathbf A \mathbf B^{-1} } 
      | r = \map \det {\mathbf B} \map \det {\mathbf A} \map \det{\mathbf B^{-1} }
      | c = Determinant of Matrix Product
}}
{{eqn | r = \map \det {\mathbf B} \map \det {\mathbf A} \map \det{\mathbf B}^{-1} 
      | c = Determinant of Inverse Matrix
}}
{{eqn | r = \map \det {\mathbf B} \map \det {\mathbf B}^{-1}  \map \det {\mathbf A}
      | c = {{Field-axiom|M2}}
}}
{{eqn | r = 1_K \map \det {\mathbf A}
      | c = {{Field-axiom|M4}}
}}
{{eqn | r = \map \det {\mathbf A}
      | c = {{Field-axiom|M1}}
}}
{{eqn | r = 1_K
      | c = {{Defof|Special Linear Group}}
}}
{{end-eqn}}
Therefore by definition of special linear group:
:$\mathbf B \mathbf A \mathbf B^{-1} \in \SL {n, K}$
Thus $\SL {n, K}$ is a normal subgroup of $\GL {n, K}$
{{qed}}
Theorem: List of Elements in Infinite Cyclic Group

Let $\gen a$ be the infinite cyclic subgroup generated by $a$.
Then:
:$\set {\ldots, a^{-2}, a^{-1}, a^0, a^1, a^2, \ldots}$
is a complete repetition-free list of the elements of $\gen a$.
That is:
:$G = \set {a^n: n \in \Z}$

Proof:
From Infinite Cyclic Group is Isomorphic to Integers:
:$\gen a \cong \struct {\Z, +}$
That is, elements of $\gen a$ correspond one-to-one with $\struct {\Z, +}$ by the bijection $\phi: \Z \to \gen a$ defined as:
:$\forall x \in \Z: \map \phi x = a^x$
Hence the result.
{{qed}}
Theorem: Galois Group Acts Faithfully on Generating Set

Let $G$ be the Galois group of a finite field extension $E/ F$.
Let $\set {\alpha_1, \ldots , \alpha_n}$ be a generating set for field $E$ as an $F$-vector space.
Let $f_1, \ldots, f_n$ be their minimal polynomials.
Let $f = f_1 \dots f_n$ be the product of them.
Then the action of $G$ on roots of $f$ is faithful.

Proof:
{{tidy|reduce the use of compound sentences}}
Suppose that $\sigma \in G$ fixes each $\alpha_i$.
Then since $\sigma$ is an $F$-linear automorphism of $E$ and since $\set {\alpha_1, \ldots, \alpha_n}$ is a generating set for $E$, we see that $\sigma$ fixes every other element of $E$.
Then $\sigma$ must be the identity map $I_E: E \to E$.
Then the action of $G$ on roots of $f$ is faithful.
{{qed}}
Category:Galois Groups of Field Extensions
7gdaoloazl9nhsb4nb6ev1q6sgujs0t
Theorem: Euler's Homogeneous Function Theorem

Let $\map f{x,y}$ be a homogeneous function of order $n$ so that:
:$\map f {t x, t y} = t^n \map f {x, y}$
Then:
:$x \dfrac {\partial f} {\partial x} + y \dfrac {\partial f} {\partial y} = n \map f {x, y}$

Proof:
Define $x' = x t$ and $y' = y t$. 
By Chain Rule for Derivatives:
{{begin-eqn}}
{{eqn | l = n t^{n - 1} \map f {x, y}
      | r = \dfrac {\partial f} {\partial x'} \dfrac {\partial x'} {\partial t} + \dfrac {\partial f} {\partial y'} \dfrac {\partial y'} {\partial t}
}}
{{eqn | r = x \dfrac {\partial f} {\partial x'} + y \dfrac {\partial f} {\partial y'}
}}
{{eqn | r = x \dfrac {\partial f} {\partial x t} + y \frac {\partial f} {\partial y t}
}}
{{end-eqn}}
Let $t = 1$.
Then:
:$x \dfrac {\partial f} {\partial x} + y \dfrac {\partial f} {\partial y} = n \map f {x, y}$
{{qed}}
Theorem: Euler Phi Function by Argument is Injective

Let $f: \N \to \N$ be the arithmetic function defined as:
:$\map f n = n \map \phi n$ 
where $\map \phi n$ is the Euler Phi function (which is not injective).
Then $f$ is injective.

