hcju/proofwikiqa huggingface dataset:
It is given by Division Theorem: Positive Divisor: Positive Dividend: Existence that such $q$ and $r$ exist. Let $a = 0$. It is immediately apparent that $q = r = 0$ is the only possible solution with $0 \le r < b$. {{qed|lemma}} Let $a > 0$ and $b = 1$. Then from the condition $0 \le r < b$ it follows that $r = 0$. Hence $r = 0, q = a$ is the only possible solution. {{qed|lemma}} Let $a > 0$ and $b > 1$. By the Basis Representation Theorem, $a$ has a unique representation to the base $b$: {{begin-eqn}} {{eqn | l = a       | r = \sum_{k \mathop = 0}^s r_k b^k       | c =  }} {{eqn | r = b \sum_{k \mathop = 0}^{s - 1} r_k b^{k - 1} + r_0       | c =  }} {{eqn | r = b q + r       | c = where $0 \le r = r_0 < b$ }} {{end-eqn}} Suppose a second pair $q', r'$ were to exist. Then there would be a representation for $q'$ to the base $b$: :$\displaystyle q' = \sum_{k \mathop = 0}^t u_k b^k$ so that: {{begin-eqn}} {{eqn | l = a       | r = b q' + r'       | c = $0 \le r < b$ }} {{eqn | r = \sum_{k \mathop = 0}^t u_k b^{k + 1} + r'       | c =  }} {{end-eqn}} But there already exists a representation of $a$ to the base $b$: :$\displaystyle a = \sum_{k \mathop = 0}^s r_k b^k$ By the Basis Representation Theorem, such a representation is unique. So: {{begin-eqn}} {{eqn | l = t       | r = s - 1 }} {{eqn | l = u_k       | r = r_{k + 1} }} {{eqn | l = r'       | r = a_0 }} {{eqn | r = r }} {{end-eqn}} and hence $q' = q$. {{qed}}
From Division Theorem: Positive Divisor: Positive Dividend: Existence: :$\forall a, b \in \Z, a \ge 0, b > 0: \exists q, r \in \Z: a = q b + r, 0 \le r < b$ That is, the result holds for positive $a$. {{qed|lemma}} It remains to be shown that the statement holds for $a < 0$. From Division Theorem: Positive Divisor: Positive Dividend: :$\exists \tilde q, \tilde r \in \Z: \size a = \tilde q b + \tilde r, 0 \le \tilde r < b$ where $\size a$ denotes the absolute value of $a$: by definition $\size a \ge 0$. As $a < 0$ it follows by definition of absolute value that $\size a = -a$. This gives: {{begin-eqn}} {{eqn | l = a       | r = -\size a       | c =  }} {{eqn | r = -\paren {\tilde q b + \tilde r}       | c =  }} {{eqn | r = b \paren {-\tilde q} + \paren {-\tilde r}       | c =  }} {{end-eqn}} Let $\tilde r = 0$. Then: : $q = -\tilde q, r = \tilde r = 0$ which gives: : $a = q b + r, 0 \le r < b$ as required. Otherwise: :$0 < \tilde r < b \implies 0 < b - \tilde r < b$ which suggests a rearrangement of the expression for $a$ above: {{begin-eqn}} {{eqn | l = a       | r = b \paren {-\tilde q} + \paren {-\tilde r}       | c =  }} {{eqn | r = b \paren {-\tilde q} - b + b - \tilde r       | c =  }} {{eqn | r = b \paren {-1 - \tilde q} + \paren {b - \tilde r}       | c =  }} {{end-eqn}} Taking: : $q = \paren {-1 - \tilde q}$ and: : $r = \paren {b - \tilde r}$ the required result follows. Uniqueness of $q$ and $r$ follow from the Uniqueness of $\tilde q$ and $\tilde r$. {{qed}}
Let: : $q = \floor {\dfrac a b}, t = \dfrac a b - \floor {\dfrac a b}$ where $\floor {\, \cdot \,}$ denotes the floor function. Thus $q \in \Z$ and $t \in \hointr 0 1$. So: :$\dfrac a b = q + t$ and so: :$(1): \quad a = q b + r$ where $r = t d$. Since $a, q, b \in \Z$, it follows from $(1)$ that: :$r = a - q b$ and so $r \in \Z$ also. Since $0 \le t < 1$ and $b > 0$, it follows that: :$0 \le t b < b$ that is: :$0 \le r < b$ {{qed}}
