hcju/proofwikiqa huggingface dataset:
Let $\map N z$ denote the field norm of $z \in \Z \sqbrk {i \sqrt 5}$. Let $z = x + i y$. Then: {{begin-eqn}} {{eqn | l = \map N z       | r = 2       | c =  }} {{eqn | ll= \leadsto       | l = x^2 + 5 y^2       | r = 2       | c = Field Norm on 5th Cyclotomic Ring }} {{eqn | ll= \leadsto       | l = x^2       | r = 2       | c = as $y$ is an integer and has to equal $0$ }} {{eqn | ll= \leadsto       | l = x       | r = \sqrt 2       | c =  }} {{end-eqn}} But Square Root of Prime is Irrational so $x \notin \Z$. Similarly for $\map N z = 3$. {{qed}}
{{TheoremWanted|For the concept of irreducibility to be defined, it needs to be demonstrated that $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$ is actually an integral domain.}} Let $z = x + i y$ be an element of $\Z \sqbrk {i \sqrt 5}$ in the set $S$, where: :$S := \set {2, 3, 1 + i \sqrt 5, 1 - i \sqrt 5}$ Let $z$ have a non-trivial factorization: :$z = z_1 z_2$ where neither $z_1$ nor $z_2$ are units of $\Z \sqbrk {i \sqrt 5}$. Let $\map N z$ denote the field norm of $z \in \Z \sqbrk {i \sqrt 5}$. Then: {{begin-eqn}} {{eqn | l = \map N z       | r = \map N {z_1 z_2}       | c =  }} {{eqn | r = \map N {z_1} \map N {z_2}       | c = {{Defof|Field Norm of Complex Number}} }} {{end-eqn}} Then we have: {{begin-eqn}} {{eqn | l = \map N 2       | r = 2^2 + 5 \times 0^2       | c = Field Norm on 5th Cyclotomic Ring }} {{eqn | r = 4       | c =  }} {{eqn | l = \map N 3       | r = 3^2 + 5 \times 0^2       | c = Field Norm on 5th Cyclotomic Ring }} {{eqn | r = 9       | c =  }} {{eqn | l = \map N {1 + i \sqrt 5}       | r = 1^2 + 5 \times 1^2       | c = Field Norm on 5th Cyclotomic Ring }} {{eqn | r = 6       | c =  }} {{eqn | l = \map N {1 - i \sqrt 5}       | r = 1^2 + 5 \times 1^2       | c = Field Norm on 5th Cyclotomic Ring }} {{eqn | r = 6       | c =  }} {{end-eqn}} From Elements of 5th Cyclotomic Ring with Field Norm 1, the only elements of $\Z \sqbrk {i \sqrt 5}$ whose field norm is $1$ are the units of $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$: $1$ and $-1$. From 5th Cyclotomic Ring has no Elements with Field Norm of 2 or 3, none of $4$, $6$ and $9$ have proper divisors which are field norms of elements of $\Z \sqbrk {i \sqrt 5}$. Thus either $z_1$ or $z_2$ is a unit of $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$. So none of the elements of $S$ has a non-trivial factorization in $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$. Hence the result, by definition of irreducible. {{qed}}
From Field Norm on 5th Cyclotomic Ring: :$\map N \alpha = a^2 + 5 b^2$ From the definition of the $5$th cyclotomic ring: :$\Z \sqbrk {i \sqrt 5} = \set {a + i \sqrt 5 b: a, b \in \Z}$ That is, both $a$ and $b$ are integers. Hence $a^2 + 5 b^2$ is also an integer. {{Qed}}
From Units of 5th Cyclotomic Ring, $1$ and $-1$ are the only units of $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$. Let $\map N z$ denote the field norm of $z \in \Z \sqbrk {i \sqrt 5}$. Let $z \in \Z \sqbrk {i \sqrt 5}$ such that $\map N z = 1$. Let $z = x + i y$. Then: {{begin-eqn}} {{eqn | l = \map N z       | r = 1       | c = Field Norm on 5th Cyclotomic Ring }} {{eqn | ll= \leadsto       | l = x^2 + 5 y^2       | r = 1       | c =  }} {{eqn | ll= \leadsto       | l = x^2       | r = 1       | c =  }} {{eqn | lo= \land       | l = y       | r = 0       | c =  }} {{end-eqn}} The result follows. {{qed}}