Proof:
By Product of Multiplicative Functions is Multiplicative and Euler Phi Function is Multiplicative, $f$ is a multiplicative function.
Suppose that $\map f {n_1} = \map f {n_2}$.
The case $n_1 = 1$ is trivial.
Let $n_1 > 1$.
Let $p$ be the largest prime factor of $n_1$.
By Largest Prime Factor of Euler Phi Function, the largest prime factor of $\map \phi {n_1}$ is less than $p$.
Thus $p$ is the largest prime factor of $\map f {n_1} = n_1 \map \phi {n_1}$.
But:
:$\map f {n_1} = \map f {n_2}$
Then $p$ is the largest prime factor of $\map f {n_2}$.
Then $p$ is the largest prime factor of $n_2$.
{{qed|lemma}}
Let $p^k$ be the largest power of $p$ dividing $n_1$.
Then $p^{k - 1}$ is the largest power of $p$ dividing $\map \phi {n_1}$.
Then $p^{2 k - 1}$ is the largest power of $p$ dividing $\map f {n_1}$.
But:
:$\map f {n_1} = \map f {n_2}$
Then $p^{2 k - 1}$ is the largest power of $p$ dividing $\map f {n_2}$.
Then $p^k$ is the largest power of $p$ dividing $n_2$.
{{qed|lemma}}
Thus, by the multiplicativity of $f$, we also have:
:$\map f {\dfrac {n_1} {p^k} } = \map f {\dfrac {n_2} {p^k} }$
{{Handwaving}}
Continuing, after having eliminated all prime factors of $n_1$, we have:
:$\map f 1 = \map f {\dfrac {n_2} {n_1} }$
hence:
:$\dfrac {n_2} {n_1} = 1$
and so:
:$n_2 = n_1$
{{qed}}
Theorem: Cyclic Group of Order 8 is not isomorphic to Group of Units of Integers Modulo n

Let $n \in \Z_{\ge 0}$ be an integer.
Let $\struct {\Z / n \Z, +, \cdot}$ be the ring of integers modulo $n$.
Let $U = \struct {\paren {\Z / n \Z}^\times, \cdot}$ denote the group of units of $\struct {\Z / n \Z, +, \cdot}$.
Let $C_8$ denote the cyclic group of order $8$
Then:
:$U$ and $C_8$ are not isomorphic.

Proof:
{{AimForCont}} $U$ and $C_8$ are isomorphic.
:$\order U = \order {C_8} = 8$
From Order of Group of Units of Integers Modulo n we have that
:$8 = \order U = \map \phi n$
where $\phi$ denotes the Euler $\phi$-function.
$\map \phi 1 = \map \phi 2 = 1$, so $n > 2$.
By Cyclicity Condition for Units of Ring of Integers Modulo n, either:
:$n = p^\alpha$
or:
:$n = 2 p^\alpha$
where $p \ge 3$ is prime and $\alpha \ge 1$.
In either case, we get:
:$8 = \map \phi n = p^{\alpha - 1} \paren{p - 1}$
so $p - 1 \divides 8$, but $p \ge 3$, so $p \in \set {3,5}$.
If $p = 3$, we get:
:$8 = 3^{\alpha - 1}\times 2$
contradiction.
If $p = 5$, we get:
:$8 = 5^{\alpha - 1}\times 4$
contradiction.
{{qed}}
Theorem: Units of Ring Direct Product are Ring Direct Product of Units

Let $R_1, R_2, \ldots, R_m$ be rings with identity.
{{MissingLinks|group isomorphism? This is rings we're talking about here.}}
Then the following group isomorphism holds:
:$\map U {R_1 \oplus R_2 \oplus \cdots \oplus R_n} \simeq \map U {R_1} \oplus \map U {R_2} \oplus \cdots \oplus \map U {R_n}$.
{{explain|what's $U$? What's $\oplus$?}}