It is given by Division Theorem: Positive Divisor: Existence that such $q$ and $r$ exist. {{:Division Theorem/Positive Divisor/Positive Dividend/Uniqueness/Proof 1}}
It is given by Division Theorem: Positive Divisor: Existence that such $q$ and $r$ exist. Let $a = q b + r$ where $q, r \in \Z$ and $0 \le r < b$. Thus: :$\dfrac a b = q + \dfrac r b$ and: :$0 \le \dfrac r b \le \dfrac {b - 1} b < 1$ So: :$q = \floor {\dfrac a b}$ and so: :$r = a - b \floor {\dfrac a b}$ Thus, given $a$ and $b$, the numbers $q$ and $r$ are unique determined. {{qed}}
Let there exist $q \in Z$ such that $a - b q = 0$. Then $a = b q$ as required, with $r = 0$. Otherwise, let $S$ be defined as the set of all positive integers of the form $a - z b$ where $z$ is an integer: :$S = \set {x \in \Z_{\ge 0}: \exists z \in \Z: x = a - z b}$ Setting $z = 0$ it is seen that $a \in S$, so $S \ne \O$. From Set of Integers Bounded Below by Integer has Smallest Element, $S$ has a smallest element. Let $r$ be the smallest element of $S$. Let $r = a - b q$. As there does not exist $q \in Z$ such that $a - b q = 0$: :$r > 0$ Suppose $r = b$. Then $a = b \paren {q + 1}$ and it has already been declared that there does not exist such a $q + 1 \in Z$. Suppose $r > b$. Then $x = a - b \paren {q + 1} \in S$ such that $x < r$, which contradicts the assumption that $r$ is the smallest element of $S$. Hence the result. {{qed}}
It is given by Division Theorem: Positive Divisor: Existence that such $q$ and $r$ exist. Suppose that: :$a = b q_1 + r_1 = b q_2 + r_2$ where both $0 \le r_1 < b$ and $0 \le r_2 < b$. {{WLOG}}, suppose $r_1 \ge r_2$. Then: :$r_1 - r_2 = b \paren {q_2 - q_1}$ That is: :$b \divides \paren {r_2 - r_1}$ where $\divides$ denotes divisibility. But: :$r_1 - r_2 < b$ while from Absolute Value of Integer is not less than Divisors: Corollary: :$r_1 - r_2 \ge b$ unless from Integer Divides Zero $r_1 - r_2 = 0$. So $r_1 = r_2$ and it follows directly that $q_1 = q_2$. {{qed}}
Let $a, b \in \Z$ such that $a \ge 0$ and $b > 0$ be given. When $a = 0$, the integers $q = r = 0$ satisfy the conditions of the theorem. Let $a > 0$. For each $k \in \left[{0 \,.\,.\, a + 1}\right]$, let $r_k = b k$. (Note that here, as elsewhere in this proof, $\left[{0 \,.\,.\, a + 1}\right]$ denotes an integer interval.) Then $\left \langle{r_k}\right \rangle_{0 \mathop \le k \mathop \le a + 1}$ is a strictly increasing sequence of positive integers. It is also the case that: :$a + 1 \in \left[{1 \,.\,.\, b \left({a + 1}\right)}\right]$ So from Strictly Increasing Sequence induces Partition there exists $q \in \left[{0 \,.\,.\, a}\right]$ such that: :$a + 1 \in \left[{r_q + 1 \,.\,.\, r_{q + 1}}\right]$ Let $r = a - b q$. Then $a = b q + r$. We have: {{begin-eqn}} {{eqn | l = b q + 1       | r = r_q + 1       | c =  }} {{eqn | o = \le       | r = a + 1       | c =  }} {{eqn | o = \le       | r = r_{q + 1}       | c =  }} {{eqn | r = b q + b       | c =  }} {{end-eqn}} and so: {{begin-eqn}} {{eqn | l = 0       | o = \le       | r = r       | c =  }} {{eqn | r = \left({a + 1}\right) - \left({b q + 1}\right)       | c =  }} {{eqn | o = \le       | r = \left({b q + b}\right) - \left({b q + 1}\right)       | c =  }} {{eqn | r = b - 1       | c =  }} {{eqn | o = <       | r = b       | c =  }} {{end-eqn}} Hence the result. {{qed}}
