hcju/proofwikiqa huggingface dataset:
Suppose $f \in D \sqbrk X \setminus \set {0_R}$ has more than one way of being expressed in the above form. Then you would be able to subtract one from the other and get a polynomial in $D \sqbrk X$ equal to zero. As $f$ is transcendental, the result follows. {{qed}}
By Ring of Polynomial Forms is Commutative Ring with Unity we know that $D \sqbrk X$ is a commutative ring with unity. Let neither $\displaystyle \map f X = \sum_{k \mathop = 0}^n a_k x^k$ nor $\displaystyle \map g X = \sum_{k \mathop = 0}^m b_k X^k$ be the null polynomial. Then their leading coefficients $a_n$ and $b_m$ are non-zero. Therefore, as $D$ is an integral domain and $a_n, b_m \in D$, so is their product $a_n b_m$. By the definition of polynomial multiplication, it follows that $f g$ is not the null polynomial. It follows that $D \sqbrk X$ has no proper zero divisors. Hence $D \sqbrk X$ is an integral domain. {{qed}}
First it is shown that the mapping $\phi: D \sqbrk {X_1} \to D \sqbrk {X_2}$ given by: :$\displaystyle \map \phi {\sum_{k \mathop = 0}^n a_k \circ X_1^k} = \sum_{k \mathop = 0}^n a_k \circ X_2^k$ is a bijection. Let $p, q \in \phi: D \sqbrk {X_1}$. Suppose $\map \phi p = \map \phi q$. Then the coefficients of $\map \phi p$ and $\map \phi q$ are equal, and $p_1 = q_1$. Thus, by definition, $\phi$ is an injection. By the same argument, the mapping $\psi: D \sqbrk {X_2} \to D \sqbrk {X_1}$ defined as: :$\displaystyle \map \psi {\sum_{k \mathop = 0}^n a_k \circ X_2^k} = \sum_{k \mathop = 0}^n a_k \circ X_1^k$ is similarly an injection. Thus by Injection is Bijection iff Inverse is Injection, $\phi$ is a bijection. It remains to show that $\phi$ is a ring homomorphism.  Let $\displaystyle p = \sum_{k \mathop = 0}^n a_k \circ X_1^k, q = \sum_{k \mathop = 0}^n b_k \circ X_2^k \in D \sqbrk {X_1}$.  For convenience we set $a_k = 0$, $k > n$ and $b_k = 0$, $k > m$. We have: {{begin-eqn}} {{eqn | l = \map \phi {p + q}       | r = \map \phi {\sum_{k \mathop = 0}^\infty \paren {a_k + b_k} X_1^k}       | c = {{Defof|Addition of Polynomials}} }} {{eqn | r = \sum_{k \mathop = 0}^\infty \left({a_k + b_k}\right) X_2^k       | c =  }} {{eqn | r = \sum_{k \mathop = 0}^m a_k X_2^k + \sum_{k \mathop = 0}^n b_k X_2^k       | c =  }} {{eqn | r = \map \phi p + \map \phi q       | c =  }} {{end-eqn}} Similarly for multiplication: {{begin-eqn}} {{eqn | l = \map \phi {p q}       | r = \map \phi {\sum_{k \mathop = 0}^{m n} \sum_{i + j = k} a_i b_j X_1^k}       | c = {{Defof|Multiplication of Polynomials}} }} {{eqn | r = \sum_{k \mathop = 0}^{m n} \sum_{i + j \mathop = k} a_i b_j X_2^k       | c =  }} {{eqn | r = \paren {\sum_{k \mathop = 0}^n a_k \circ X_1^k} \paren {\sum_{k \mathop = 0}^n b_k \circ X_2^k}       | c =  }} {{eqn | r = \map \phi p \, \map \phi q       | c =  }} {{end-eqn}} This completes the proof. {{qed}}
=== Necessary Condition === Let $\phi$ be a bijection. Then from Bijection iff Inverse is Bijection, its inverse $\phi^{-1}$ is also a bijection and therefore by definition an injection. {{qed|lemma}} === Sufficient Condition === Let $\phi$ be an injection such that $\phi^{-1}$ is also an injection. By Cardinality of Surjection, and the Cantor-Bernstein-Schröder Theorem, the result follows. {{qed}} Category:Injections Category:Bijections ieo6rvg42igne5veer4re8to6lotq4m