Proof:
It suffices to prove the case $m = 2$.
{{explain|no it doesn't, we need an inductive argument}}
By {{Defof|Ring Direct Sum}} the finite direct sum 
{{explain|Is ring direct sum the same thing as finite direct sum? At least be consistent.}}
is isomorphic to the finite direct product.
{{mistake|I believe the wrong link has been implemented, because it does not match the text. Is this the RDP being linked to? Linking is of paramount importance on this site.}}
{{begin-eqn}}
{{eqn | l = \tuple {r_1, r_2}
      | o = \in
      | r = \map U {R_1 \times R_2}
      | c = {{Defof|Unit of Ring}}
}}
{{eqn | ll= \leadstoandfrom
      | l = \tuple {r_1, r_2}
      | o = 
      | r = 
      | c = is invertible in $R_1 \times R_2$
}}
{{eqn | ll= \leadstoandfrom
      | q = \exists \tuple {s_1, s_2} \in R_1 \times R_2
      | l = \tuple {r_1, r_2} \tuple {s_1, s_2}
      | r = \tuple {s_1, s_2} \tuple {r_1, r_2}
      | rr= = \tuple {1_{R_1}, 1_{R_2} }
      | c = 
}}
{{eqn | ll= \leadstoandfrom
      | l = \tuple {r_1 s_1, r_2 s_2}
      | r = \tuple {s_1 r_1, s_2 r_2}
      | rr= = \tuple {1_{R_1}, 1_{R_2} }
      | c = 
}}
{{eqn | ll= \leadstoandfrom
      | l = r_1 s_1
      | r = s_1 r_1
      | rr= = 1_{R_1}
      | c = 
}}
{{eqn | lo= \land
      | l = r_2 s_2
      | r = s_2 r_2
      | rr= = 1_{R_2}
      | c = 
}}
{{eqn | ll= \leadstoandfrom
      | l = r_1
      | o = 
      | r = 
      | c = is invertible in $R_1$
}}
{{eqn | lo= \land
      | l = r_2
      | o = 
      | r = 
      | c = is invertible in $R_2$
}}
{{eqn | ll= \leadstoandfrom
      | l = r_1
      | o = \in
      | r = \map U {R_1}
      | c = {{Defof|Unit of Ring}}
}}
{{eqn | lo= \land
      | l = r_2
      | o = \in
      | r = \map U {R_2}
      | c = {{Defof|Unit of Ring}}
}}
{{end-eqn}}
{{qed}}
Category:Ring Direct Products
li3syaocmwvgnhxfwab0r0vrkkvdefv
Theorem: Phi is 8 has only 5 solutions

$\phi$ denotes the Euler $\phi$-function.
There are $5$ numbers $n$ with the property that $\map \phi n = 8$, and they are $15$, $16$, $20$, $24$ and $30$.

Proof:
Suppose $\phi(n)=8$.
By Fundamental Theorem of Arithmetic, $n$ has a prime factorization: $$n=\prod_{i=1}^k p_i^{\alpha_i}$$
where $p_i$ are all distinct primes. Since $\phi(1)=1\neq8$, we assume $k>0$. Say that $n$ has $k$ distinct prime factors, so that each $\alpha_i > 0$.
By Euler Phi Function is Multiplicative, $$\phi(n) = 8 =\prod_{i=1}^k \phi(p_i^{\alpha_i})$$
By Euler Phi Function of Prime Power, $$8 = \prod_{i=1}^k (p_i-1)p_i^{\alpha_i-1}$$
Because each $p_i-1$ is a factor of $8$, it follows that $p_i-1\leq 8$. Thus, each $p_i\leq 9$. The only primes less than $9$ are $2$, $3$, $5$, and $7$. Further, $7-1$ is not a factor of $8$, so $7$ is not a prime factor of $n$.
Thus, $n$ is of the form $2^{a}3^{b}5^{c}$, for nonnegative integers $a, b, c$.
By Euler Phi Function is Multiplicative and Euler Phi Function of Prime Power, if $b$ or $c$ are greater than $1$, then $\phi(n)$ is divisible by $3$ or $5$, respectively. This is not the case, so $3$ and $5$ each divide $n$ at most once. 
This splits the analysis into four cases. Because $\phi(n)\leq n$ for all $n\in\mathbb{N}$, $n$ cannot be $1$, $3$, or $5$. Thus, in the first three cases, $n$ must have at least one factor of $2$.
Theorem: Ramaré's Theorem

Every positive integer can be expressed as the sum of no more than $6$ prime numbers.

