hcju/proofwikips huggingface dataset:
Modulo Arithmetic/Examples/n(n^2-1)(3n-2) Modulo 24
Modulo Arithmetic/Examples/n(n^2-1)(3n-2) Modulo 24 0 74319 390579 2019-01-29T21:13:22Z Prime.mover 59 Created page with "== Example of Modulo Arithmetic ==  :$n \paren {n^2 - 1} \paren {3 n + 2} \equiv 0 \pmod {24}$   == Proof ==  The p..." wikitext text/x-wiki == Example of Modulo Arithmetic ==  :$n \paren {n^2 - 1} \paren {3 n + 2} \equiv 0 \pmod {24}$  == Proof == The proof proceeds by induction. For all $n \in \Z_{> 0}$, let $\map P n$ be the proposition: :$n \paren {n^2 - 1} \paren {3 n + 2} \equiv 0 \pmod {24}$ === Basis for the Induction === $\map P 1$ is the case: {{begin-eqn}} {{eqn | l = 1 \paren {1^2 - 1} \paren {3 \times 1 + 2}       | r = 0       | c =  }} {{eqn | o = \equiv       | r = 0       | rr= \pmod {24}       | c =  }} {{end-eqn}} Thus $\map P 1$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true. So this is the induction hypothesis: :$k \paren {k^2 - 1} \paren {3 k + 2} \equiv 0 \pmod {24}$ from which it is to be shown that: :$\paren {k + 1} \paren {\paren {k + 1}^2 - 1} \paren {3 \paren {k + 1} + 2} \equiv 0 \pmod {24}$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | l = \paren {k + 1} \paren {\paren {k + 1}^2 - 1} \paren {3 \paren {k + 1} - 2}       | r = k \paren {k + 1} \paren {k + 2} \paren {3 k + 5}       | c =  }} {{eqn | r = k \paren {k + 1} \paren {k - 1 + 3} \paren {3 k + 2 + 3}       | c =  }} {{eqn | r = k \paren {k + 1} \paren {k - 1} \paren {3 k + 2 + 3} + 3 k \paren {k + 1} \paren {3 k + 2 + 3}       | c =  }} {{eqn | r = k \paren {k + 1} \paren {k - 1} \paren {3 k + 2} + 3 k \paren {k + 1} \paren {k - 1} + 3 k \paren {k + 1} \paren {3 k + 5}       | c =  }} {{eqn | r = k \paren {k^2 - 1} \paren {3 k + 2} + 3 k \paren {k + 1} \paren {k - 1 + 3 k + 5}       | c =  }} {{eqn | r = k \paren {k^2 - 1} \paren {3 k + 2} + 3 k \paren {k + 1} \paren {4 k + 4}       | c =  }} {{eqn | r = k \paren {k^2 - 1} \paren {3 k + 2} + 12 k \paren {k + 1}^2       | c =  }} {{end-eqn}} By the induction hypothesis: :$k \paren {k^2 - 1} \paren {3 k + 2} = 24 r$ for some $r \in \Z$. Take $12 k \paren {k + 1}^2$. If $k$ is even, then $12 k$ and so $12 k \paren {k + 1}^2$ is divisible by $24$. If $k$ is odd, then $k + 1$ is even and so $12 k \paren {k + 1}^2$ is again divisible by $24$. Thus: :$12 k \paren {k + 1}^2 = 24 s$ for some $s \in \Z$. Thus: :$k \paren {k^2 - 1} \paren {3 k + 2} + 12 k \paren {k + 1}^2 = 24 \paren {r + s}$ So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n \in \Z_{>0}: n \paren {n^2 - 1} \paren {3 n + 2} \equiv 0 \pmod {24}$  == Sources == * {{BookReference|Number Theory|1971|George E. Andrews|prev = Sum of Sequence of Product of Lucas Numbers with Powers of 2|next = Sum of Odd Positive Powers}}: $\text {1-1}$ Principle of Mathematical Induction: Exercise $18$
Euclidean Algorithm/Examples/341 and 527
== Examples of Use of Euclidean Algorithm ==  The GCD of $341$ and $527$ is found to be: :$\gcd \set {341, 527} = 31$ ==== Integer Combination ==== {{:Euclidean Algorithm/Examples/341 and 527/Integer Combination}} == Proof == {{begin-eqn}} {{eqn | n = 1       | l = 527       | r = 1 \times 341 + 186 }} {{eqn | n = 2       | l = 341       | r = 1 \times 186 + 155 }} {{eqn | n = 3       | l = 186       | r = 1 \times 155 + 31 }} {{eqn | n =        | l = 155       | r = 5 \times 31 }} {{end-eqn}} Thus: :$\gcd \set {341, 527} = 31$  == Sources == * {{BookReference|Number Theory|1971|George E. Andrews|prev = GCD of Integer and Divisor|next = Euclidean Algorithm}}: $\text {2-2}$ Divisibility: Example $\text {2-7}$
Lowest Common Multiple/Examples/n and n+1
== Example of Lowest Common Multiple of Integers == Let $n \in \Z_{>0}$ be a strictly positive integer.  The lowest common multiple of $n$ and $n + 1$ is: :$\lcm \set {n, n + 1} = n \paren {n + 1}$  == Proof == We find the greatest common divisor of $n$ and $n + 1$ using the Euclidean Algorithm: {{begin-eqn}} {{eqn | n = 1       | l = n + 1       | r = 1 \times n + 1 }} {{eqn | n = 2       | l = n       | r = n \times 1 }} {{end-eqn}} Thus $\gcd \set {n, n + 1} = 1$. Hence by definition $n$ and $n + 1$ are coprime. The result follows from LCM of Coprime Integers.  == Sources == * {{BookReference|Number Theory|1971|George E. Andrews|prev = Lowest Common Multiple/Examples/28 and 29|next = Lowest Common Multiple of Consecutive Odd Integers}}: $\text {2-2}$ Divisibility: Exercise $5 \ \text {(e)}$