From Integer Combination of Coprime Integers: :$\forall j \in \set {1, 2, \ldots, r}: a x_j + b_j y_j = 1$ for some $x_j, y_j \in \Z$. Thus: :$\displaystyle \prod_{j \mathop = 1}^r b_j y_j = \prod_{j \mathop = 1}^r \paren {1 - a x_j}$ But $\displaystyle \prod_{j \mathop = 1}^r \paren {1 - a x_j}$ is of the form $1 - a z$. Thus: {{begin-eqn}} {{eqn | l = \prod_{j \mathop = 1}^r b_j \prod_{j \mathop = 1}^r y_j       | r = 1 - a z       | c =  }} {{eqn | ll= \leadsto       | l = a z + b y       | r = 1       | c = where $\displaystyle y = \prod_{j \mathop = 1}^r y_j$ }} {{end-eqn}} and so, by Integer Combination of Coprime Integers, $a$ and $b$ are coprime. {{qed}}
''This proof follows the structure of {{AuthorRef|Euclid}}'s proof, if not its detail.'' Let $a, b, c \in \Z$ such that $c$ is coprime to each of $a$ and $b$. Let $d = a b$. {{AimForCont}} $c$ and $d$ are not coprime. Then: :$\exists e \in \Z_{>1}: e \divides c, e \divides d$ We have that $c \perp a$ and $e \divides c$ From {{EuclidPropLink|book=VII|prop=23|title=Divisor of One of Coprime Numbers is Coprime to Other}}: :$e \perp a$ As $e \divides d$ it follows by definition of divisibility that: :$\exists f \in \Z: e f = d$ We also have that: :$a b = d$ So: :$a b = e f$ But from {{EuclidPropLink|book=VII|prop=19|title=Relation of Ratios to Products}}: :$e : a = b : f$ or in more contemporary language: :$\dfrac a e = \dfrac b f$ But $a \perp e$. From {{EuclidPropLink|book = VII|prop = 21|title = Coprime Numbers form Fraction in Lowest Terms}}, $\dfrac a e$ is in canonical form. From {{EuclidPropLink|book = VII|prop = 20|title = Ratios of Fractions in Lowest Terms}}: :$e \divides b$ But we already have that: :$e \divides c$ So $e$ is a divisor of both $b$ and $c$. But this is a contradiction of the assertion that $c$ and $b$ are coprime. Hence the result. {{qed}}
Proof by induction: In the following, it is assumed at all times that $S = \set {a_1, a_2, \ldots, a_r} \subseteq \Z$ is pairwise coprime. For all $r \in \N_{> 1}$, let $\map P r$ be the proposition: :$\displaystyle \prod_{j \mathop = 1}^r a_j \divides b$ === Basis for the Induction === $\map P 2$ is the case: :$a_1 a_2 \divides b$ By definition of divisibility: :$b = a_1 q_1 = a_2 q_2$ where $q_1, q_2 \in \Z$. By Integer Combination of Coprime Integers: :$\exists x, y \in \Z: a_1 x + a_2 y = 1$ So: {{begin-eqn}} {{eqn | l = b       | r = a_2 q_2 \cdot a_1 x + a_1 q_1 \cdot a_2 y       | c =  }} {{eqn | r = a_1 a_2 \paren {q_2 x + q_1 y}       | c =  }} {{end-eqn}} and so $a_1 a_2 \divides b$. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true. So this is our induction hypothesis: :$\displaystyle \prod_{j \mathop = 1}^k a_j \divides b$ Then we need to show: :$\displaystyle \prod_{j \mathop = 1}^{k + 1} a_j \divides b$ === Induction Step === This is our induction step: By hypothesis: :$a_{k + 1} \divides b$ By Integer Coprime to all Factors is Coprime to Whole: :$\displaystyle a_{k + 1} \perp \prod_{j \mathop = 1}^k a_j$ By the induction hypothesis: :$\displaystyle \prod_{j \mathop = 1}^k a_j \divides b$ So by the basis for the induction: :$\displaystyle \prod_{j \mathop = 1}^{k + 1} a_j \divides b$ So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: $\displaystyle r \in \N_{> 1}: \prod_{j \mathop = 1}^r a_j \divides b$ {{qed}}
