hcju/proofwikiqa huggingface dataset:
Recall that Real Numbers form Field. Thus by definition, $\R$ is also a division ring. Thus we only need to show that $\R$-module $\C$ is a unitary module, by demonstrating the module properties: $\forall x, y, \in \C, \forall \lambda, \mu \in \R$: :$(1): \quad \lambda \paren {x + y} = \paren {\lambda x} + \paren {\lambda y}$ :$(2): \quad \paren {\lambda + \mu} x = \paren {\lambda x} + \paren {\mu x}$ :$(3): \quad \paren {\lambda \mu} x = \lambda \paren {\mu x}$ :$(4): \quad 1 x = x$ As $\lambda, \mu \in \R$ it follows that $\lambda, \mu \in \C$. Thus from Complex Multiplication Distributes over Addition, $(1)$ and $(2)$ immediately follow. $(3)$ follows from Complex Multiplication is Associative. $(4)$ follows from Complex Multiplication Identity is One, as $1 + 0 i$ is the unity of $\C$. {{qed}}
Let $S$ be a non-zero subspace of $\left({\R^2, +, \times}\right)_\R$. Then $S$ contains a non-zero vector $\left({\alpha_1, \alpha_2}\right)$. Hence $S$ also contains $\left\{{\lambda \times \left({\alpha_1, \alpha_2}\right), \lambda \in \R}\right\}$. From Equation of Straight Line in Plane, this set may be described as a line through the origin. Suppose $S$ also contains a non-zero vector $\left({\beta_1, \beta_2}\right)$ which is not on that line. Then $\alpha_1 \times \beta_2 - \alpha_2 \times \beta_1 \ne 0$. Otherwise $\left({\beta_1, \beta_2}\right)$ would be $\zeta \times \left({\alpha_1, \alpha_2}\right)$, where either $\zeta = \beta_1 / \alpha_1$ or $\zeta = \beta_2 / \alpha_2$ according to whether $\alpha_1 \ne 0$ or $\alpha_2 \ne 0$. But then $S = \left({\R^2, +, \times}\right)_\R$. Because, if $\left({\gamma_1, \gamma_2}\right)$ is any vector at all, then: : $\left({\gamma_1, \gamma_2}\right) = \lambda \times \left({\alpha_1, \alpha_2}\right) + \mu \times \left({\beta_1, \beta_2}\right)$ where $\lambda = \dfrac {\gamma_1 \times \beta_2 - \gamma_2 \times \beta_1} {\alpha_1 \times \beta_2 - \alpha_2 \times \beta_1}, \mu = \dfrac {\alpha_1 \times \gamma_2 - \alpha_2 \times \gamma_1} {\alpha_1 \times \beta_2 - \alpha_2 \times \beta_1}$ which we get by solving the simultaneous eqns: {{begin-eqn}} {{eqn | l=\alpha_1 \times \lambda + \beta_1 \times \mu       | r=0       | c= }} {{eqn | l=\alpha_2 \times \lambda + \beta_2 \times \mu       | r=0       | c= }} {{end-eqn}} The result follows. {{qed}}
Follows directly from Dimension of Proper Subspace is Less Than its Superspace. {{qed}}
By definition, $\map \CC {\mathbb J} \subseteq \R^{\mathbb J}$. Let $f, g \in \map \CC {\mathbb J}$. By Two-Step Vector Subspace Test, it needs to be shown that: :$\paren 1: \quad f + g \in \map \CC {\mathbb J}$ :$\paren 2: \quad \lambda f \in \map \CC {\mathbb J}$ for any $\lambda \in \R$ $\paren 1$ follows by Sum Rule for Continuous Functions. $\paren 2$ follows by Multiple Rule for Continuous Functions. Hence $\struct {\map \CC {\mathbb J}, +, \times}_\R$ is a subspace of the $\R$-vector space $\struct {\R^{\mathbb J}, +, \times}_\R$. {{qed}}
The only sequence of distinct terms in $\set x$ is the one that goes: $x$. Suppose $\exists \lambda \in K: \lambda \circ x = e$. From Zero Vector Space Product iff Factor is Zero it follows that $\lambda = 0$. Hence the result from definition of linearly independent set. {{qed}}
