hcju/proofwikiqa huggingface dataset:
By Cauchy Sequence Equivalence, for all sequences $\sequence {x_n}$ in $R$: :$\sequence {x_n}$ is a Cauchy sequence in $\norm{\,\cdot\,}_1$ {{iff}} $\sequence {x_n}$ is a Cauchy sequence in $\norm{\,\cdot\,}_2$. By Convergent Equivalence, for all sequences $\sequence {x_n}$ in $R$: :$\sequence {x_n}$ converges in $\norm{\,\cdot\,}_1$ {{iff}} $\sequence {x_n}$ converges in $\norm{\,\cdot\,}_2$. Hence: :every Cauchy sequence in $\norm{\,\cdot\,}_1$ converges in $\norm{\,\cdot\,}_1$ {{iff}} every Cauchy sequence in $\norm{\,\cdot\,}_2$ converges in $\norm{\,\cdot\,}_2$. The result follows. {{qed}} Category:Normed Division Rings Category:Complete Metric Spaces ekkypuxvfwisc4mndifp6qmzk0hcwwq
From Ring of Polynomial Forms over Field is Vector Space we note that $\struct {F, +, \times}$ is a vector space over $F$. The remaining question is that $S$ remains closed under polynomial addition and scalar multiplication. Let $\mathbf x, \mathbf y \in S$ such that $\map \deg {\mathbf x} = m$ and $\map \deg {\mathbf y} = n$. We have: {{begin-eqn}} {{eqn | l = \mathbf x + \mathbf y       | r = \sum_{j \mathop = 0}^m x_j X^j + \sum_{k \mathop = 0}^n y_k X^k       | c =  }} {{eqn | r = \sum_{k \mathop = 0}^{\max \set {m, n} } \paren {x_k + y_k} X^k       | c = {{Defof|Polynomial Addition}} }} {{end-eqn}} As both $m < d$ and $n < d$ by dint of them belonging to $S$, it follows that $\max \set {m, n} < d$. Hence $\mathbf x + \mathbf y \in S$ and so $S$ is closed under polynomial addition. Let the operation $\times': F \to F \sqbrk X$ be defined as follows. Let $\lambda \in F$. Let $\mathbf x \in F \sqbrk X$ be defined as: :$\mathbf x = \displaystyle \sum_{k \mathop = 0}^n x_k X^k$ where $n = \map \deg {\mathbf x}$ denotes the degree of $\mathbf x$. Thus: :$\lambda \times' \mathbf x := \displaystyle \lambda \times' \sum_{k \mathop = 0}^n x_k X^k = \displaystyle \sum_{k \mathop = 0}^n \paren {\lambda \times x_k} X^k$ We have that $\times': F \to F \sqbrk X$ is an instance of polynomial multiplication where the multiplier $\lambda$ is a polynomial of degree $0$. By definition of polynomial multiplication: {{begin-eqn}} {{eqn | l = \map \deg {\lambda \times' \mathbf x}       | r = \map \deg \lambda + \map \deg {\mathbf x}       | c =  }} {{eqn | r = 0 + \map \deg {\mathbf x}       | c =  }} {{eqn | o = <       | r = d       | c = as $\mathbf x \in S$ }} {{end-eqn}} The result follows. {{qed}}
By definition of the $p$-adic numbers $\Q_p$, the rational numbers $\Q$ are dense in $\Q_p$. By the definition of a dense subset then $\map \cl \Q = \Q_p$. By Closure of Subset of Metric Space by Convergent Sequence then: :there exists a sequence $\sequence {x_n} \subseteq \Q$ that converges to $x$. That is: :$\displaystyle \lim_{n \mathop \to \infty} x_n = x$ By Convergent Sequence in Normed Division Ring is Cauchy Sequence, $\sequence {x_n}$ is a Cauchy sequence. Since $\sequence {x_n}$ does not converge to $0$, by Non-Null Cauchy Sequence in Non-Archimedean Norm is Eventually Stationary then: :$\exists N \in \N: \forall n, m > N: \norm {x_n}_p = \norm {x_m}_p$ By definition of the $p$-adic norm on $\Q$: :$\exists v \in \Z: \norm {x_{N + 1} }_p = p^{-v}$ Hence: :$\forall n > N: \norm {x_n}_p = p^{-v}$ Let $\sequence {y_n}$ be the subsequence of $\sequence {\norm {x_n}}$ defined by: :$\forall n: y_n = \norm {x_{N + n} }$ Then $\sequence {y_n}$ is the constant