Proof:
{{ProofWanted}}
{{Namedfor|Olivier Ramaré|cat = Ramaré}}
Theorem: Dirichlet's Integral Form of Digamma Function

:$\ds \map \psi z = \int_0^\infty \paren {\frac {e^{-t} } t - \frac 1 {t \paren {1 + t}^z } } \rd t$

Proof:
We have:
{{begin-eqn}}
{{eqn | l = \map \psi z 
      | r = \int_0^\infty \paren {\frac {e^{-t} } t - \frac {e^{-z t} } {1 - e^{-t} } } \rd t
      | c = Gauss's Integral Form of Digamma Function
}}
{{eqn | r = \int_0^\infty \frac {e^{-t} } t  \rd t - \int_0^\infty \frac {\paren {e^{-t} }^z } {1 - e^{-t} } \rd t
      | c = Linear Combination of Integrals
}}
{{end-eqn}}
Let:
{{begin-eqn}}
{{eqn | l = \frac 1 {\paren {1 + x} }
      | r = e^{-t}
      | c = Integration by Substitution
}}
{{eqn | ll= \leadsto
      | l = \paren {1 + x}
      | r = e^t
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \map \ln {1 + x}
      | r = t
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac 1 {\paren {1 + x} } \rd x
      | r = \d t
      | c = 
}}
{{end-eqn}}
We also have:
{{begin-eqn}}
{{eqn | l = \lim_{x \mathop \to 0} \map \ln {1 + x}
      | r = 0
      | c = Natural Logarithm of 1 is 0
}}
{{eqn | l = \lim_{x \mathop \to \infty} \map \ln {1 + x}
      | r = \infty
      | c = Logarithm Tends to Infinity
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \map \psi z
      | r = \int_0^\infty \frac {e^{-t} } t \rd t - \int_0^\infty \frac {\paren {e^{-t} }^z} {1 - e^{-t} } \rd t
      | c = 
}}
{{eqn | r = \int_0^\infty \frac {e^{-t} } t \rd t - \int_0^\infty \frac {\paren {\frac 1 {\paren {1 + x} } }^z} {1 - \frac 1 {\paren {1 + x} } } \paren {\frac 1 {\paren {1 + x} } \rd x}
      | c = substituting in second integral
}}
{{eqn | r = \int_0^\infty \frac {e^{-t} } t \rd t - \int_0^\infty \frac {\paren {\frac 1 {\paren {1 + x} } }^z} {\paren {1 + x} - 1} \rd x
      | c = 
}}
{{eqn | r = \int_0^\infty \frac {e^{-t} } t \rd t - \int_0^\infty \frac 1 {x \paren {1 + x}^z} \rd x
      | c = 
}}
{{eqn | r = \int_0^\infty \frac {e^{-t} } t \rd t - \int_0^\infty \frac 1 {t \paren {1 + t}^z} \rd t
      | c = $x \to t$
}}
{{eqn | r = \int_0^\infty \paren {\frac {e^{-t} } t - \frac 1 {t \paren {1 + t}^z} } \rd t
      | c = Linear Combination of Integrals
}}
{{end-eqn}}
{{qed}}
{{Namedfor|Johann Peter Gustav Lejeune Dirichlet|cat = Dirichlet}}
Theorem: Lüroth's Theorem

{{MissingLinks|throughout}}
Let $K$ be a field.
Let $\map K X$ be the rational function field, for some indeterminate $X$.
Let $M$ be an intermediate field between $K$ and $\map K X$.
Then there exists a rational function $\map f X \in \map K X$ such that:
{{explain|Domain and codomain of $\map f X$}}
:$M = \map K {\map f X}$
In other words, every intermediate extension between $K$ and $\map K X$ is a simple extension.