=== Necessary Condition === Suppose $\sequence {a_k}_{1 \mathop \le k \mathop \le n}$ is linearly dependent. By hypothesis, the set of all integers $r \in \closedint 1 n$ such that $\sequence {a_k}_{1 \mathop \le k \mathop \le r}$ is linearly dependent is not empty. Let $p$ be its smallest element. Then from Singleton is Linearly Independent, $p \ge 2$, as $a_1 \ne \mathbf 0$ and hence $\set {a_1}$ is linearly independent. Also, there exist scalars $\lambda_1, \ldots, \lambda_p$, not all of which are zero, such that $\displaystyle \sum_{k \mathop = 1}^p \lambda_k \circ a_k = \mathbf 0$. Suppose $\lambda_p = 0$. Then not all of $\lambda_1, \ldots, \lambda_{p-1}$ can be zero. Then $\sequence {a_k}_{1 \mathop \le k \mathop \le p-1}$ is linearly dependent. That contradicts the definition of $p$, so $\lambda_p \ne 0$. So, because: :$\displaystyle \lambda_p \circ a_p = - \sum_{k \mathop = 1}^{p - 1} \lambda_k \circ a_k$ we must have: :$\displaystyle a_p = \sum_{k \mathop = 1}^{p - 1} \paren {-{\lambda_p}^{-1} \lambda_k} \circ a_k$ and thus $a_p$ is a linear combination of $\sequence {a_k}_{1 \mathop \le k \mathop \le p - 1}$. {{qed|lemma}} === Sufficient Condition === Now suppose that $a_p$ is a linear combination of $\sequence {a_k}_{1 \mathop \le k \mathop \le p - 1}$. Then: :$\displaystyle a_p = \sum_{k \mathop = 1}^{p - 1} \mu_k \circ a_k$ So we can assign values to $\lambda_k$ as follows: :$\forall k \in \closedint 1 n: \lambda_k = \begin{cases} \mu_k & : k < p \\ -1 & : k = p \\ 0 & : k > p \\ \end{cases}$ Then: :$\displaystyle \sum_{k \mathop = 1}^n \lambda_k \circ a_k = \mathbf 0$ Hence the result. {{qed}}
Let $\mathbb S$ be the set of all $S \subseteq G$ such that $S$ is a generator for $G$ and that $H \subseteq S \subseteq F$. Because $F \in \mathbb S$, it follows that $\mathbb S \ne \O$. Because $F$ is finite, then so is every element of $\mathbb S$. Let $R = \set {r \in \Z: r = \card S \in \mathbb S}$. That is, $R$ is the set of all the integers which are the number of elements in generators for $G$ that are subsets of $F$. Let $n$ be the smallest element of $R$. Let $B$ be an element of $\mathbb S$ such that $\card B = n$. We note that as $H$ is a linearly independent set, it does not contain $0$ by Subset of Module Containing Identity is Linearly Dependent. Then $0 \notin B$, or $B \setminus \set 0$ would be a generator for $G$ with $n - 1$ elements. This would contradict the definition of $n$. Let $m = \card H$. Let $\sequence {a_n}$ be a sequence of distinct vectors such that $H = \set {a_1, \ldots, a_m}$ and $B = \set {a_1, \ldots, a_n}$. Suppose $B$ were linearly dependent. By Linearly Dependent Sequence of Vector Space, there would exist $p \in \closedint 2 n$ and scalars $\mu_1, \ldots, \mu_{p - 1}$ such that $\displaystyle a_p = \sum_{k \mathop = 1}^{p - 1} \mu_k a_k$. As $H$ is linearly independent, $p > m$ and therefore $B' = B \setminus \set {a_p}$ would contain $H$. {{explain|why $H$ being L.I. implies $p > m$}} Now if $\displaystyle x = \sum_{k \mathop = 1}^n \lambda_k a_k$, then: :$\displaystyle x = \sum_{k \mathop = 1}^{p - 1} \paren {\lambda_k + \lambda_p \mu_k} a_k + \sum_{k \mathop = p + 1}^n \lambda_k a_k$ Hence $B'$ would be a generator for $G$ containing $n - 1$ elements, which contradicts the definition of $n$. Thus $B$ must be linearly independent and hence is a basis. {{qed}} {{Proofread}}