For any $d \in F \sqbrk X$, let $\ideal d$ denote the principal ideal of $F \sqbrk X$ generated by $d$. Let $J$ be any ideal of $F \sqbrk X$. What we need to prove is that $J$ is a principal ideal. Let us first distinguish the following two cases for $J$: :If $J = \set {0_F}$, then by Zero Element Generates Null Ideal $J = \ideal {0_F}$, and hence is a principal ideal. :If $J = F \sqbrk X$, then by Ideal of Unit is Whole Ring: Corollary $J = \ideal {1_F}$, and hence is a principal ideal.  Now suppose $J \ne \set {0_F}$ and $J \ne F \sqbrk X$. Then $J$ necessarily contains a non-zero element. By the Well-Ordering Principle, we can introduce the lowest degree of a non-zero element of $J$. Denote this degree by $n$. If $n = 0$, then $J$ contains a polynomial of degree $0$. This is a non-zero element of $F$. As $F$ is a field, this is therefore a unit of $F$, and thus by Ideal of Unit is Whole Ring, $J = F \sqbrk X$. Because the degree of a non-zero element is a natural number, we conclude that $n \ge 1$. Now let $d$ be a polynomial of degree $n$ in $J$, and let $f \in J$. By Division Theorem for Polynomial Forms over Field, $f = q \circ d + r$ for some $q, r \in F \sqbrk X$ where either: :$r = 0_F$ or: :$r$ is a polynomial of degree smaller than $n$. Because $J$ is an ideal and $d \in J$, it follows that: :$q \circ d \in J$ Since $f \in J$, we also conclude: :$r = f - q \circ d \in J$ From the construction of $d$, it follows that we must have $r = 0_F$. Therefore: :$f = q \circ d$ and thus: :$f \in \ideal d$. This reasoning shows that: :$J \subseteq \ideal d$ From property $(3)$ of the principal ideal $\ideal d$, we conclude that: :$\ideal d \subseteq J$ as $d \in J$. Hence $J = \ideal d$. These $2$ distinguished cases cover all of the possible ideals of $F \sqbrk X$.  Hence $F \sqbrk X$ is a principal ideal domain. {{qed}}
We have that Polynomial Forms over Field is Euclidean Domain. We also have that Euclidean Domain is Principal Ideal Domain. Hence the result. {{qed}}
Suppose there were two consecutive gaps between $3$ consecutive prime numbers that were equal, but not divisible by $6$. Then the difference is $2 k$ where $k$ is not divisible by $3$. Therefore the (supposed) prime numbers will be: :$p, p + 2 k, p + 4 k$ But then $p + 4 k$ is congruent modulo $3$ to $p + k$. That makes the three numbers congruent modulo $3$ to $p, p + k, p + 2k$. One of those is divisible by $3$ and so can not be prime. So two consecutive gaps must be divisible by $3$ and therefore (as they have to be even) by $6$. {{Qed}} Category:Prime Numbers 3d5frank10lwfirx7e0rhzfct00dpq0
Let $d: S \times S \to \R$ denote the standard discrete metric on the underlying set $S$ of some space $\left({S, d}\right)$. By definition: :$\forall x, y \in S: d \left({x, y}\right) = \begin{cases} 0 & : x = y  \\  1 & : x \ne y \end{cases}$ === Proof of $M1$ === {{begin-eqn}} {{eqn | l = d \left({x, x}\right)       | r = 0       | c = Definition of Standard Discrete Metric }} {{end-eqn}} So axiom $M1$ holds for $d$. {{qed|lemma}} === Proof of $M2$ === Let $x = z$. {{begin-eqn}} {{eqn | l = d \left({x, z}\right)       | r = 0       | c =  }} {{eqn | ll= \implies       | l = d \left({x, y}\right) + d \left({y, z}\right)       | o = \ge       | r = d \left({x, z}\right)       | c = Definition of Standard Discrete Metric }} {{end-eqn}} Let $x \ne z$. Either $x \ne y$ or $y \ne z$, or both. So: {{begin-eqn}} {{eqn | l = d \left({x, y}\right) + d \left({y, z}\right)       | o = \ge       | r = 1       | c = Definition of Standard Discrete Metric }} {{eqn | o = \ge       | r = d \left({x, z}\right)       | c = Definition of Standard Discrete Metric }} {{end-eqn}} So in either case: : $d \left({x, y}\right) + d \left({y, z}\right) \ge d \left({x, z}\right)$ and axiom $M2$ holds for $d$. {{qed|lemma}} === Proof of $M3$ === Let $x \ne y$. {{begin-eqn}} {{eqn | l = d \left({x, y}\right)       | r = 1       | c = Definition of Standard Discrete Metric }} {{eqn | r = d \left({y, x}\right)       | c = Definition of $d$ }} {{end-eqn}} So axiom $M3$ holds for $d$. {{qed|lemma}} === Proof of $M4$ === {{begin-eqn}} {{eqn | l = x       | o = \ne       | r = y       | c =  }} {{eqn | ll= \implies       | l = d \left({x, y}\right)       | o = >       | r = 0       | c = Definition of Standard Discrete Metric }} {{end-eqn}} So axiom $M4$ holds for $d$. {{qed}}
{{begin-eqn}} {{eqn | l = \map {f'} \xi       | r = \lim_{h \mathop \to 0} \frac {\map f {\xi + h} - \map f \xi} h       | c = }} {{eqn | r = \lim_{h \mathop \to 0} \frac {\map j {\xi + h} \map k {\xi + h} - \map j \xi \map k \xi} h       | c = }} {{eqn | r = \lim_{h \mathop \to 0} \frac {\map j {\xi + h} \map k {\xi + h} - \map j {\xi + h} \map k \xi + \map j {\xi + h} \map k \xi - \map j \xi \map k \xi} h       | c = }} {{eqn | r = \lim_{h \mathop \to 0} \paren {\map j {\xi + h} \frac {\map k {\xi + h} - \map k \xi} h + \frac {\map j {\xi + h} - \map j \xi} h \map k \xi}       | c = }} {{eqn | r = \map j \xi \map {k'} \xi + \map {j'} \xi \map k \xi       | c = }} {{end-eqn}} Note that $\map j {\xi + h} \to \map j \xi$ as $h \to 0$ because, from Differentiable Function is Continuous, $j$ is continuous at $\xi$. {{qed}}
Let $\map g x = y$, and let: {{begin-eqn}} {{eqn | l = \map g {x + \delta x}       | r = y + \delta y       | c =  }} {{eqn | ll= \leadsto       | l = \delta y       | r = \map g {x + \delta x} - \map g x       | c =  }} {{end-eqn}} Thus: :$\delta y \to 0$ as $\delta x \to 0$ and: :$(1): \quad \dfrac {\delta y} {\delta x} \to \map {g'} x$ There are two cases to consider: === Case 1 === Suppose $\map {g'} x \ne 0$ and that $\delta x$ is small but non-zero. Then $\delta y \ne 0$ from $(1)$ above, and: {{begin-eqn}} {{eqn | l = \lim_{\delta x \mathop \to 0} \frac {\map h {x + \delta x} - \map h x} {\delta x}       | r = \lim_{\delta x \mathop \to 0} \frac {\map f {\map g {x + \delta x} } - \map f {\map g x} } {\map g {x + \delta x} - \map g x} \frac {\map g {x + \delta x} - \map g x} {\delta x}       | c =  }} {{eqn | r = \lim_{\delta x \mathop \to 0} \frac {\map f {y + \delta y} - \map f y} {\delta y} \frac {\delta y} {\delta x}       | c =  }} {{eqn | r = \map {f'} y \, \map {g'} x       | c =  }} {{end-eqn}} hence the result. {{qed|lemma}} === Case 2 === Now suppose $\map {g'} x = 0$ and that $\delta x$ is small but non-zero. Again, there are two possibilities: === Case 2a === If $\delta y = 0$, then $\dfrac {\map h {x + \delta x} - \map h x} {\delta x} = 0$. Hence the result. {{qed|lemma}} === Case 2b === If $\delta y \ne 0$, then: :$\dfrac {\map h {x + \delta x} - \map h x} {\delta x} = \dfrac {\map f {y + \delta y} - \map f y} {\delta y} \dfrac {\delta y} {\delta x}$ As $\delta y \to 0$: :$(1): \quad \dfrac {\map f {y + \delta y} - \map f y} {\delta y} \to \map {f'} y$ :$(2): \quad \dfrac {\delta y} {\delta x} \to 0$ Thus: :$\displaystyle \lim_{\delta x \mathop \to 0} \frac {\map h {x + \delta x} - \map h x} {\delta x} \to 0 = \map {f'} y \, \map {g'} x$ Again, hence the result. {{qed|lemma}} All cases have been covered, so by Proof by Cases, the result is complete. {{Qed}}