By definition, a unique factorization domain $D$ is an integral domain with the properties that: :For all $x \in D$ such that $x$ is non-zero and not a unit of $D$: ::$(1): \quad x$ possesses a complete factorization in $D$ ::$(2): \quad$ Any two complete factorizations of $x$ are equivalent. A complete factorization is a tidy factorization  :$x = u \circ y_1 \circ y_2 \circ \cdots \circ y_n$ such that: :$u$ is a unit of $D$ :all of $y_1, y_2, \ldots, y_n$ are irreducible. {{TheoremWanted|It needs to be demonstrated that $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$ is actually an integral domain.}} From Units of 5th Cyclotomic Ring, the only units of $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$ are $1$ and $-1$. From Irreducible Elements of 5th Cyclotomic Ring, all of the elements of the set $S$, where: :$S := \set {2, 3, 1 + i \sqrt 5, 1 - i \sqrt 5}$ are irreducible in $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$ Then we have: {{begin-eqn}} {{eqn | l = 6       | r = 1 \times 2 \times 3       | c =  }} {{eqn | r = 1 \times \paren {1 + i \sqrt 5} \paren {1 - i \sqrt 5}        | c =  }} {{end-eqn}} So there are two tidy factorizations of $6$ in $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$ which are not equivalent. Hence the result. {{qed}}
Let $n \in \Z_{> 0}$ be arbitrary. Let $P = a_n X^n + a_{n - 1} X^{n - 1} + \dotsb + a_1 X + a_0$ be a polynomial over $D$ in $X$. Consider the mapping $\phi: D \sqbrk X \to D$ defined as: :$\forall P \in D \sqbrk X: \map \phi P = a_0$ Let: :$P_1 = a_m X^m + a_{m - 1} X^{m - 1} + \dotsb + a_1 X + a_0$ :$P_2 = b_m X^m + b_{m - 1} X^{m - 1} + \dotsb + b_1 X + b_0$ We have that: {{begin-eqn}} {{eqn | l = \map \phi {P_1 \times P_2}       | r = a_0 \times b_0       | c =  }} {{eqn | r = \map \phi {P_1} \times \map \phi {P_2}       | c =  }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = \map \phi {P_1 + P_2}       | r = a_0 + b_0       | c =  }} {{eqn | r = \map \phi {P_1} + \map \phi {P_2}       | c =  }} {{end-eqn}} Thus it is seen that $\phi$ is a homomorphism. The First Isomorphism Theorem for Rings can then be used. {{finish}}
For example, $X + 2$ is a polynomials over $\Z$ in $X$ with an even constant term. So $S$ is not empty. Let $P_1 = \displaystyle \sum_{k \mathop = 0}^n a_k X^k$ and $P_2 = \displaystyle \sum_{k \mathop = 0}^n b_k X^k$ be elements of $S$. We have: {{begin-eqn}} {{eqn | l = P_1 - P_2       | r = \sum_{k \mathop = 0}^n a_k X^k + \sum_{k \mathop = 0}^n b_k X^k       | c =  }} {{eqn | r = \sum_{k \mathop = 0}^{\max \set {m, n} } \paren {a_k - b_k} X^k       | c =  }} {{end-eqn}} The constant term of $P_1 - P_2$ is $a_0 - b_0$ which is even. Thus $P_1 - P_2 \in S$. Let $P_3 = \sum_{k \mathop = 0}^s c_k X^k \in \Z \sqbrk X$. Then the constant term of $P_3 \times P_1$ is $c_0 \times a_0$. As $a_0$ is even, so is $c_0 \times a_0$. The result follows by Test for Ideal. {{qed}}
Let $J$ be the ideal formed from the set of polynomials over $\Z$ in $X$ which have a constant term which is even. From Polynomials in Integers with Even Constant Term forms Ideal, $J$ is indeed an ideal. {{AimForCont}} $J$ is a principal ideal of $\Z \sqbrk X$ such that $J = \ideal f$. But $2 \in J$, and so $2$ is a multiple of $f$ in $\Z \sqbrk X$. So $f = \pm 2$ or $f = \pm 1$. But this contradicts the fact that $J = \ideal f$. Hence the result by Proof by Contradiction. {{qed}}