This follows from Vector Space has Basis Between Linearly Independent Set and Finite Spanning Set. It suffices to find: :A linearly independent subset $L\subset V$ :A finite generator $S\subset V$ with $L\subset S$. By Empty Set is Linearly Independent, we make take $L = \O$ and $S$ any finite generator, which exists because $V$ is finitely generated. {{qed}}
Since a basis is, by definition, both linearly independent and a generator, the result follows directly from Size of Linearly Independent Subset is at Most Size of Finite Generator. {{qed}}
Suppose that: :$\displaystyle \sum_{k \mathop = 1}^n \lambda_k x_k + \lambda b = 0$ where $\sequence {x_n}$ is a sequence of distinct vectors of $S$. If $\displaystyle \lambda \ne 0$, then $\displaystyle b = -\lambda^{-1} \paren {\sum_{k \mathop = 1}^n \lambda_k x_k} \in M$ which contradicts the definition of $b$. Hence $\lambda = 0$, and so: :$\displaystyle \sum_{k \mathop = 1}^n \lambda_k x_k = 0$ Therefore $\lambda_1 = \lambda_2 = \cdots = \lambda_n = \lambda = 0$ as $S$ is linearly independent. {{qed}}
=== 1 implies 2 and 3 === Let $B$ be a basis of $E$. Then conditions $(2)$ and $(3)$ follow directly by the definition of basis. {{qed|lemma}} === 2 implies 1 === Let $B$ be linearly independent. Suppose $B$ does not generate $E$. Then, because $\card B = n$, by Linearly Independent Subset also Independent in Generated Subspace there would be a linearly independent subset of $n + 1$ vectors of $E$. But this would contradict Size of Linearly Independent Subset is at Most Size of Finite Generator. Thus condition $(2)$ implies $(1)$. {{qed|lemma}} === 3 implies 1 === Let $B$ be a generator for $E$. By Vector Space has Basis Between Linearly Independent Set and Finite Spanning Set, $B$ contains a basis $B'$ of $E$. But $B'$ has $n$ elements and hence $B' = B$ by Bases of Finitely Generated Vector Space have Equal Cardinality. Thus condition $(3)$ implies $(1)$. {{Qed}}
Let $H$ be a subspace of $G$. Every linearly independent subset of the vector space $H$ is a linearly independent subset of the vector space $G$. Therefore, it has no more than $n$ elements by Size of Linearly Independent Subset is at Most Size of Finite Generator. So the set of all natural numbers $k$ such that $H$ has a linearly independent subset of $k$ vectors has a largest element $m$, and $m \le n$. Now, let $B$ be a linearly independent subset of $H$ having $m$ vectors. If the subspace generated by $B$ were not $H$, then $H$ would contain a linearly independent subset of $m + 1$ vectors. This follows by Linearly Independent Subset also Independent in Generated Subspace. This would contradict the definition of $m$. Hence $B$ is a generator for $H$ and is thus a basis for $H$. Thus $H$ is finite dimensional and $\map \dim H \le \map \dim G$. Now, if $\map \dim H = \map \dim G$, then a basis of $H$ is a basis of $G$ by Sufficient Conditions for Basis of Finite Dimensional Vector Space, and therefore $H = G$. {{Qed}}
If $\phi = 0$ then the assertion is clear. Let $\phi$ be a non-zero linear transformation. By Dimension of Proper Subspace is Less Than its Superspace and Generator of Vector Space Contains Basis, there is an ordered basis $\sequence {a_n}$ of $G$ such that: :$\exists r \in \N_n: \set {a_k: r + 1 \le k \le n}$ is a basis of $\map \ker \phi$ As a consequence: :$\map \nu \phi = n - r$ and by Unique Linear Transformation Between Vector Spaces: :$\map \rho \phi = r$ The result follows. {{qed}}