From Derivative of Monotone Function, it follows that $f$ is either: :strictly increasing on $I$ (if $\forall x \in I^o: D \, \map f x > 0$) or: :strictly decreasing on $I$ (if $\forall x \in I^o: D \, \map f x < 0$). Therefore from Inverse of Strictly Monotone Function it follows that $f^{-1}: J \to I$ exists. As $f$ is continuous, from Image of Interval by Continuous Function is Interval it follows that $J$ is an interval. By the Corollary to Limit of Increasing Function and the Corollary to Limit of Decreasing Function, $f^{-1}: J \to I$ is continuous. Next its derivative is to be considered. Suppose $f$ is strictly increasing. Let $y \in J^o$. Then $\map {f^{-1} } y \in I^o$. Let $k = \map {f^{-1} } {y + h} - \map {f^{-1} } y$. Thus: :$\map {f^{-1} } {y + h} = \map {f^{-1} } y + k = x + k$ Thus: :$y + h = \map f {x + k}$ and hence: :$h = \map f {x + k} - y = \map f {x + k} - \map f x$ Since $f^{-1}$ is continuous on $J$, it follows that $k \to 0$ as $h \to 0$. Also, $f^{-1}$ is strictly increasing from Inverse of Strictly Monotone Function and so $k \ne 0$ unless $h = 0$. So by Limit of Composite Function we get: :$\dfrac {\map {f^{-1} } {y + h} - \map {f^{-1} } y} h = \dfrac k {\map f {x + k} - \map f x}$ Thus: :$\dfrac {\map {f^{-1} } {y + h} - \map {f^{-1} } y} h \to \dfrac 1 {D \map f x}$ as $h \to 0$. Suppose $f$ is strictly decreasing. Exactly the same argument applies. {{qed}}
For all $\nu \in 1, 2, \ldots, n$, let $\closedint {x_{\nu - 1} } {x_\nu}$ be a closed subinterval of $\closedint a b$. As $f$ is bounded on $\closedint a b$, it is bounded on $\closedint {x_{\nu - 1} } {x_\nu}$. So, let $m_\nu$ be the infimum and $M_\nu$ be the supremum of $\map f x$ on the interval $\closedint {x_{\nu - 1} } {x_\nu}$. By definition, $m_\nu \le M_\nu$. So $m_{\nu} \paren {x_\nu - x_{\nu - 1} } \le M_{\nu} \paren {x_\nu - x_{\nu - 1} }$. It follows directly that $\displaystyle \sum_{\nu \mathop = 1}^n m_\nu \paren {x_\nu - x_{\nu - 1} } \le \sum_{\nu \mathop = 1}^n M_\nu \paren {x_\nu - x_{\nu - 1} }$. {{qed}} Category:Real Analysis 9cugcptjnm98xa2lsslxmftu3hmjg09
Suppose $P$ and $Q$ are two properties which are mutually exclusive. Suppose that one of either of $P$ and $Q$ are possessed by every $x \in \R$. Suppose that any number having $P$ is less than any which have $Q$. Let us call the numbers with $P$ the ''left hand set'' $L$, and the ones with $Q$ the ''right hand set'' $R$. There are two possibilities, as follows. * $L$ has a greatest element, or * $R$ has a least element. '''It is not possible that both of the above can happen.''' Because suppose $l$ is the greatest element of $L$ and $r$ is the least element of $R$. Then the number $\displaystyle \frac {l + r} 2$ is greater than $l$ and less than $r$, so it could not be in either class. '''However, one of the above ''must'' occur.''' Because, suppose the following. Let $L_1$ and $R_1$ be the subsets of $L$ and $R$ respectively consisting of only the rational numbers in $L$ and $R$. Then $L_1$ and $R_1$ form a section of the set of rational numbers $\Q$. There are two cases to think about: '''Maybe $L_1$ has a greatest element $\alpha$.''' In this case, $\alpha$ must also be the