We have that Integers form Unique Factorization Domain. The result follows from Gauss's Lemma on Unique Factorization Domains. {{qed}}
Let $f_1$ and $f_2$ be generators of $J$. Then $f_1$ and $f_2$ are unit multiples of each other. The units of $F \sqbrk X$ are the non-zero elements of $F$. {{explain|The above needs to be properly derived.}}
From Gaussian Integers form Integral Domain, we have that $\struct {\Z \sqbrk i, +, \times}$ is an integral domain. Let $a, d \in \Z \sqbrk i$ such that $d \ne 0$. Suppose $\cmod a \ge \cmod d$. Reference to an Argand diagram shows that one of: :$a + d, a - d, a + i d, a - i d$ is closer to the origin than $a$ is. So it is possible to subtract Gaussian integer multiples of $d$ from $a$ until the square of the modulus of the remainder drops below $\cmod d^2$. That remainder can only take integer values. Thus a Division Theorem result follows: :$\exists q, r \in \Z \sqbrk i: a = q d + r$ where $\cmod r < \cmod d$. Let $J$ be an arbitrary non-null ideal of $\Z \sqbrk i$. Let $d$ be an element of minimum modulus in $J$. Then the Division Theorem can be used to prove that $J = \ideal d$. {{finish|The above is the outline only.}}
{{AimForCont}} $x^2 + 1$ has a non-trivial factorization in $\R \sqbrk X$. Then: :$\exists \alpha, \beta \in \R: \paren {X - \alpha} \paren {X - \beta}$ and from the Polynomial Factor Theorem: :$\alpha^2 + 1 = 0$ But that means: :$\alpha^2 = -1$ and such an $\alpha$ does not exist in $\R$. Hence the result by Proof by Contradiction. {{qed}}
Let $\mathbf A = \sqbrk a_n$ and $\mathbf B = \sqbrk b_n$ be order $n$ square matrices over $R$. By definition of matrix multiplication, $\mathbf A \mathbf B = \mathbf C = \sqbrk c_n$ where: :$\displaystyle \forall i \in \closedint 1 n, j \in \closedint 1 n: c_{i j} = \sum_{k \mathop = 1}^n a_{i k} \circ b_{k j}$ But by definition of the trivial ring: :$\forall a, b \in R: a \circ b = 0_R$ where $0_R$ is the zero of $R$. Thus $\mathbf A \mathbf B$ is the zero $n \times n$ matrix. The same applies to $\mathbf B \mathbf A$, which is also the zero $n \times n$ matrix. That is: :$\mathbf A \mathbf B = \mathbf B \mathbf A = \bszero_n$ and the result follows by definition of commutative operation. {{qed}} Category:Conventional Matrix Multiplication Category:Trivial Rings 1t7trbnhmhdi2g676rctto7cmkgtb9k
{{begin-eqn}} {{eqn | l = a - b       | r = 0_R       | c =  }} {{eqn | ll= \leadstoandfrom       | l = a + \paren {-b}       | r = 0_R       | c = {{Defof|Ring Subtraction}} }} {{eqn | ll= \leadstoandfrom       | l = \paren {a + \paren {-b} } + b       | r = 0_R + b       | c = Cancellation Laws }} {{eqn | ll= \leadstoandfrom       | l = a + \paren {b^{-1} + b}       | r = 0_R \circ b       | c = {{GroupAxiom|1}} }} {{eqn | ll= \leadstoandfrom       | l = a       | r = b       | c = {{GroupAxiom|2}} and {{GroupAxiom|3}} }} {{end-eqn}} {{qed}}
From Ring of Integers Modulo m is Ring, $\struct {\Z_m, +, \times}$ is a ring. It remains to be shown that $\struct {\Z_m, +, \times}$ has no proper zero divisors {{iff}} $m$ is prime. === $m$ Composite === Let $m$ be composite. Then: :$m = m_1 m_2$ where $0 < m_1 < m, 0 < m_2 < m$. Then: :$\eqclass {m_1} m \eqclass {m_2} m = \eqclass 0 m$ and so both $\eqclass {m_1} m$ and $\eqclass {m_2} m$ are proper zero divisors. Hence if $m$ is not prime then $\struct {\Z_m, +, \times}$ is not an integral domain by definition. {{qed|lemma}} === $m$ Prime === Let $m$ be prime. Let $\eqclass a m \eqclass b m = \eqclass 0 m$. Then by Modulo Multiplication is Well-Defined: :$\eqclass {a b} m = \eqclass 0 m$ Thus either :$m \divides a$ or $m \divides b$ where $\divides$ denotes the divisibility relation. If $m \divides a$ then $\eqclass a m = 0$. If $m \divides b$ then $\eqclass b m = 0$. Thus neither $a$ nor $b$ is a proper zero divisor. Hence there are no proper zero divisors of $\struct {\Z_m, +, \times}$. Hence, by definition, $\struct {\Z_m, +, \times}$ is an integral domain {{qed}}