sequence $\tuple {p^{-v}, p^{-v}, p^{-v}, \dotsc}$. Hence: {{begin-eqn}} {{eqn | l = \norm x_p       | r = \displaystyle \lim_{n \mathop \to \infty} \norm {x_n}       | c = Modulus of Limit }} {{eqn | r = \displaystyle \lim_{n \mathop \to \infty} y_n       | c = Limit of Subsequence equals Limit of Real Sequence }} {{eqn | r = p^{-v}       | c = Constant Sequence Converges to Constant in Normed Division Ring }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | lo= \neg \forall x:       | l = \exists y       | o = :       | r = \map P {x, y}       | c =  }} {{eqn | ll= \leadstoandfrom       | lo= \exists x:       | l = \neg \exists y       | o = :       | r = \map P {x, y}       | c = Denial of Universality }} {{eqn | ll= \leadstoandfrom       | lo= \exists x:       | l = \forall y       | o = :       | r = \neg \map P {x, y}       | c = Denial of Existence }} {{end-eqn}} {{qed}}
By Principal Ideal from Element in Center of Ring then $p \Z_p$ is a principal ideal. Now: {{begin-eqn}} {{eqn | l = \norm x_p       | o = <       | r = 1 }} {{eqn | ll= \leadstoandfrom        | l = \norm x_p       | o = \le       | r = \dfrac 1 p       | c = P-adic Norm of p-adic Number is Power of p }} {{eqn | ll= \leadstoandfrom        | l = p \norm x_p       | o = \le       | r = 1 }} {{eqn | ll= \leadstoandfrom        | l = \dfrac {\norm x_p} {\norm p_p}       | o = \le       | r = 1       | c = as ${\norm p_p} = \dfrac 1 p$ }} {{eqn | ll= \leadstoandfrom        | l = \norm {\dfrac x p}_p       | o = \le       | r = 1       | c = Norm of quotient }} {{eqn | ll= \leadstoandfrom        | l = \dfrac x p       | o = \in       | r = \Z_p       | c = {{Defof|P-adic Integers|$p$-adic integers}} }} {{eqn | ll= \leadstoandfrom        | l = x       | o = \in       | r = p \Z_p }} {{end-eqn}} Hence: :$p \Z_p = \set {x \in \Q_p: \norm x_p < 1}$ {{qed}}
Let $n \in \N$. By definition of the $p$-adic numbers, the rational numbers are dense in $\Q_p$. So there exists: :$\dfrac a b \in \Q: \norm {x - \dfrac a b}_p \le p^{-n}$ From Unique Integer Close to Rational in Valuation Ring of P-adic Norm,  there exists  unique $\alpha \in \Z$ such that: :$\norm {\dfrac a b - \alpha}_p \le p^{-n}$ :$0 \le \alpha \le p^n - 1$ Then: {{begin-eqn}} {{eqn | l = \norm {x - \alpha}_p       | r = \norm {\paren {x - \dfrac a b} + \paren {\dfrac a b - \alpha} }_p }} {{eqn | o = \le       | r = \max \set {\norm {x - \dfrac a b}_p, \: \norm {\dfrac a b - \alpha}_p }       | c = Norm axiom (N4) (Ultrametric Inequality) }} {{eqn | o = \le       | r = p^{-n} }} {{end-eqn}} Now suppose $\beta \in \Z$ also satisfies conditions $(1)$ and $(2)$, that is: :$0 \le \beta \le p^n - 1$ :$\norm { x - \beta}_p \le p^{-n}$ Then: {{begin-eqn}} {{eqn | l = \norm{\dfrac a b - \beta}_p       | r = \norm{\paren{\dfrac a b - x} + \paren{x - \beta} }_p }} {{eqn | o = \le       | r = \max \set{\norm{\dfrac a b - x}_p,\:\norm{x - \beta}_p}       | c = Norm axiom (N4) (Ultrametric Inequality) }} {{eqn | o = \le       | r = \max \set{\norm{x - \dfrac a b}_p,\:\norm{x - \beta}_p}       | c = Norm of Negative }} {{eqn | o = \le       | r = p^{-n} }} {{end-eqn}} From Unique Integer Close to Rational in Valuation Ring of P-adic Norm,  $\alpha \in \Z$ was unique, so: :$\beta = \alpha$ The result follows. {{qed}}