greatest element of $L$. Because if not, then there's a greater one, which we can call $\beta$. There are always rational numbers between $\alpha$ and $\beta$ from Rational Numbers are Close Packed. These are less than $\beta$ and thus belong to $L$ and (because they're rational) also to $L_1$. This is a contradiction, so if $\alpha$ is the greatest element of $L_1$, it's also the greatest element of $L$. '''On the other hand, $L_1$ may ''not'' have a greatest element.''' In this case, the section of the rational numbers formed by $L_1$ and $R_1$ is a real number $\alpha$. {{Explain|Justify the above assertion.}} It must belong to either $L$ or $R$. If it belongs to $L$ we can show, like we did before, that it is the greatest element of $L$. Similarly, if it belongs to $R$ we can show it is the least element of $R$. So in any case, either $L$ has a greatest element or $R$ has a least element. Thus, any section of the real numbers corresponds to a real number. {{qed}}
=== Proof of Uniqueness === {{AimForCont}} both $\alpha$ and $\beta$ produce $\tuple {L, R}$. By the Trichotomy Law for Real Numbers either $\beta < \alpha$ or $\alpha < \beta$. Suppose that $\beta < \alpha$. From Real Numbers are Close Packed, there exists at least one real number $c$ such that $\beta < c$ and $c < \alpha$. Because $c < \alpha$, it must be the case that $c \in L$. Because $\beta < c$, it must be the case that $c \in R$. That is: :$c \in L \cap R$ But by the definition of Dedekind Cut, $\tuple {L, R}$ is a partition of $\R$. That is, $L$ and $R$ are disjoint. That is: :$L \cap R = \O$ This is a contradiction. Similarly, $\alpha < \beta$ also leads to a contradiction. It follows that $\alpha$ is unique. {{qed|lemma}} === Proof of Existence === Let $\alpha = \sup L$. Let $u_1 \in \R$ such that $u_1 < \alpha$. {{AimForCont}} $u_1 \in R$. Suppose there is no real number $s \in L$ such that $u_1 < s \le \alpha$. Then $u_1$ would be an upper bound of $L$ which is less than $\sup {L}$. This is a contradiction. So there exists at least one real number $s \in L$ such that $u_1 < s \le \alpha$. But this is also a contradiction because all the elements of $R$ are greater than all the elements of $L$. It follows that $u_1 \in L$. Let $u_2 \in \R$ such that $\alpha < u_2$. {{AimForCont}} $u_2 \in L$. There exists $s' \in \R$ such that $\alpha < s' < u_2$. But $s' \notin L$ and thus $s' \in R$. This is a contradiction because all the elements of $R$ are greater than all the elements of $L$. It follows that $u_2 \in R$. Hence, $\alpha$ produces the cut $\tuple {L, R}$. {{qed}}
=== Proof of Uniqueness === {{AimForCont}} both $\alpha$ and $\beta$ produce $\tuple {L, R}$. By the Trichotomy Law for Real Numbers either $\beta < \alpha$ or $\alpha < \beta$. Suppose that $\beta < \alpha$. From Real Numbers are Close Packed, there exists at least one real number $c$ such that $\beta < c$ and $c < \alpha$. Because $c < \alpha$, it must be the case that $c \in L$. Because $\beta < c$, it must be the case that $c \in R$. That is: :$c \in L \cap R$ But by the definition of Dedekind Cut, $\tuple {L, R}$ is a partition of $\R$. That is, $L$ and $R$ are disjoint. That is: :$L \cap R = \O$ This is a contradiction. Similarly, $\alpha < \beta$ also leads to a contradiction. It follows that $\alpha$ is unique. {{qed|lemma}} === Proof of Existence === Let $\gamma$ be the set of all rational numbers $p$ such that $p \in \alpha$ for some $\alpha \in L$. It is to be verified that $\gamma$ is a cut. Because $L$ is not empty, neither is $\gamma$. Suppose $\beta \in \R$ and $q \notin \beta$. Then because $\alpha < \beta$, we have that $q \notin \alpha$ for any $\alpha \in L$. Thus $q \notin \gamma$. Thus $\gamma$ satisfies criterion $(1)$ for being a cut. Suppose $p \in \gamma$ and $q < p$. Then $p \in \alpha$ for some $\alpha \in L$. Hence $q \in \alpha$. Hence $q \in \gamma$. Thus $\gamma$ satisfies criterion $(2)$ for being a cut. Suppose $p \in \gamma$. Then $p \in \alpha$ for some $\alpha \in L$. Hence there exists $q > p$ such that $q \in \alpha$. Hence $q \in \gamma$. Thus $\gamma$ satisfies criterion $(3)$ for being a cut. Thus, by definition of the real numbers by identifying them with cuts, $\gamma$ is a real number. We have that: :$\alpha \le \gamma$ for all $\alpha \in L$. {{AimForCont}} there exists $\beta \in R$ such that $\beta < \gamma$. Then there exists some rational number $p \in\ Q$ such that $p \in \gamma$ and $p \notin beta$. But if $p \in \gamma$ then $p \in \alpha$ for some $\alpha\ in L$. This implies that $\beta < \alpha$. But this contradicts the definition of Dedekind cut: :$(3): \quad \forall x \in L: \forall y \in R: x < y$ Thus $\gamma \le \beta$ for all $\beta \in R$. That is, $\gamma$ is a producer of $\tuple {L, R}$. {{qed}}
From the informal definition of complex numbers, we define the following: :$z = x_1 + i y_1$ :$w = x_2 + i y_2$ where $i = \sqrt {-1}$ and $x_1, x_2, y_1, y_2 \in \R$. Then from the definition of complex addition: :$z + w = \paren {x_1 + x_2} + i \paren {y_1 + y_2}$ From Real Numbers under Addition form Abelian Group, real addition is closed. So: :$\paren {x_1 + x_2} \in \R$ and $\paren {y_1 + y_2} \in \R$ Hence the result. {{qed}}
From the formal definition of complex numbers, we have: :$z = \tuple {x_1, y_1}$ :$w = \tuple {x_2, y_2}$ where $x_1, x_2, y_1, y_2 \in \R$. Then from the definition of complex addition: :$z + w = \tuple {x_1 + x_2, y_1 + y_2}$ From Real Numbers under Addition form Abelian Group, real addition is closed. So: :$\paren {x_1 + x_2} \in \R$ and $\paren {y_1 + y_2} \in \R$ and hence the result. {{qed}}
From the definition of complex numbers, we define the following: {{begin-eqn}} {{eqn | l = z_1       | o = :=       | r = \tuple {x_1, y_1} }} {{eqn | l = z_2       | o = :=       | r = \tuple {x_2, y_2} }} {{eqn | l = z_3       | o = :=       | r = \tuple {x_3, y_3} }} {{end-eqn}} where $x_1, x_2, x_3, y_1, y_2, y_3 \in \R$. Thus: {{begin-eqn}} {{eqn | l = z_1 + \paren {z_2 + z_3}       | r = \tuple {x_1, y_1} + \paren {\tuple {x_2, y_2} + \tuple {x_3, y_3} }       | c = {{Defof|Complex Number|index = 2}} }} {{eqn | r = \tuple {x_1, y_1} + \paren {\tuple {x_2 + x_3, y_2 + y_3} }       | c = {{Defof|Complex Number/Definition 2/Addition|Complex Addition}} }} {{eqn | r = \tuple {x_1 + \paren {x_2 + x_3}, y_1 + \paren {y_2 + y_3} }       | c = {{Defof|Complex Number/Definition 2/Addition|Complex Addition}} }} {{eqn | r = \tuple {\paren {x_1 + x_2} + x_3, \paren {y_1 + y_2} + y_3}       | c = Real Addition is Associative }} {{eqn | r = \paren {\tuple {x_1 + x_2, y_1 + y_2} } + \tuple {x_3, y_3}       | c = {{Defof|Complex Number/Definition 2/Addition|Complex Addition}} }} {{eqn | r = \paren {\tuple {x_1, y_1} + \tuple {x_2, y_2} } + \tuple {x_3, y_3}       | c = {{Defof|Complex Number/Definition 2/Addition|Complex Addition}} }} {{eqn | r = \paren {z_1 + z_2} + z_3       | c = {{Defof|Complex Number|index = 2}} }} {{end-eqn}} {{qed}}