Let $\struct {D, +, \times}$ be a integral domain whose zero is $0_D$ and whose unity is $1_D$. === $(1)$ implies $(2)$ === Let $\struct {D, +, \times \le}$ be a ordered integral domain by definition 1. By Strict Positivity Property induces Total Ordering, $P$ induces this total ordering $\le$ on $D$. Thus $\struct {D, +, \times \le}$ is a ordered integral domain by definition 2. {{qed|lemma}} === $(2)$ implies $(1)$ === Let $\struct {D, +, \times \le}$ be a ordered integral domain by definition 2. That is, $\struct {D, +, \times}$ has a relation $\le$ which is compatible with the ring structure of $D$: {{begin-axiom}} {{axiom | n = \text {OR} 1         | lc= $\le$ is compatible with $+$:         | q = \forall a, b, c \in  D         | ml= a \le b         | mo= \implies         | mr= \paren {a + c} \le \paren {b + c} }} {{axiom | n = \text {OR} 2         | lc= Product of Positive Elements is Positive:         | q = \forall a, b \in D         | ml= 0_D \le x, 0_D \le y         | mo= \implies         | mr= 0_D \le x \times y }} {{end-axiom}} Let $P$ be the set of elements of $D$ which fulfil the conditions: :$P = \set {x \in D: 0_D \le x \land 0_D \ne x}$ We check the (strict) positivity property axioms as follows. Let $x, y \in P$. That is: :$0_D \le x \land 0_D \ne x$ :$0_D \le y \land 0_D \ne y$ $(\text P 1)$: Because $\le$ is an ordering, it is ''a fortiori'' a preordering. We have: {{begin-eqn}} {{eqn | l = 0_D + 0_D       | o = \le       | r = x + y       | c = $\text {OR} 1$: {{Defof|Ordering Compatible with Ring Structure}} }} {{eqn | ll= \leadsto       | l = 0_D       | o = \le       | r = x + y       | c = Preordering of Products under Operation Compatible with Preordering }} {{end-eqn}} But as $x, y \ne 0_D$ it follows that: :$0_D \ne x + y$ That is: :$x + y \in P$ So it is seen that $P$ fulfils (strict) positivity property $\text P 1$. $(\text P 2)$: From $\text {OR} 2$: Product of Positive Elements is Positive: :$0_D \le x \times y$ We have that: :$0_D \ne x$ and: :$0_D \ne y$ As $\struct {D, +, \times}$ is an integral domain, it has no proper zero divisors by definition. It follows that: :$0_D \ne x \times y$ and so: :$x \times y \in P$ So it is seen that $P$ fulfils (strict) positivity property $\text P 2$. $(\text P 3)$: {{finish|The page exists that provides this result, but it needs to be dug out and straightened up.}} Thus $\struct {D, +, \times \le}$ is a ordered integral domain by definition 1. {{qed}} Category:Ordered Integral Domains kvzh53zop4ceor5mntpykdtafbvq3hp