By Open Ball Characterization of Denseness it is sufficient to show that every open ball of $\struct {\Z_p, d_p}$ contains an element of $\Z$. Let $x \in \Z_p$ and $\epsilon \in \R_{>0}$. By definition the open ball $\map {B_\epsilon} x$ is: :$\map {B_\epsilon} x = \set {y \in \Z_p: \norm y_p < \epsilon}$ By Sequence of Powers of Number less than One then: :$\displaystyle \lim_{n \mathop \to \infty} p^{-n} = 0$ Hence there exists $N \in \N$: :$\forall n \ge N: p^{-n} < \epsilon$ Consider the open ball $\map {B_{p^{-N} } } x$. Since $0 < p^{-n} < \epsilon$ then: :$\map {B_{p^{-N} } } x \subseteq \map {B_\epsilon} x$. By Integers are Arbitrarily Close to P-adic Integers then: :$\exists \alpha \in \Z: \alpha \in \map {B_{p^{-N} } } x$ Hence $\alpha \in \map {B_\epsilon} x$. Since $x$ and $\epsilon$ were arbitrary then every open ball of $\struct {\Z_p, d_p}$ contains an element of $\Z$. By Open Ball Characterization of Denseness then $\Z$ is dense in the metric space $\struct {\Z_p, d_p}$. {{qed}}
By the definition of a coherent sequence it needs to be proved that there exists a unique integer sequence $\sequence {\alpha_n}$: :$(1): \quad \forall n \in \N: \alpha_n \in \Z$ and $0 \le \alpha_n \le p^{n + 1} - 1$ :$(2): \quad \forall n \in \N: \alpha_{n + 1} \equiv \alpha_n \pmod {p^{n + 1}}$ :$(3): \quad \displaystyle \lim_{n \mathop \to \infty} \alpha_n = x$ By Integers are Arbitrarily Close to P-adic Integers then for all $n \in \N$ there exists $\alpha_n \in \Z$: ::$\text {(a)}: \quad 0 \le \alpha_n \le p^{n + 1} - 1$ ::$\text {(b)}: \quad \norm {x - \alpha_n}_p \le p^{-\paren{n + 1} }$ Hence the sequence $\sequence {\alpha_n}$ satisfies $(1)$ above. === Lemma 1 === {{:P-adic Integer is Limit of Unique Coherent Sequence of Integers/Lemma 1}} Hence the sequence $\sequence {\alpha_n}$ satisfies $(2)$ above. === Lemma 2 === {{:P-adic Integer is Limit of Unique Coherent Sequence of Integers/Lemma 2}} Hence the sequence $\sequence {\alpha_n}$ satisfies $(3)$ above. === Lemma 3 === {{:P-adic Integer is Limit of Unique Coherent Sequence of Integers/Lemma 3}}{{qed}}
Let $\triangle ABC$ be such that $\angle C$ is a right angle. Let: :$A$ be opposite the side $a$ :$B$ be opposite the side $b$ :$C$ be opposite the side $c$ :500px From Solutions of Pythagorean Equation, we have that: :$c = m^2 + n^2$ for some $m, n \in \Z_{>0}$, and that the other sides are $m^2 - n^2$ and $2 m n$ {{WLOG}}, let the sides $a$ and $b$ be such that: {{begin-eqn}} {{eqn | l = a       | r = m^2 - n^2 }} {{eqn | l = b       | r = 2 m n }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = c       | r = \paren {a - r} + \paren {b - r}       | c =  }} {{eqn | r = a + b - 2 r       | c =  }} {{eqn | ll= \leadsto       | l = r       | r = \dfrac {a + b - c} 2       | c =  }} {{eqn | ll= \leadsto       | l = r       | r = \dfrac {m^2 - n^2 + 2 m n - \paren {m^2 + n^2} } 2       | c =  }} {{eqn | ll= \leadsto       | l = r       | r = \dfrac {2 m n - 2 n^2} 2       | c =  }} {{eqn | ll= \leadsto       | l = r       | r = n \paren {m - n}       | c =  }} {{end-eqn}} As $m$ and $n$ are both integers, it follows that $r$ is also an integer. {{qed}}