From the definition of complex numbers, we define the following: {{begin-eqn}} {{eqn | l = z       | o = :=       | r = \tuple {x_1, y_1} }} {{eqn | l = w       | o = :=       | r = \tuple {x_2, y_2} }} {{end-eqn}} where $x_1, x_2, y_1, y_2 \in \R$. Then: {{begin-eqn}} {{eqn | l = z + w       | r = \tuple {x_1, y_1} + \tuple {x_2, y_2}       | c = {{Defof|Complex Number|index = 2}} }} {{eqn | r = \tuple {x_1 + x_2, y_1 + y_2}       | c = {{Defof|Complex Number/Definition 2/Addition|Complex Addition}} }} {{eqn | r = \tuple {x_2 + x_1, y_2 + y_1}       | c = Real Addition is Commutative }} {{eqn | r = \tuple {x_2, y_2} + \tuple {x_1, y_1}       | c = {{Defof|Complex Number/Definition 2/Addition|Complex Addition}} }} {{eqn | r = w + z       | c = {{Defof|Complex Number|index = 2}} }} {{end-eqn}} {{qed}}
Define the inclusion mapping $i: \N \to \Z$. From Inclusion Mapping is Injection, $i: \N \to \Z$ is an injection. Thus there exists an injection from $\N$ to $\Z$. Hence $\Z$ is infinite. Next, let us arrange $\Z$ in the following order: :$\Z = \set {0, 1, -1, 2, -2, 3, -3, \ldots}$ Then we can directly see that we can define a mapping $\phi: \Z \to \N$ as follows: :$\forall x \in \Z: \map \phi x = \begin{cases} 2 x - 1 & : x > 0 \\ -2 x & : x \le 0 \end{cases}$ This is shown to be an injection as follows: Let $\map \phi x = \map \phi y$. Then one of the following applies: : $(1): \quad -2 x = -2 y$ in which case $x = y$ : $(2): \quad 2 x - 1 = 2 y - 1$ in which case $2 x = 2 y$ and so $x = y$ : $(3): \quad 2 x - 1 = -2 y$ in which case $y = -x + \frac 1 2$ and therefore $y \notin \Z$ : $(4): \quad 2 y - 1 = -2 x$ in which case $x = -y + \frac 1 2$ and therefore $x \notin \Z$. So $2 x - 1 = -2 y$ and $2 y - 1 = -2 x$ can't happen and so $x = y$. Thus $\phi$ is injective. The result follows from Domain of Injection to Countable Set is Countable. {{qed}}
First it is shown that: :$\displaystyle \map \det {\mathbf A} = \sum_{\mu \mathop \in S_n} \paren {\map \sgn \mu \map \sgn \lambda \prod_{k \mathop = 1}^n a_{\map \lambda k, \map \mu k} }$ Let $\nu: \N_{> 0} \to \N_{> 0}$ be a permutation on $\N_{> 0}$ such that $\lambda \circ \nu = \mu$. The product can be rearranged as: :$\displaystyle \prod_{k \mathop = 1}^n a_{\map \lambda k, \map \mu k} = a_{\map \lambda 1, \map \mu 1} a_{\map \lambda 2, \map \mu 2} \cdots a_{\map \lambda n, \map \mu n} = a_{1, \map \nu 1} a_{2, \map \nu 2} \cdots a_{n, \map \nu n} = \prod_{k \mathop = 1}^n a_{k, \map \nu k}$ from {{Field-axiom|M2}}. By Parity Function is Homomorphism: :$\map \sgn \mu \map \sgn \lambda = \map \sgn \lambda \map \sgn \nu \map \sgn \lambda = \map {\sgn^2} \lambda \map \sgn \nu = \map \sgn \nu$ and so: :$\displaystyle \map \det {\mathbf A} = \sum_{\nu \mathop \in S_n} \paren {\map \sgn \nu \prod_{k \mathop = 1}^n a_{k, \map \nu k} }$ which is the usual definition for the determinant. Next it is to be shown: :$\displaystyle \map \det {\mathbf A} = \sum_{\mu \mathop \in S_n} \paren {\map \sgn \mu \map \sgn \lambda \prod_{k \mathop = 1}^n a_{\map \mu k, \map \lambda k} }$ Let $\nu: \N_{> 0} \to \N_{> 0}$ be a permutation on $\N_{> 0}$ such that $\mu \circ \nu = \lambda$. The result follows via a similar argument. {{Qed}} Category:Determinants ncgpkk0vowuqgjdvv37pev1rssrws1s