By definition, $\precsim$ is compatible with $\circ$ {{iff}}: {{begin-eqn}} {{eqn | ll= \forall x, y, z \in S:       | l = x \precsim y       | o = \implies       | r = \paren {x \circ z} \precsim \paren {y \circ z} }} {{eqn | l = x \precsim y       | o = \implies       | r = \paren {z \circ x} \precsim \paren {z \circ y} }} {{end-eqn}} === Sufficient Condition === Let $\precsim$ be compatible with $\circ$. Then for all $x_1, x_2, y_1, y_2 \in S$: {{begin-eqn}} {{eqn | l = x_1 \precsim x_2       | o = \implies       | r = \paren {x_1 \circ y_1} \precsim \paren {x_2 \circ y_1}       | c = {{Defof|Relation Compatible with Operation}} }} {{eqn | l = y_1 \precsim y_2       | o = \implies       | r = \paren {x_2 \circ y_1} \precsim \paren {x_2 \circ y_2}       | c = {{Defof|Relation Compatible with Operation}} }} {{end-eqn}} As $\precsim$ is a preordering it is by definition transitive. Thus it follows that: :$x_1 \precsim x_2 \land y_1 \precsim y_2 \implies \paren {x_1 \circ y_1} \precsim \paren {x_2 \circ y_2}$ {{qed|lemma}} === Necessary Condition === Let $\precsim$ fulfil the property that: :$\forall x_1, x_2, y_1, y_2 \in S: x_1 \precsim x_2 \land y_1 \precsim y_2 \implies \paren {x_1 \circ x_2} \precsim \paren {y_1 \circ y_2}$ As $\precsim$ is a preordering it is by definition reflexive. That is: :$\forall z \in S: z \precsim z$ Make the substitutions: {{begin-eqn}} {{eqn | l = x_1       | o = \to       | r = x }} {{eqn | l = x_2       | o = \to       | r = y }} {{eqn | l = y_1       | o = \to       | r = z }} {{eqn | l = y_2       | o = \to       | r = z }} {{end-eqn}} It follows that: :$\forall x, y, z \in S: x \precsim y \implies \paren {x \circ z} \precsim \paren {y \circ z}$ Similarly, make the substitutions: {{begin-eqn}} {{eqn | l = x_1       | o = \to       | r = z }} {{eqn | l = x_2       | o = \to       | r = z }} {{eqn | l = y_1       | o = \to       | r = x }} {{eqn | l = y_2       | o = \to       | r = y }} {{end-eqn}} It follows that: :$\forall x, y, z \in S: x \precsim y \implies \paren {z \circ x} \precsim \paren {z \circ y}$ So for all $x, y, z \in S$, both conditions are fulfilled: {{begin-eqn}} {{eqn | l = x \precsim y       | o = \implies       | r = \paren {x \circ z} \precsim \paren {y \circ z} }} {{eqn | l = x \precsim y       | o = \implies       | r = \paren {z \circ x} \precsim \paren {z \circ y} }} {{end-eqn}} for $\precsim$ to be compatible with $\circ$. {{qed|lemma}} The result follows. {{qed}} Category:Preorder Theory Category:Compatible Relations 9pogqcvs165m94664h0ve6uy63qtwsu
By assumption: :$\forall n \in \N: p^n \divides \paren {x_n^k - a}$ By the definition of the $p$-adic norm: :$\forall n \in \N: \norm {x_n^k - a}_p \le \dfrac 1 {p^n}$ By Sequence of Powers of Number less than One: :$\displaystyle \lim_{n \mathop \to \infty} \dfrac 1 {p^n} = 0$ By Squeeze Theorem for Real Sequences: :$\displaystyle \lim_{n \mathop \to \infty} \norm {x_n^k - a}_p = 0$. By the definition of convergence in $\struct {\Q, \norm {\,\cdot\,}_p}$ then: :$\displaystyle \lim_{n \mathop \to \infty} x_n^k = a$ {{qed}} Category:P-adic Norm not Complete on Rational Numbers orpqibd6mumv5ml5n87nontx9ivpf76
By assumption: :$\forall n \in \N: p^n \divides \paren {x_{n + 1} - x_n}$ By the definition of the $p$-adic norm: :$\forall n \in \N: \norm {x_{n + 1} - x_n}_p \le \dfrac 1 {p^n}$ By Sequence of Powers of Number less than One: :$\displaystyle \lim_{n \mathop \to \infty} \dfrac 1 {p^n} = 0$ By the Squeeze Theorem for Real Sequences: :$\displaystyle \lim_{n \mathop \to \infty} \norm{x_{n+1} - x_n}_p = 0$. From P-adic Norm is Non-Archimedean Norm, the $p$-adic norm is  non-Archimedean. By Characterisation of Cauchy Sequence in Non-Archimedean Norm: :$\sequence {x_n}$ is a Cauchy sequence in $\struct {\Q, \norm {\,\cdot\,}_p}$. {{qed}} Category:Metric Spaces Category:Normed Spaces Category:P-adic Number Theory hec0ra2qr4h55bn1py1cwb2l55mknq0