Let $f$ be a piecewise continuous function with one-sided limits. By definition, there exists a finite subdivision $\set {x_0, x_1, \ldots, x_n}$ of $\closedint a b$, where $x_0 = a$ and $x_n = b$, such that $f$ is continuous on $\openint {x_{i − 1} } {x_i}$ for every $i \in \set {1, 2, \ldots, n}$. For every $i \in \set {1, 2, \ldots, n}$, we define a function $f_i$ with domain $\closedint {x_{i − 1} } {x_i}$, as follows: :$\map {f_i} x := \begin{cases} \displaystyle \lim_{x \mathop \to x_{i − 1}^+} \map f x & : x = x_{i - 1} \\ \map f x & : x \in \openint {x_{i - 1} } {x_i} \\ \displaystyle \lim_{x \mathop \to x_i^-} \map f x & : x = x_i \end{cases}$ The one-sided limits in this definition exist because $f$ is piecewise continuous with one-sided limits. We have that $f$ is continuous on $\openint {x_{i - 1} } {x_i}$. We also have that $f_i = f$ on $\openint {x_{i - 1} } {x_i}$. Therefore $f_i$ is also continuous on $\openint {x_{i - 1} } {x_i}$. By definition, $f_i$ is right-continuous at $x_{i − 1}$ and left-continuous at $x_i$. Therefore, $f_i$ is continuous throughout its domain $\closedint {x_{i − 1} } {x_i}$. By Continuous Function on Compact Subspace of Euclidean Space is Bounded and Closed Real Interval is Compact in Metric Space, $f_i$ is bounded. Therefore, $f_i$ is bounded on $\closedint {x_{i − 1} } {x_i}$. We have that $\openint {x_{i - 1} } {x_i}$ constitutes a subset of $\closedint {x_{i − 1} } {x_i}$. Thus $f_i$ is also bounded on $\openint {x_{i - 1} } {x_i}$. As $f_i = f$ on $\openint {x_{i - 1} } {x_i}$, $f$ is also bounded on $\openint {x_{i - 1} } {x_i}$. We have that: :$f$ is bounded on the intervals $\openint {x_{i - 1} } {x_i}$ and :the set of these intervals is finite. Hence the set of bounds of $f$ on these intervals is itself bounded. The bound of this set of bounds serves as a bound for $f$ on each of the intervals $\openint {x_{i - 1} } {x_i}$. Therefore, this bound is a bound for $f$ on the union of these intervals. That is, $f$ is bounded on the union of $\openint {x_{i - 1} } {x_i}$ for all $i \in \set {1, 2, \ldots, n}$. The only points left to consider are the points in the set $\set {x_0, x_1, \ldots, x_n}$.  Since this set is finite, the maximum $\map \max {\size {\map f {x_0} }, \size {\map f {x_1} }, \ldots, \size {\map f {x_n} } }$ is finite and serves as a bound for $f$ on $\set {x_0, x_1, \ldots, x_n}$. Thus $f$ is bounded on the whole of its domain $\closedint a b$. Hence $f$ is a bounded piecewise continuous function. {{qed}}
Proof by strong induction: Let us write $p_n = \map p n$. For all $n \in \N_{>0}$, let $\map P n$ be the proposition: :$\map p n \le 2^{2^{n - 1} }$ === Basis for the Induction === $\map P 1$ is true, as this just says $2 \le 2^{2^0} = 2$. This is our basis for the induction. === Induction Hypothesis === Suppose that, for some $k \in \N$, each of $\map P 1, \map P 2, \ldots, \map P k$ is true. It remains to show that it logically follows that $\map P {k + 1}$ is true. That is, that: :$\map p {k + 1} \le 2^{2^k}$ This is our induction