Let $\triangle ABC$ be such that $\angle C$ is a right angle. Let: :$A$ be opposite the side $a$ :$B$ be opposite the side $b$ :$C$ be opposite the side $c$ :500px Let $I$ denote the incenter of $\triangle ABC$. From Solutions of Pythagorean Equation, we have that: :$c = m^2 + n^2$ for some $m, n \in \Z_{>0}$, and that the other sides are $m^2 - n^2$ and $2 m n$ {{WLOG}}, let the sides $a$ and $b$ be such that: {{begin-eqn}} {{eqn | l = a       | r = m^2 - n^2 }} {{eqn | l = b       | r = 2 m n }} {{end-eqn}} We have that: {{begin-eqn}} {{eqn | l = \map \Area {\triangle ABC}       | r = \map \Area {\triangle BIC} + \map \Area {\triangle AIC} + \map \Area {\triangle AIB}       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac {a b} 2       | r = \dfrac {a r} 2 + \dfrac {b r} 2 + \dfrac {c r} 2       | c = Area of Triangle in Terms of Side and Altitude }} {{eqn | r = \dfrac {r \paren {a + b + c} } 2       | c =  }} {{eqn | ll= \leadsto       | l = r       | r = \dfrac {a b} {a + b + c}       | c =  }} {{eqn | r = \dfrac {\paren {m^2 - n^2} \paren {2 m n} } {\paren {m^2 - n^2} + \paren {2 m n} + \paren {m^2 + n^2} }       | c =  }} {{eqn | r = \dfrac {2 m n \paren {m - n} \paren {m + n} } {2 m \paren {m + n} }       | c =  }} {{eqn | r = \dfrac {2 m n \paren {m - n} \paren {m + n} } {2 m \paren {m + n} }       | c =  }} {{eqn | r = n \paren {m - n}       | c =  }} {{end-eqn}} As $m$ and $n$ are both integers, it follows that $r$ is also an integer. {{qed}}
Let the unity of $K$ be $1$, and the zero of $K$ be $0$. Consider the vectors: {{begin-eqn}} {{eqn | l = \mathbf e_1       | o = :=       | r = \underbrace {\tuple {1, 0, \ldots, 0} }_{n \text { coordinates} }       | c =  }} {{eqn | l = \mathbf e_2       | o = :=       | r = \underbrace {\tuple {0, 1, \ldots, 0} }_{n \text { coordinates} }       | c =  }} {{eqn | o = \vdots       | c =  }} {{eqn | l = \mathbf e_n       | o = :=       | r = \underbrace {\tuple {0, 0, \ldots, 1} }_{n \text { coordinates} }       | c =  }} {{end-eqn}} Thus $\tuple {\mathbf e_1, \mathbf e_2, \ldots, \mathbf e_n}$ is the standard ordered basis of $\mathbf V$. From Standard Ordered Basis is Basis, $\tuple {\mathbf e_1, \mathbf e_2, \ldots, \mathbf e_n}$ is a basis of $\mathbf V$. The result follows from Linearly Independent Set is Basis iff of Same Cardinality as Dimension. {{qed}}
Let $B$ be the set of all the identity functions $I^k$ on $F_n \sqbrk X$ where $n \in \Z_{\ge 0}$. By definition, every element of $F_n \sqbrk X$ is a linear combination of $B$. Suppose: :$\displaystyle \sum_{k \mathop = 0}^m \alpha_k I^k = 0, \alpha_m \ne 0$ Then by differentiating $m$ times, we obtain from Nth Derivative of Nth Power: :$m! \alpha_m = 0$ whence $\alpha_m = 0$ which is a contradiction. Hence $B$ is linearly independent and therefore is a basis for $F_n \sqbrk X$. The result follows from Linearly Independent Set is Basis iff of Same Cardinality as Dimension. {{qed}}
A vector space is by definition a unitary module over a division ring.  $S$ is a division ring by assumption. $\struct {R, +, \circ_S}_S$ is a unitary module by Subring Module/Special Case. {{qed}}
We have that $L$ is a  By definition, $L / K$ is a field extension over $K$. Thus, by definition, $K$ is a subfield of $L$. Thus, also by definition, $K$ is a division subring of $L$. The result follows by Vector Space over Division Subring is Vector Space. {{qed}} Category:Examples of Vector Spaces 1xsw69v077em48g99y2v1buy7i14d4k