Since $x, p > 0$ then $a > 0$. {{AimForCont}} for some $c \in \Z:c^k = a$. Since $c^k \in \Z$, by Nth Root of Integer is Integer or Irrational then: :$c \in \Z$ Suppose $k$ is odd. Since $a > 0$, by Odd Power Function is Strictly Increasing then $c > 0$ Hence $a = \size c^k$ On the other hand, suppose $k$ is even, that is $k = 2l$ for some $l \in Z_{> 0}$. Then: {{begin-eqn}} {{eqn | l = a       | r = c^{2l} }} {{eqn | r = \paren {c^2}^l }} {{eqn | r = \paren {\size c^2}^l       | c = Equivalence of Definitions of Absolute Value Function }} {{eqn | r = \size c^{2l} }} {{eqn | r = \size c^k }} {{end-eqn}} In either case $\size c \in \Z_{> 0}$ and $\size c^k = a$ Let $d = \size c$   By the definition of $a$ it follows that $d^k = x^k + p$ Hence: {{begin-eqn}} {{eqn | l = p        | r = d^k - x^k }} {{eqn | r = \paren {d - x} \paren {d^{k - 1} + d^{k - 2} x + d^{k - 3} x^2 + \dotsb + c x^{k - 2} + x^{k - 1} }       | c = Difference of Two Powers }} {{end-eqn}} Let $y = d^{k - 1} + d^{k - 2} x + d^{k - 3} x^2 + \dotsb + d x^{k - 2} + x^{k - 1}$ Since $d, x \in \Z_{> 0}$ then $d - x \in \Z$ and $y \in \Z$ So $d - x$ and $y$ are factors of $p$ The factors of $p$ by definition are: :$\pm 1$ and $\pm p$ Since $d, x \in \Z_{> 0}$ then: {{begin-eqn}} {{eqn | l = d^{k - 1} + d^{k - 2} x + d^{k - 3} x^2 + \dotsb + d x^{k - 2} + x^{k - 1}       | o = \ge        | r = d^{k - 1} + x^{k - 1} }} {{eqn | o = \ge        | r = d + x }} {{eqn | o = \ge        | r = 1 + 1 }} {{eqn | o = \ge        | r = 2 }} {{end-eqn}} Hence $y = p$ Then: {{begin-eqn}} {{eqn | l = p        | r = d^{k - 1} + d^{k - 2} x + d^{k - 3} x^2 + \dotsb + d x^{k - 2} + x^{k - 1} }} {{eqn | o = \ge        | r = d^{k - 1} + x^{k - 1} }} {{eqn | o = \ge        | r = d + x }} {{end-eqn}} It also follows that $d - x = 1$, that is, $d = x + 1$ Then {{begin-eqn}} {{eqn | l = d + x       | r = \paren {x + 1} + x }} {{eqn | r = 2x + 1 }} {{eqn | o = >       | r = 2x }} {{eqn | o = >       | r = p + 1 }} {{eqn | o = >       | r = p }} {{end-eqn}} This contradicts the previous conclusion that $p \ge d + x$ So: :$\nexists \,c \in \Z : c^k = a$ {{qed}} Category:P-adic Norm not Complete on Rational Numbers phaadqtr1ve7gbf201hybvqikj8ieqs
By Euclid's Lemma for Prime Divisors: :$p \nmid k x_1^{k - 1}$ Hence: :$k x_1^{k - 1} \not \equiv 0 \mod p$ The formal derivative $\map {f'} X \in \Z \sqbrk X$ of $\map f X$ is by definition: :$k X^{k - 1}$ Then: :$\map {f'} {x_1} = k x_1^{k - 1} \not \equiv 0 \pmod p$ {{qed}} Category:P-adic Norm not Complete on Rational Numbers nvzcpl2u0gv7vtca4wdgk9kegsflq64
Let $g: T \to \N$ be a bijection from $T$ to $\N$. By Inverse of Bijection is Bijection, $g^{-1}: \N \to T$ is a bijection from $\N$ to $T$. === Necessary Condition === Let $f: T \to S$ be a surjection. By Composite of Surjections is Surjection, $f \circ g^{-1}:\N \to S$ is a surjection. By Surjection from Natural Numbers iff Countable, $S$ is a countable set. {{qed|lemma}} === Sufficient Condition === Suppose that $S$ is countable and non-empty. By Surjection from Natural Numbers iff Countable, there exists a surjection $f: \N \to S$. By Composite of Surjections is Surjection, $f \circ g:T \to S$ is a surjection. {{qed}} Category:Countable Sets Category:Surjections c07x8v8ecbjipe10fv8564tehnr3ojm
By Corollary 1 no mapping from $T$ to $S$ is a surjection. {{qed}} Category:Countable Sets Category:Surjections 220auysxelec96dq78xe8ymn2ex26jk