Suppose $\phi$ is a monomorphism. Let $\sequence {a_n}$ be a linearly independent sequence. Let: :$\displaystyle \sum_{k \mathop = 1}^n \lambda_k \map \phi {a_k} = 0$ Then: :$\displaystyle \map \phi {\sum_{k \mathop = 1}^n \lambda_k a_k} = 0$ So by hypothesis: :$\displaystyle \sum_{k \mathop = 1}^n \lambda_k a_k = 0$ Hence: :$\forall k \in \closedint 1 n: \lambda_k = 0$ Suppose that for every linearly independent sequence $\sequence {a_n}$ of vectors of $G$, $\sequence {\map \phi {a_n} }$ is a linearly independent sequence of vectors of $H$. Let $\map \phi {a_1} = 0$. Then $a_1 = 0$, otherwise the sequence $\sequence {a_1}$ of one term would be linearly independent but $\sequence {\map \phi {a_1} }$ would not. Thus $\map \ker \phi = \set 0$ and by the Quotient Theorem for Group Epimorphisms $\phi$ is an isomorphism and therefore a monomorphism. {{Qed}}
* $(1)$ implies $(2)$ by definition. * $(2)$ implies $(4)$ by Linear Transformation of Vector Space Monomorphism and Results concerning Generators and Bases of Vector Spaces. * $(4)$ implies $(5)$ by basic logic. * Suppose $\phi \left({B}\right)$ is a basis of $H$. Then the image of $\phi$ is a subspace of $H$ generating $H$ and hence is $H$ itself. Thus $(5)$ implies $(3)$. * Finally, $(3)$ implies that $\phi$ is injective. If $\phi$ is surjective, the dimension of its kernel is $0$ by Rank Plus Nullity Theorem. Hence $\phi$ is an isomorphism and therefore $(3)$ implies $(1)$. {{qed}}
:$(1): \quad M^\circ$ is an $\paren {n - m}$-dimensional subspace of $G^*$, and $M^{\circ \circ} = \map J M$ Let $\sequence {a_n}$ be an ordered basis of $G$ such that $\sequence {a_m}$ is an ordered basis of $M$. Let $\sequence {a'_n}$ be the ordered dual basis of $G^*$. Let $\ds t' = \sum_{k \mathop = 1}^n \lambda_k a'_k \in M^\circ$. Then: {{begin-eqn}} {{eqn | lo= \forall j \in \closedint 1 m:       | l = \lambda_j       | r = \sum_{k \mathop = 1}^n \lambda_k a'_k \paren {a_j}       | c =  }} {{eqn | r = \paren {\sum_{k \mathop = 1}^n \lambda_k a'_k} \paren {a_j}       | c =  }} {{eqn | r = t' \paren {a_j}       | c =  }} {{eqn | r = 0       | c =  }} {{end-eqn}} So $t'$ is a linear combination of $\set {a'_k: m + 1 \le k \le n}$. But $a'_k$ clearly belongs to $M^\circ$ for each $k \in \closedint {m + 1} n$. Therefore $M^\circ$ has dimension $n - m$. {{qed|lemma}} When we apply this result to $M^\circ$ instead of $M$, it is seen that the annihilator $M^{\circ \circ}$ of $M^\circ$ has dimension $n - \paren {n - m} = m$. But clearly $\map J M \subseteq M^{\circ \circ}$. As $J$ is an isomorphism, $\map J M$ has dimension $m$. So by Dimension of Proper Subspace is Less Than its Superspace, $\map J M = M^{\circ \circ}$. As a consequence, $\map {J^{-1} } {M^{\circ \circ} } = M$. Hence the result: $M^{\circ \circ} = \map J M$ {{qed|lemma}} :$(2) \quad \map {J^{-1} } {N^\circ}$ is an $\paren {n - p}$-dimensional subspace of $G$ If $N$ is a $p$-dimensional subspace of $G$, then $N^\circ$ and hence also $\map {J^{-1} } {N^\circ}$ have dimension $n - p$ by what has just been proved. {{qed|lemma}} By definition: $\paren {\map {J^{-1} } {N^\circ} }^\circ = \set {z' \in G: \forall x \in G: \forall t' \in N: \map {t'} x = 0: \map {z'} x = 0}$ Thus $N \subseteq \paren {\map {J^{-1} } {N^\circ} }^\circ$. But as $\paren {\map {J^{-1} } {N^\circ} }^\circ$ has dimension $n - \paren {n - p} = p$, it follows that $N = \paren {\map {J^{-1} } {N^\circ} }^\circ$ by Dimension of Proper Subspace is Less Than its Superspace. {{explain|Where in the above assertion is $(3)$ proved?}} :$(4) \quad$ Its inverse is the bijection $N \to \map {J^{-1} } {N^\circ}$. The final assertion follows by the definition of an inverse mapping. {{qed}}