Proof:
By Gauss's Lemma (Polynomial Theory),
{{ProofWanted}}
{{Namedfor|Jacob Lüroth|cat = Lüroth}}
Theorem: Abel-Plana Formula

:$\ds \sum_{n \mathop = 0}^\infty \map f n = \int_0^\infty \map f x \rd x + \dfrac 1 2 \map f 0 + i \int_0^\infty \dfrac {\map f {i t} - \map f {-i t} } {e^{2 \pi t} - 1} \rd t$

Proof:
{{ProofWanted}}
{{Namedfor|Niels Henrik Abel|name2 = Giovanni Antonio Amedeo Plana|cat = Abel|cat2 = Plana}}
Theorem: Isomorphism between Group of Units Ring of Integers Modulo 2^n and C2 x C2^(n-2)

Let $n \in \Z_{\ge 0}$ be a positive integer and $n \ge 2$.
Let $R = \struct {\Z / 2^n \Z, +, \times}$ be the ring of integers modulo $2^n$.
Let $U = \struct {\paren {\Z / 2^n \Z}^\times, \times}$ denote the group of units of $R$.
Let $C_2$ be cyclic group of order $2$.
Let $C_{2^{n - 2} }$ be the cyclic group of order $2^{n - 2}$.
Then $U$ is isomorphic to the group direct product of $C_2$ and $C_{2^{n - 2} }$.

Proof:
{{tidy|coherence needs to be improved}}
Theorem: Isomorphism between Group of Units of Ring of Integers Modulo p^n and C((p-1)p^(n-1))

Let $n \in \Z_{\ge 0}$ be a positive integer.
Let $p$ be an odd prime.
Let $R = \struct {\Z / p^n \Z, +, \times}$ be the ring of integers modulo $p^n$.
Let $U = \struct {\paren {\Z / p^n \Z}^\times, \times}$ denote the group of units of $R$.
Let $C_{(p - 1)p^{n - 1}}$ be be cyclic group of order $(p - 1)p^{n - 1}$.
Then $U$ is isomorphic to $C_{(p - 1)p^{n - 1}}$.

Proof:
The case $n = 1$ is proved in Ring of Integers Modulo Prime is Field.
Suppose $n \ge 2$.
{{ProofWanted}}
Theorem: Kernel of Linear Transformation is Linear Subspace

Let $V, W$ be normed vector spaces on a field $F$.
{{explain|Why do they need to be normed?}}
Let $T: V \to W$ be a linear transformation.
Then the kernel of $T$ is a linear subspace of $V$.
{{explain|We need to decide whether we want to use the term "linear subspace" in the first place. The term has a number of different meanings depending on the universe of discourse.}}

Proof:
By Kernel of Linear Transformation contains Zero Vector, $0_V \in \ker T$.
Theorem: Complete Graph K5 is not Planar

The complete graph $K_5$ is not planar.

Proof:
Recall the definition of planar graph:
{{:Definition:Planar Graph}}
First we note that the complete graph $K_4$ is planar by demonstrating its planarity:
:200px
Now we attempt to create $K_5$ by building it from its subgraph $K_4$.
Each of the faces of $K_4$ is in the same form as each of the others.
That is, each face consists of a triangle of $3$ vertices which is not incident to the $4$th vertex.
So exactly where in the plane we place the $5$th vertex does not matter.
{{WLOG}} let us place vertex $E$ outside the triangle $ABC$ as shown below:
:300px
Edges $AE$, $BE$ and $CE$ can be constructed without crossing an existing edge of $K_4$, as indicated by dotted black lines.
However, it is not possible to construct edge $DE$ without crossing one of $AB$, $AC$ or $BC$.
This is shown in dotted $\color{red} {\text {red} }$.
Hence the result.
{{qed}}
Theorem: Thomsen Graph is not Planar

The Thomsen graph is not planar.