hypothesis. === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = \map p {k + 1}       | o = \le       | r = p_1 p_2 \cdots p_k + 1       | c = Euclid's Theorem }} {{eqn | o = \le       | r = 2 \times 2^2 \times 2^{2^2} \times \cdots \times 2^{2^{k - 1} } + 1       | c = Induction Hypothesis }} {{eqn | r = 2^{1 + 2 + 2^2 + 2^3 + \cdots + 2^{k - 1} } + 1       | c = Exponent Combination Laws }} {{eqn | r = 2^{2^k - 1} + 1       | c = Sum of Geometric Sequence }} {{eqn | o = \le       | r = 2^{2^k - 1} + 2^{2^k - 1}       | c = since $1 \le 2^{2^k - 1}$ for all $k$ }} {{eqn | r = 2^{2^k}       | c =  }} {{end-eqn}} The result follows by the Second Principle of Mathematical Induction. {{qed}}
Let us write $p_n = \map p n$. Let us take $N = p_1 p_2 \cdots p_n + 1$. By the same argument as in Euclid's Theorem, we have that either $N$ is prime, or it is not. If $N$ is prime, then either $N = p_{n + 1}$ or not, in which case $N > p_{n + 1}$. In the second case, $N$ has a prime factor not in $\set {p_1, p_2, \ldots, p_n}$ Therefore it must have a prime factor greater than any of $\set {p_1, p_2, \ldots, p_n}$. In any case, the next prime after $p_n$ can be no greater than $p_1 p_2 \cdots p_n + 1$. Hence the result. {{qed}}
Let us write $p_n = p \left({n}\right)$. From Bertrand's Conjecture, for each $n \ge 2$ there exists a prime $p$ such that $n < p < 2 n$. For all $n \in \N_{>0}$, let $P \left({n}\right)$ be the proposition: : $p_n < 2^n$ $P(1)$ is the statement: :$p_1 = 2 = 2^1$ As this does not fulfil the criterion: :$p \left({n}\right) < 2^n$ it is not included in the result. === Basis for the Induction === $P(2)$ is true, as this just says: :$p_2 = 3 < 2^2 = 4$ This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k+1}\right)$ is true. So this is our induction hypothesis: :$p_k < 2^k$ Then we need to show: :$p_{k+1} < 2^{k+1}$ === Induction Step === This is our induction step: Suppose that $p_k < 2^k$ for some $k \ge 2$. By Bertrand's Conjecture, there exists a prime $q$ such that $2^k < q < 2^{k+1}$. So: : $p < 2^k < q < 2^{k+1}$ But as $q$ is a prime exceeding $p_k$, it means: : $p_{k+1} \le q < 2^{k+1}$ So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: : $\forall n \in \N_{>1}: p \left({n}\right) < 2^n$ {{qed}} Category:Upper Bounds for Prime Numbers gshqsisc4nz55ru10a6lfuqs81lt1vg
{{AimForCont}} the contrary. That is, suppose there is a finite number of prime numbers of the form $4 n - 1$. Let there be $k$ of them: $p_1, p_2, \ldots, p_k$. Let $S = \set {p_1, p_2, \ldots, p_k}$. Let $N$ be constructed as: :$\displaystyle N = 4 \prod_{i \mathop = 1}^k p_i - 1$ If $N$ is a prime number, then it is of the form $4 n - 1$. But then we have that $N \notin S$, which means that $S$ is not the complete set of prime numbers of the form $4 n - 1$. Therefore $N$ must be composite. Suppose all the prime factors of $N$ are of the form $4 n + 1$. Then from Product of Integers of form $4 n + 1$ it follows that $N$ itself is of the form $4 n + 1$. Therefore $N$ must have at least one prime factor $p$ of the form $4 n - 1$. Suppose $p \in S$. We have that: : $\displaystyle p \divides 4 \prod_{i \mathop = 1}^k p_i$ and so: : $\displaystyle p \divides 4 \prod_{i \mathop = 1}^k p_i - p$ So: :$N = q p + \left({p - 1}\right)$ where: :$q = \dfrac {\displaystyle 4 \prod_{i \mathop = 1}^k p_i - p} p$ So applying the Euclidean Algorithm: {{begin-eqn}} {{eqn | l = \gcd \set {N, p}       | r = \gcd \set {p - 1, p}       | c =  }} {{eqn | r = \gcd \set {p - 1, 1}       | c =  }} {{eqn | r = 1       | c =  }} {{end-eqn}} So if $p \in S$, then it cannot be a divisor of $N$. Therefore there must be a prime numbers of the form $4 n - 1$ which is not in $S$. Therefore $S$ is not the complete set of prime numbers of the form $4 n - 1$. Therefore the assumption that $S$ is finite was false. Hence the result. {{qed}}
{{begin-eqn}} {{eqn | l = a x       | o = \equiv       | r = 1       | rr= \pmod m       | c =  }} {{eqn | lll= \iff       | ll= \exists y \in \Z:       | l = a x + m y       | r = 1       | c = Definition of congruence }} {{end-eqn}} From Integer Combination of Coprime Integers: : $a \perp m \iff \exists x, y \in \Z: a x + m y = 1$ That is, such an $x$ exists {{iff}} $a$ and $m$ are coprime. {{qed}}
By definition of congruence: :$a \not \equiv 0 \pmod p \iff p \nmid a$ where $p \nmid a$ denotes that $p$ is not a divisor of $a$. From Prime not Divisor implies Coprime: :$p \nmid a \iff p \perp a$ where $p \perp a$ denotes that $p$ and $a$ are coprime. The result follows from Integer Coprime to Modulus iff Linear Congruence to 1 exists. {{qed}}
Consider the linear congruence $a x \equiv b \pmod n$. Suppose $\exists x_0 \in \Z: a x_0 \equiv b \pmod n$. Then $\exists y_0 \in Z: a x_0 - b = n y_0$ by definition of congruence. Thus $x = x_0, y = y_0$ is a solution to the linear Diophantine equation $a x - n y = b$. On the other hand, if $x = x_0, y = y_0$ is a solution to the linear Diophantine equation $a x - n y = b$, then it follows that $a x \equiv b \pmod n$. Hence: : the problem of finding all integers satisfying the linear congruence $a x \equiv b \pmod n$  is the same problem as: : the problem of finding all the $x$ values in the linear Diophantine equation $a x - n y = b$. From Solution of Linear Diophantine Equation: The linear Diophantine equation $a x - n y = b$ has at least one solution {{iff}}: :$\gcd \set {a, n} \divides b$ Hence the result. {{qed}} Category:Solution of Linear Congruence i2q79zyugpt7691hpi24jfc0wf67dz4
From Solution of Linear Congruence: Existence: :the problem of finding all integers satisfying the linear congruence $a x \equiv b \pmod n$  is the same problem as: :the problem of finding all the $x$ values in the linear Diophantine equation $a x - n y = b$. From Integers Divided by GCD are Coprime: :$\gcd \set {\dfrac a d, \dfrac n d} = 1$ So the {{RHS}} has a unique solution modulo $\dfrac n d$, say: :$x \equiv x_1 \paren {\bmod \dfrac n d}$ So the integers $x$ which satisfy $a x \equiv b \pmod n$ are exactly those of the form $x = x_1 + k \dfrac n d$ for some $k \in \Z$. Consider the set of integers: : $\set {x_1, x_1 + \dfrac n d, x_1 + 2 \dfrac n d, \ldots, x_1 + \paren {d - 1} \dfrac n d}$ None of these are congruent modulo $n$ and none differ by as much as $n$. Further, for any $k \in Z$, we have that $x_1 + k \dfrac n d$ is congruent modulo $n$ to one of them. To see this, write $k = d q + r$ where $0 \le r < d$ from the Division Theorem. Then: {{begin-eqn}} {{eqn | l = x_1 + k \frac n d       | r = x_1 + \paren {d q + r} \frac n d       | c =  }} {{eqn | r = x_1 + n q + r \frac n d       | c =  }} {{eqn | o = \equiv       | r = x_1 + r \frac n d       | rr= \pmod n       | c =  }} {{end-eqn}} So these are the $d$ solutions of $a x \equiv b \pmod n$. {{qed}} Category:Solution of Linear Congruence npnsey8asyo4dq5g5ub0fbwfcrigw4m