Let $M^\circ$ be the annihilator of $M$. Let $N = M^{\circ}$. By Results Concerning Annihilator of Vector Subspace, $N$ is one-dimensional and $M = \map {J^{-1} } {N^\circ}$. Let $\phi \in N: \phi \ne 0$. Then $N$ is the set of all scalar multiples of $\phi$. Because: :$\map {J^{-1} } {N^\circ} = \set {x \in K^n: \forall \psi \in N: \map \psi x = 0}$ it follows that $\map {J^{-1} } {N^\circ}$ is simply the kernel of $\phi$. Hence $(1)$ implies $(2)$. By Rank Plus Nullity Theorem, $(2)$ also implies $(1)$. {{qed|lemma}} Suppose $\sequence {\alpha_n}$ is any sequence of scalars. Let $\sequence {e'_n}$ be the ordered basis of $\paren {K^n}^*$ dual to the standard ordered basis of $K^n$. Let $\displaystyle \phi = \sum_{k \mathop = 1}^n \alpha_k e'_k$. Then, by simple calculation: :$\map \ker \phi = \set {\tuple {\lambda_1, \ldots, \lambda_n}: \alpha_1 \lambda_1 + \cdots + \alpha_n \lambda_n = 0}$ {{explain|Prove the above.}} It follows that: :$\phi \ne 0 \iff \exists k \in \closedint 1 n: \alpha_k \ne 0$ Thus $(2)$ and $(3)$ are equivalent. {{qed|lemma}} Suppose $M = \map \ker \psi$, where $\displaystyle \psi = \sum_{k \mathop = 1}^n \beta_k e'_k$. Then $\psi = M^\circ$. As $M^\circ$ is one-dimensional and since $\psi \ne 0$, it follows that: :$\exists \gamma \ne 0: \psi = \gamma \phi$ Therefore: :$\forall k \in \closedint 1 n: \beta_k = \gamma \alpha_k$ {{qed}}
From the definitions of the transpose $u^t$ and the annihilator $\left({u \left({G}\right)}\right)^\circ$, it follows that: :$u^t \left({y'}\right) = 0 \iff y' = \left({u \left({G}\right)}\right)^\circ$ Thus: :$\ker \left({u^t}\right) = \left({u \left({G}\right)}\right)^\circ$. Let $x \in \ker \left({u}\right)$. Let $H^*$ be the algebraic dual of $H$. Let $\left \langle {x, t'} \right \rangle$ be the evaluation linear transformation. Then: :$\forall y' \in H^*: \left \langle {x, u^t \left({y'}\right)} \right \rangle = \left \langle {u \left({x}\right), y'} \right \rangle = \left \langle {0, y'} \right \rangle = 0$ So: :$u^t \left({H^*}\right) \subseteq \left({\ker \left({u}\right)}\right)^\circ$ From Rank Plus Nullity Theorem and Results Concerning Annihilator of Vector Subspace: {{begin-eqn}} {{eqn | l = \dim \left({u^t \left({H^*}\right)}\right)       | r = n - \dim \left({\ker \left({u^t}\right)}\right)       | c =  }} {{eqn | r = n - \dim \left({\left({u \left({G}\right)}\right)^\circ}\right)       | c =  }} {{eqn | r = \dim \left({u \left({G}\right)}\right)       | c =  }} {{eqn | r = n - \dim \left({\ker \left({u}\right)}\right)       | c =  }} {{eqn | r = \dim \left({\left({\ker \left({u}\right)}\right)^\circ}\right)       | c =  }} {{end-eqn}} So it follows that $u$ and $u^t$ have the same rank and nullity, and that: :$u^t \left({H^*}\right) = \left({\ker \left({u}\right)}\right)^\circ$ {{Qed}}