By definition, $\struct {D, +, \times, \le}$ is an integral domain endowed with a strict positivity property. From Strict Positivity Property induces Total Ordering, the ordering $\le$ on $\struct {D, +, \times, \le}$ is a total ordering. Hence the result by definition of totally ordered ring. {{qed}} Category:Ordered Integral Domains Category:Totally Ordered Rings aunzqytajll4lim3grktfz3nxvzq9dr
{{begin-eqn}} {{eqn | l = \map N a       | o = \leadstoandfrom       | r = \map P {-a}       | c = {{Defof|Strict Negativity Property}} }} {{eqn | o = \leadstoandfrom       | r = \map P {-a + 0}       | c =  }} {{eqn | o = \leadstoandfrom       | r = a < 0       | c = Strict Positivity Property induces Total Ordering }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \map N a, \map N b       | o = \leadsto       | r = \map P {-a}, \map P {-b}       | c = {{Defof|Strict Negativity Property}} }} {{eqn | o = \leadsto       | r = \map P {\paren {-a} + \paren {-b} }       | c = Strict Positivity Property: $(P \, 1)$ }} {{eqn | o = \leadsto       | r = \map P {-\paren {a + b} }       | c =  }} {{eqn | o = \leadsto       | r = \map N {a + b}       | c = {{Defof|Strict Negativity Property}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \map N a, \map N b       | o = \leadsto       | r = \map P {-a}, \map P {-b}       | c = {{Defof|Strict Negativity Property}} }} {{eqn | o = \leadsto       | r = \map P {\paren {-a} \times \paren {-b} }       | c = Strict Positivity Property: $(P \, 2)$ }} {{eqn | o = \leadsto       | r = \map P {a \times b}       | c = Product of Ring Negatives }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \map N a, \map P b       | o = \leadsto       | r = \map P {-a}, \map P b       | c = {{Defof|Strict Negativity Property}} }} {{eqn | o = \leadsto       | r = \map P {\paren {-a} \times b}       | c = Strict Positivity Property: $(P \, 2)$ }} {{eqn | o = \leadsto       | r = \map P {-\paren {a \times b} }       | c = Product with Ring Negative }} {{eqn | o = \leadsto       | r = \map N {a \times b}       | c = {{Defof|Strict Negativity Property}} }} {{end-eqn}} {{qed}}
=== $(1)$ implies $(2)$ === Let $\struct {D, +, \times \le}$ be a well-ordered integral domain by definition 1. Then by definition the ordering $\le$ is a well-ordering on the set $P$ of (strictly) positive elements of $D$. Let $S \subseteq P$. Thus by definition of well-ordering, $S$ has a minimal element. Thus $\struct {D, +, \times \le}$ is a well-ordered integral domain by definition 2. {{qed|lemma}} === $(2)$ implies $(1)$ === Let $\struct {D, +, \times \le}$ be a well-ordered integral domain by definition 2. Then by definition every subset $S$ of the set $P$ of (strictly) positive elements of $D$ has a minimal element: :$\forall S \subseteq D_{\ge 0_D}: \exists x \in S: \forall a \in S: x \le a$ where $D_{\ge 0_D}$ denotes all the elements $d \in D$ such that $\map P d$. Thus by definition $\le$ is a well-ordering of $P$. Thus $\struct {D, +, \times \le}$ is a well-ordered integral domain by definition 1. {{qed}} Category:Well-Ordered Integral Domains 3li6rwnx3jgfsghvnaajbh37busnvzg
Consider $\N$ defined as a Peano structure. The result follows from Principle of Mathematical Induction for Peano Structure. {{qed}}
Consider $\N$ defined as a naturally ordered semigroup. The result follows directly from Principle of Mathematical Induction for Naturally Ordered Semigroup. {{qed}}