From the Little Bézout Theorem, the remainder of $\map P x$ when divided by $x - a$ is equal to $\map P a$. === Sufficient Condition === Let $x - a$ be a divisor of $\map P x$. From the Little Bézout Theorem, the remainder of $\map P x$ when divided by $x - a$ is equal to $\map P a$. By definition of divisor, the remainder of $\map P x$ when divided by $x - a$ equals $0$. That is: :$\map P a = 0$ It follows by definition that $a$ is a root of $P$. {{qed|lemma}} === Necessary Condition === Let $a$ be a root of $P$. From the Little Bézout Theorem, and by Polynomial Long Division, we have: :$\map P x = \paren {x - a} \map Q x + \map P a$ where $\map Q x$ is a polynomial in $x$ of degree one less than $\map P x$. By definition of root of polynomial: :$\map P a = 0$ So we have: :$\map P x = \paren {x - a} \map Q x$ and the result follows by definition of divisor. {{qed}}
Let $z = a + i b$. Let $z = a - i b$ be the complex conjugate of $z$. From Product of Complex Number with Conjugate: :$z \overline z = a^2 + b^2$ From Sum of Complex Number with Conjugate: :$z + \overline z = 2 a$ Thus from Viète's Formulas, both $z$ and $\overline z$ are roots of the polynomial: :$z^2 - 2 a + \paren {a^2 + b^2}$ Hence the result by definition of algebraic over $\R$. {{qed}}
For $z$ to be a root of $F$, $z$ must be algebraic over $F$. Let us write: :$\map P x = \map m x \, \map q x + \map r x$ where $\map q x$ and $\map r x$ are polynomials in $F$. Then either $\map r x = 0$ or $\map \deg {\map r x} < \map \deg {\map m x}$. Then: :$\map P z = \map m z \, \map q z + \map r z$ But as $z$ is a root of both $\map P x$ and $\map m x$, we have that: :$\map P z = \map m z = 0$ and so: :$\map r z = 0$ So if $\map r x \ne 0$ we have that $\map r x$ is a polynomial of smaller degree than $\map m x$. This contradicts the minimality of $\map m x$. Thus $\map r x = 0$ and so $\map P x$ is a multiple of $\map m x$. {{qed}}
Let $H$ be the set of all numbers which can be written in the form $\map f \theta$. We have that: :$H$ is closed under addition and multiplication. :$H$ contains $0$ and $1$ :For every element of $H$, $H$ also contains its negative. Let $\map f \theta \ne 0$. Then $\theta$ is not a root of $\map f x$. Hence from Polynomial with Algebraic Number as Root is Multiple of Minimal Polynomial: :the minimal polynomial $\map m x$ in $\theta$ does not divide $\map f x$. From Minimal Polynomial is Irreducible, the GCD of $\map m x$ and $\map f x$ is $1$. Therefore: :$\exists \map s x, \map t x: \map s x \map m x + \map t x \map f x = 1$ Substituting for $\theta$: :$\map s \theta \, \map m \theta + \map t \theta \, \map f \theta = 1$ Because $\map m \theta = 0$ it follows that: :$\map t \theta \, \map f \theta = 1$ We have that $\map t \theta \in H$. Thus $\map t \theta$ is the product inverse of $\map f x$ in $H$. Thus $H$ is a field. A field containing $F$ and $\theta$ must contain $1$ and all the powers of $\theta$ for positive integer index. Hence such a field also contains all linear combinations of these, with coefficients in $F$. So a field containing $F$ and $\theta$ contains all the elements of $H$: :$H \subseteq \map F \theta$ But by definition, $\map F \theta$ is the smallest field containing $F$ and $\theta$. That is: :$\map F \theta \subseteq H$ Thus: :$\map F \theta = H$ and the result follows. {{Qed}}
From Simple Algebraic Field Extension consists of Polynomials in Algebraic Number, an arbitrary element of $\map F \theta$ can be written as $\map f \theta$. But: :$\map f x = \map m x \, \map q x + \map r x$ where: :$\map m x$ is minimal polynomial in $\theta$ :$\map q x$ is a polynomial in $\map F \theta$ :$\map r x$ is a polynomial in $\map F \theta$ such that either: ::$\map \deg {\map r x} < \map \deg {\map m x}$ :or: ::$\map r x = 0$ Thus: :$\map f \theta = \map m \theta \, \map q \theta + \map r \theta$ and as $\map m \theta = 0$ we have: :$\map f \theta = \map r \theta$ So $\map f \theta$ can be expressed as $\map r \theta$ instead, which is of degree strictly less than that of $\map m \theta$. Hence the result. {{Qed}}