Let $\phi$ be a linear operator on $\R^2$. Let $\alpha_{11}, \alpha_{12}, \alpha_{21}, \alpha_{22} \in \R$ be the real numbers which satisfy the equations: {{begin-eqn}} {{eqn | l = \phi \left({e_1}\right)       | r = \alpha_{11} e_1 + \alpha_{21} e_2       | c =  }} {{eqn | l = \phi \left({e_2}\right)       | r = \alpha_{12} e_1 + \alpha_{22} e_2       | c =  }} {{end-eqn}} where $\left({e_1, e_2}\right)$ is the standard ordered basis of $\R^2$. Then, by linearity: {{begin-eqn}} {{eqn | l = \phi \left({\lambda_1, \lambda_2}\right)       | r = \phi \left({\lambda_1 e_1 + \lambda_2 e_2}\right)       | c =  }} {{eqn | r = \lambda_1 \phi \left({e_1}\right) + \lambda_2 \phi \left({e_2}\right)       | c =  }} {{eqn | r = \left({\lambda_1 \alpha_{11} + \lambda_2 \alpha_{12} }\right) e_1 + \left({\lambda_1 \alpha_{21} + \lambda_2 \alpha_{22} }\right) e_2       | c =  }} {{eqn | r = \left({\lambda_1 \alpha_{11} + \lambda_2 \alpha_{12}, \lambda_1 \alpha_{21} + \lambda_2 \alpha_{22} }\right)       | c =  }} {{end-eqn}} Conversely, if $\alpha_{11}, \alpha_{12}, \alpha_{21}, \alpha_{22} \in \R$ are ''any'' real numbers, then we can define the mapping $\phi$ as: : $\phi \left({\lambda_1, \lambda_2}\right) = \left({\lambda_1 \alpha_{11} + \lambda_2 \alpha_{12}, \lambda_1 \alpha_{21} + \lambda_2 \alpha_{22}}\right)$ which is easily verified as being a linear operator on $\R^2$: {{begin-eqn}} {{eqn | l = b \cdot \phi \left({\lambda_1, \lambda_2}\right) + c \cdot \phi \left({\lambda_3, \lambda_4}\right)       | r = b \left({\lambda_1 \alpha_{11} + \lambda_2 \alpha_{12}, \lambda_1 \alpha_{21} + \lambda_2 \alpha_{22} }\right) + c \left({\lambda_3 \alpha_{11} + \lambda_4 \alpha_{12}, \lambda_3 \alpha_{21} + \lambda_4 \alpha_{22} }\right)       | c =  }} {{eqn | r = \left({b \lambda_1 \alpha_{11} + b \lambda_2 \alpha_{12}, b \lambda_1 \alpha_{21} + b \lambda_2 \alpha_{22} }\right) + \left({c \lambda_3 \alpha_{11} + c\lambda_4 \alpha_{12}, c\lambda_3 \alpha_{21} + c\lambda_4 \alpha_{22} }\right)       | c =  }} {{eqn | r = \left({b \lambda_1 \alpha_{11} + b \lambda_2 \alpha_{12} + c \lambda_3 \alpha_{11} + c \lambda_4 \alpha_{12}, b \lambda_1 \alpha_{21} + b \lambda_2 \alpha_{22} + c \lambda_3 \alpha_{21} + c \lambda_4 \alpha_{22} }\right)       | c =  }} {{eqn | r = \left({\left({b \lambda_1 + c \lambda_3}\right) \alpha_{11} + \left({b \lambda_2 + c \lambda_4}\right) \alpha_{12}, \left({b \lambda_1 + c \lambda_3}\right) \alpha_{21} + \left({c \lambda_2 + c \lambda_4}\right) \alpha_{22} }\right)       | c =  }} {{eqn | r = \left({b \lambda_1 + c \lambda_3, b \lambda_2 + c \lambda_4}\right)       | c =  }} {{end-eqn}} Thus, by Condition for Linear Transformation, $\phi$ is a linear operator on $\R^2$. Thus each linear operator on $\R^2$ is completely determined by the ordered tuple: :$\left({\alpha_{11}, \alpha_{12}, \alpha_{21}, \alpha_{22}}\right)$ of real numbers. {{Qed}}
=== Step 1 === We will show that there exists a mapping from the infinite straight line $L$ to the set of real numbers $\R$. Let us establish a relation $h$ between points on $L$ and elements of $\R$. We allow the Axiom of Choice to set up a choice function to allow the points of $L$ to be selected systematically. Pick any point on $L$, and label it $z$. This point can be referred to informally as the origin. Map $z$ to zero, that is, by allowing $\tuple {z, 0} \in h$.  