Proof:
{{Recall|Thomsen Graph|Thomsen graph}}
{{:Definition:Thomsen Graph}}
{{Recall|Planar Graph|planar graph}}
{{:Definition:Planar Graph}}
Recall that the complete bipartite graph $K_{2, 3}$ is planar:
:280px
Let $A = \set {A_1, A_2, A_3}$ and $B = \set {B_1, B_2, B_3}$ be the two subsets which partition the vertex set of $K_{3, 3}$ which we are to demonstrate non-planarity.
Now we attempt to create $K_{3, 3}$ by building it from its subgraph $K_{2, 3}$, using $\set {A_1, A_2}$ (black) and $\set {B_1, B_2, B_3}$ (white) as the partition of the vertex set of $K_{2, 3}$.
Each of the faces of $K_{2, 3}$ is in the same form as each of the others.
That is, each face consists of $4$ vertices, which is not incident to the $5$th vertex of $K_{2, 3}$.
So exactly where in the plane we place the $6$th vertex does not matter.
{{WLOG}} let us place vertex $A_3$ outside the face $A_1 B_1 A_2 B_3$ as shown below:
:500px
This face is not incident to $B_2$.
Edges $A_3 B_1$ and $A_3 B_3$ can be constructed without crossing an existing edge of $K_{2, 3}$, as indicated by dotted black lines.
However, it is not possible to construct edge $A_3 B_2$ without crossing one of $A_1 B_1$ or $A_1 B_3$, $A_2 B_1$ or $A_2 B_3$.
This is shown in dotted $\color{red} {\text {red} }$.
Hence the result.
{{qed}}
Theorem: Sobolev Space is Banach Space

Let $k \in \Z_{\ge 0}$.
Let $p \in \R_{\ge 1}$ or $p = \infty$.
Let $U \subset \R^n$ be an open set.
Then the Sobolev space $\map {W^{k, p} } U$ equipped with the Sobolev norm is a Banach space.

Proof:
{{tidy}}
{{MissingLinks|still some missing}}
By Sobolev Norm is Norm, it remains to show that $\map {W^{k, p} } U$ is complete.
Assume $\sequence {u_m}$ is a Cauchy sequence in $\map {W^{k, p} } U$.
Then for each multiindex $\size \alpha \le k$, $\sequence {D^\alpha u_m}$ is a Cauchy sequence in $\map {L^p} U$.
From the Riesz-Fischer Theorem, $\map {L^p} U$ is complete.
Hence $\sequence {D^\alpha u_m}$ converges in $L^p$ norm:
:$D^\alpha u_m \to u_\alpha\in \map {L^p} U$
for each $\size \alpha \le k$.
In particular, $\sequence{u_m}$ converges in $L^p$ norm:
:$u_m \to u_{\tuple {0, \ldots, 0} } =: u$
We now claim:
{{begin-eqn}}
{{eqn | n = 1
      | q = \forall u \in \map {W^{k, p} } U: \forall \size \alpha \le k
      | l = D^\alpha u
      | r = u_\alpha
}}
{{end-eqn}}
To verify this assertion, fix $\phi \in \map {C_c^\infty} U$.
Then:
{{begin-eqn}}
{{eqn | l = \int_U u D^\alpha \phi \rd x
      | r = \lim_{m \mathop \to \infty} \int_U u_m D^\alpha \phi \rd x
      | c = Definition of $u$
}}
{{eqn | r = \lim_{m \mathop \to \infty} \paren {-1}^{\size \alpha} \int_U D^\alpha u_m \phi \rd x
      | c = Integration by Parts
}}
{{eqn | r = \paren {-1}^{\size \alpha} \int_U u_\alpha \phi \rd x
      | c = Definition of $u_\alpha$
}}
{{end-eqn}}
By the definition of distributional derivative, $(1)$ holds.
Therefore:
:$D^\alpha u_m \to D^\alpha u$ in $\map {L^p} U$
for all $\size \alpha \le k$.
We see that:
:$u_m \to u$ in $\map {W^{k, p} } U$
as required.
{{qed}}
Theorem: Sobolev Norm is Norm

Let $k \in \Z_+$ and $1 \le p \le \infty$.
Let $U \subset \R^n$ be an open set.
Then the Sobolev norm is a norm on the Sobolev space $\map {W^{k, p} } U$.

Proof:
=== {{NormAxiomVector|1|nolink}} ===
Let $u, v \in \map {W^{k, p} } U$.
From P-Seminorm of Function Zero iff A.E. Zero: 
:$\norm u_{\map {W^{k, p} } U} = 0$
for $u \in \map {W^{k, p} } U$ {{iff}} $u = 0$ almost everywhere.
{{MissingLinks|The above is stated as a fact without reference to the question in hand. }}
So {{NormAxiomVector|1}} is satisfied.
{{qed|lemma}}