From Solution of Linear Congruence: Existence: :the problem of finding all integers satisfying the linear congruence $a x \equiv b \pmod n$  is the same problem as: :the problem of finding all the $x$ values in the linear Diophantine equation $a x - n y = b$. Let: :$\gcd \set {a, n} = 1$ Let $x = x_0, y = y_0$ be one solution to the linear Diophantine equation: :$a x - n y = b$ From Solution of Linear Diophantine Equation, the general solution is: :$\forall k \in \Z: x = x_0 + n k, y = y_0 + a k$ But: :$\forall k \in \Z: x_0 + n k \equiv x_0 \pmod n$ Hence $x \equiv x_0 \pmod n$ is the only solution of $a x \equiv b \pmod n$. {{qed}}
Proving each of the norm axioms one by one: === Proving $(N1) : \forall x \in R: \norm x = 0 \iff x = 0_R$ === This follows directly from the definition of the trivial norm. {{qed|lemma}} === Proving $(N2) : \forall x, y \in R: \norm {x \circ y} = \norm x \times \norm y$ === If $x = 0_R$: {{begin-eqn}} {{eqn | l = \norm {x \circ y}       | r = \norm {0_R \circ y}       | c = by substitution }} {{eqn | r = \norm {0_R}       | c = Ring Product with Zero }} {{eqn | r = 0       | c = {{Defof|Trivial Norm on Division Ring}} }} {{eqn | r = 0 \times \norm y       | c =  }} {{eqn | r = \norm x \times \norm y       | c = since $x = 0_R$ }} {{end-eqn}} The reasoning is similar if $y = 0_R$. If $x,y \ne 0_R$, then $x \circ y \ne 0_R$ by alternative definition $(3)$ of division rings. We get: {{begin-eqn}} {{eqn | l = \norm {x \circ y}       | r = 1       | c = since $x \circ y \ne 0_R$ }} {{eqn | r = 1 \times 1       | c =  }} {{eqn | r = \norm x \times \norm y       | c = since $x, y \ne 0_R$ }} {{end-eqn}} {{qed|lemma}} === Proving $(N3) : \forall x, y \in R: \norm {x + y} \le \norm x + \norm y$ === If $x = 0_R$: {{begin-eqn}} {{eqn | l = \norm {x + y}       | r = \norm {0_R + y}       | c =  }} {{eqn | r = \norm y       | c =  }} {{eqn | r = 0 + \norm y       | c =  }} {{eqn | r = \norm x + \norm y       | c = since $x = 0_R$ }} {{end-eqn}} The reasoning is similar if $y = 0_R$. If $x, y \ne 0_R$: {{begin-eqn}} {{eqn | l = \norm {x + y}       | o = \le       | r = 1       | c =  }} {{eqn | o = <       | r = 1 + 1       | c =  }} {{eqn | r = \norm x + \norm y       | c = since $x, y \ne 0_R$ }} {{end-eqn}} {{qed}} Category:Division Rings Category:Trivial Norms maljxp1yu5tbbkzkk6jv4dfoj081df2
First we check the ring axioms: :$\text A$: The Integers Modulo $m$ under Addition form Abelian Group. :From Modulo Addition has Identity, $\eqclass 0 m$ is the identity of the additive group $\struct {\Z_m, +_m}$. From Integers Modulo m under Multiplication form Commutative Monoid: :$\text M 0$: $\struct {\Z_m, \times_m}$ is closed. :$\text M $: $\struct {\Z_m, \times_m}$ is associative. :$\text M 2$: $\struct {\Z_m, \times_m}$ has an identity $\eqclass 1 m$. :$\text C$: $\struct {\Z_m, \times_m}$ is commutative. Then: :$\text D$: $\times_m$ distributes over $+_m$ in $\Z_m$. {{qed}}