Considered as a vector space over $F$, $\map F \theta$ is generated by the set $B$, where: :$B := \set {1, \theta, \theta^2, \ldots, \theta^{n - 1} }$ But $S$ is linearly independent over $F$, because otherwise: :$c_0 1 + c_1 \theta + c_2 \theta^2 + \dotsb + c_{n - 1} \theta^{n - 1} = 0$ with all the $c$s non-zero. That would mean $\theta$ was the root of a polynomial whose degree was less than that of the minimal polynomial $\map m x$ of $\theta$, whose degree equals $n$. Therefore $B$ is a basis of $\map F \theta$ over $F$. Thus $\map F \theta$ is of dimension $\size B = n$. Hence the $\index {\map F \theta} F = n$. {{Qed}}
Let $\alpha \in K$. The dimension of $K / F$ considered as a vector space is $n$. Let $S$ be the set defined as: :$S := \set {1, \alpha, \alpha^2, \ldots, \alpha_n}$ We have that: :$\card S = n + 1$ From Cardinality of Linearly Independent Set is No Greater than Dimension: :$S$ is linearly dependent on $F$. Hence there are scalars $c_0, c_1, \ldots, c_n \in F$, not all zero, such that: :$c_0 + c_1 \alpha + c_2 \alpha^2 + \dotsb + c_n \alpha_n = 0$ which is a polynomial of degree $n$. So $\alpha$ is algebraic over $F$ with degree no greater than $n$. Consider $\map F \alpha$, the simple algebraic extension of $F$ on $\alpha$. We have that: :$F \subseteq \map F \alpha \subseteq K$ Let $\alpha$ be algebraic over $F$ with degree $m$. Then $\map F \alpha$ is a finite extension of $F$ with degree $\index {\map F \alpha} F = m$. We also have that $K$ is a finite extension of $F$ with degree $n$. So by Degree of Field Extensions is Multiplicative: :$\index K F = \index K {\map F \alpha} \index {\map F \alpha} F$ But $\index K F = n$ and $\index {\map F \alpha} F = m$ so it follows that: :$m \divides n$ The result follows. {{Qed}}
{{AimForCont}} $\alpha \in K$. From Degree of Element of Finite Field Extension divides Degree of Extension: :$\map \deg \alpha \divides \map \deg {K / F}$ But: :$3 \nmid 2^m$ From this contradiction, it follows that $\alpha \notin K$. {{qed}}
Let $x = \dfrac m n$ where $m, n \in \Z_{\ne 0}$ are non-zero integers. Let $O$ denote the point $\tuple {0, 0}$. Let $A$ denote the point $\tuple {1, 0}$. Let $M$ denote the point $\tuple {0, m}$. Let $N$ denote the point $\tuple {0, n}$. The $x$-axis is identified with the straight line through $O$ and $A$. The $y$-axis is constructed as the line perpendicular to $OA$ through $O$. Using Construction of Lattice Point in Cartesian Plane, the points $M$ and $\N$ are constructed. The line $NA$ is drawn. Using Construction of Parallel Line, $MQ$ is drawn, where $Q$ is the point of intersection of $MQ$ with the $x$-axis. We have that $\dfrac {OM} {ON} = \dfrac m n$ As $\triangle ONA$ is similar to $\triangle OMQ$, it follows that $\dfrac {OM} {ON} = \dfrac {OQ} {OA}$ Therefore $Q$ is the point $\tuple {\dfrac m n, 0}$. :400px Let $y = \dfrac r s$ where $r, s \in \Z_{\ne 0}$ are non-zero integers. Using Construction of Lattice Point in Cartesian Plane, the points $\tuple {r, 0}$ and $\tuple {s, 0}$ are constructed. Using the same technique as above, mutatis mutandis, the point $\tuple {0, \dfrac r s}$ is constructed. Using Construction of Parallel Line, a straight line is drawn through $\tuple {\dfrac m n, 0}$ parallel to the $y$-axis. Using Construction of Parallel Line, a straight line is drawn through $\tuple {0, \dfrac r s}$ parallel to the $x$-axis. By definition of Cartesian plane, their intersection is at $\tuple {\dfrac m n, \dfrac r s}$, which is the required point $P = \tuple {x, y}$. {{qed}}