Using the choice function pick any other point on $L$. From Euclid's first postulate, we can draw a line segment between the two points. If the second point is to the right of the origin, let its length be positive. If the second point is to the left of the origin, let its length be negative. The existence of the choice function allows that this process can be done for any point on $L$ Hence: :$\forall p \in L: \exists x \in \R: \tuple {p, x} \in h$ That is, $h$ is left-total. The method of construction of $h$ is such that every point on $L$ is assigned to exactly one element of $\R$. Thus it is seen that $h$ is many-to-one. Hence, by definition, $h$ is a mapping. Since every point mapped to is associated to exactly one element of $L$, $h$ is injective. {{qed|lemma}} === Step 2 === Now we need to demonstrate that there exists a mapping from $\R$ to $L$. Let us establish a relation $h$ between elements of $\R$ and points on $L$. We allow the Axiom of Choice to set up a choice function to allow the elements of $\R$ to be selected systematically. Pick any point on $L$, and label it $z$. Map zero to $z$ that is, by allowing $\tuple {0, z} \in g$.  Using the choice function pick any other element $x$ of $\R$. Suppose $x$ is positive.  Then let the image of $x$ be a point to the right of the origin. Define the magnitude of its displacement be the absolute value of $x$. Suppose $x$ is negative.  Then let the image of $x$ be a point to the left of the origin. Define the magnitude of its displacement be the absolute value of $x$ multiplied by $-1$. The union of all such images, with the zero vector from $z$ to $z$, is the infinite straight line. By the method of construction, there exists a point on $L$ for all elements of $\R$, the relation $g$ is left-total. The method of construction of $g$ is also such that every element of $\R$ is assigned to exactly one point on $L$. Thus it is seen that $g$ is many-to-one. Hence, by definition, $g$ is a mapping. Since every element of $L$ mapped to is associated to exactly one point on $\R$, $g$ is injective. {{qed|lemma}} === Step 3 === Since: :$h: L \hookrightarrow \R$ is an injection  :$g: \R \hookrightarrow L$ is an injection by the Cantor-Bernstein-Schröder Theorem, there exists a one-to-one mapping between them. Hence by definition, $L$ and $\R$ are equinumerous. {{qed}}{{proofread}} {{explain|From a careful study of the discussion in the talk page, I suspect that we may need to assume and invoke an axiom of {{AuthorRef|Alfred Tarski}}'s.}} {{AoC}}
By definition, $1$ '''radian''' is the angle which sweeps out an arc on a circle whose length is the radius $r$ of the circle. From Perimeter of Circle, the length of the circumference of a circle of radius $r$ is equal to $2 \pi r$. Therefore, $1$ radian sweeps out $\dfrac 1 {2 \pi}$ of a circle. It follows that $2 \pi$ radians sweeps out the entire circle, or one full angle. {{qed}}
Let $\tuple {\lambda_1, \lambda_2} = \tuple {\rho \cos \sigma, \rho \sin \sigma}$. Then: {{begin-eqn}} {{eqn | l = r_\alpha \left({\lambda_1, \lambda_2}\right)       | r = \tuple {\rho \cos \alpha \cos \sigma - \rho \sin \alpha \sin \sigma, \rho \sin \alpha \cos \sigma + \rho \cos \alpha \sin \sigma}       | c =  }} {{eqn | r = \tuple {\lambda_1 \cos \alpha - \lambda_2 \sin \alpha, \lambda_1 \sin \alpha + \lambda_2 \cos \alpha}       | c =  }} {{end-eqn}} The result follows from Linear Operator on the Plane. {{ProofWanted|This definition requires to be approached from several conceptual directions.}}