Let $AB$ be given. Let the straight line through $AC$ be constructed. Let $D_1$ be constructed on $AC$ such that $A D_1 = AB$ by constructing a circle whose center is at $A$ and whose radius is $B$. Let $D_0$ be identified as the point $A$. For each $k \in \set {1, 2, \ldots, n - 1}$, construct a circle whose center is at $D_k$ and whose radius is $D_{k - 1} D_k$. We have that $D_k D_{k + 1} = D_{k - 1} D_k = AB$. Thus $D_0 D_{k + 1}$ is ${k + 1}$ times $D_0 D_1 = AB$. When $k = n - 1$ we have that $D_{k + 1} = D_n = D$ is $n$ times $AB$. The diagram below illustrates the construction for $n = 4$. :800px {{qed}} Category:Euclidean Geometry tle54cgwus4m4tar0cwy8pqxhhhf5ah
A point $P$ is constructed in a compass and straightedge construction from one of $3$ basic operations: :$(1): \quad$ the intersection of $2$ straight lines :$(2): \quad$ the intersection of a straight line and the circumference of a circle :$(3): \quad$ the intersection of the circumferences of $2$ circle. Let $A$, $B$, $C$ and $D$ be points in $S$. In case $(1)$, the equations defining the straight lines $AB$ and $CD$ are polynomials of degree $1$. Hence the coordinates of $P$ can be found by solving the linear simultaneous equations defining $AB$ and $CD$. It follows that the solution is in $F$. {{qed|lemma}} In case $(2)$: :the equation defining the circle whose center is at $A$ and whose radius is $AB$ is of the form $x^2 + y^2 + 2 f x + 2 g y + c = 0$, where $c, f, g \in F$. while: :the equation defining the straight line $CD$ is a polynomial of degree $1$. The $x$-coordinate of $P$ is then the root of a quadratic equation with coefficients in $F$. The $y$-coordinate of $P$ is then found by solving a linear equation. Thus the coordinates of $P$ are elements of either a field extension of $F$ of degree $2$ or of $F$ itself. In case $(3)$: :the equation defining the circle whose center is at $A$ and whose radius is $AB$ is of the form $x^2 + y^2 + 2 f x + 2 g y + c = 0$, where $c, f, g \in F$ :the equation defining the circle whose center is at $C$ and whose radius is $CD$ is of the form $x^2 + y^2 + 2 f' x + 2 g' y + c' = 0$, where $c', f', g' \in F$. The solution of these equations is the same as the solution of either circle and the straight line $2 \paren {f - f'} x + 2 \paren {g - g'} y + \paren {c - c'} = 0$, which is the same as case $(2)$. We have that the coordinates of the various points belong to the field $F = F_0$, say. Suppose $P$ is constructed using $k$ of the basic constructions $(1)$, $(2)$ and $(3)$. Let $F_i$ be the smallest field containing all the points so far obtained by $i$ such constructions, for $i = 1, 2, \ldots, k$. Either $F_i = F_{i - 1}$ or $F_i$ is an finite extension of $F_i$ of degree $2$. Thus for each $i$, either: :$\index {F_i} {F_{i - 1} } = 1$ or: :$\index {F_i} {F_{i - 1} } = 2$ Therefore: :$\index {F_k} {F_0} = \index {F_k} {F_{k - 1} } \index {F_{k - 1} } {F_{k - 2} } \dotsm  {F_1} {F_0} = 2^m$ where $0 \le m \le k$. Hence the result. {{qed}}