Area of Triangle in Determinant Form with Vertex at Origin
== Example of Area of Triangle in Determinant Form ==  Let $A = \tuple {0, 0}, B = \tuple {b, a}, C = \tuple {x, y}$ be points in the Cartesian plane. Let $T$ the triangle whose vertices are at $A$, $B$ and $C$. Then the area $\mathcal A$ of $T$ is: :$\map \Area T = \dfrac {\size {b y - a x} } 2$  == Proof == {{begin-eqn}} {{eqn | l = \map \Area T       | r = \dfrac 1 2 \size {\paren {\begin{vmatrix} 0 & 0 & 1 \\ b & a & 1 \\ x & y & 1 \end{vmatrix} } }       | c = Area of Triangle in Determinant Form }} {{eqn | r = \dfrac 1 2 \size {b y - a x}       | c = {{Defof|Determinant of Order 3}} }} {{end-eqn}}  == Example == {{:Area of Triangle in Determinant Form with Vertex at Origin/Example}} == Sources == * {{BookReference|Number Theory|1971|George E. Andrews|prev = Linear Diophantine Equation/Examples/15x + 51y = 41/Graph|next = Area of Triangle in Determinant Form with Vertex at Origin/Example}}: $\text {2-3}$ The Linear Diophantine Equation: Exercise $4$
Square Modulo 3/Corollary 3
== Corollary to Square Modulo 3 ==  Let $n \in \Z$ be an integer such that: :$3 \nmid n$ where $\nmid$ denotes non-divisibility. Then: :$3 \divides n^2 - 1$ where $\divides$ denotes divisibility.  == Proof == From Square Modulo 3: :$n \equiv 0 \pmod 3 \iff n^2 \equiv 0 \pmod 3$ Hence also from Square Modulo 3: :$n \not \equiv 0 \pmod 3 \iff n^2 \equiv 1 \pmod 3$ That is: $3 \nmid n \iff 3 \divides n^2 - 1$  == Sources == * {{BookReference|Number Theory|1971|George E. Andrews|prev = Integer and Fifth Power have same Last Digit|next = Sufficient Condition for 5 to divide n^2+1}}: $\text {3-2}$ Fermat's Little Theorem: Exercise $5$
Quadratic Residue/Examples/3
== Example of Quadratic Residues ==  There exists exactly $1$ quadratic residue modulo $3$, and that is $1$.  == Proof == To list the quadratic residues of $3$ it is enough to work out the squares $1^2$ and $2^2$ modulo $3$. {{begin-eqn}} {{eqn | l = 1^2       | o = \equiv       | r = 1       | rr= \pmod 3 }} {{eqn | l = 2^2       | o = \equiv       | r = 1       | rr= \pmod 3 }} {{end-eqn}} So the set of quadratic residues modulo $3$ is: :$\set 1$ The set of quadratic non-residues of $3$ therefore consists of all the other non-zero least positive residues: :$\set 2$  == Sources == * {{BookReference|Number Theory|1971|George E. Andrews|prev = Quadratic Residue/Examples/11|next = Quadratic Residue/Examples/5}}: $\text {3-5}$ The Use of Computers in Number Theory: Exercise $6$
Quadratic Residue/Examples/5
== Example of Quadratic Residues ==  The set of quadratic residues modulo $5$ is: :$\set {1, 4}$  == Proof == To list the quadratic residues of $5$ it is enough to work out the squares $1^2, 2^2, 3^2, 4^2$ modulo $5$. {{begin-eqn}} {{eqn | l = 1^2       | o = \equiv       | r = 1       | rr= \pmod 5 }} {{eqn | l = 2^2       | o = \equiv       | r = 4       | rr= \pmod 5 }} {{eqn | l = 3^2       | o = \equiv       | r = 4       | rr= \pmod 5 }} {{eqn | l = 4^2       | o = \equiv       | r = 1       | rr= \pmod 5 }} {{end-eqn}} So the set of quadratic residues modulo $5$ is: :$\set {1, 4}$ The set of quadratic non-residues of $5$ therefore consists of all the other non-zero least positive residues: :$\set {2, 3}$  == Sources == * {{BookReference|Number Theory|1971|George E. Andrews|prev = Quadratic Residue/Examples/3|next = Quadratic Residue/Examples/7}}: $\text {3-5}$ The Use of Computers in Number Theory: Exercise $6$
Quadratic Residue/Examples/7
== Example of Quadratic Residues ==  The set of quadratic residues modulo $7$ is: :$\set {1, 2, 4}$  == Proof == To list the quadratic residues of $7$ it is enough to work out the squares $1^2, 2^2, \dotsc, 6^2$ modulo $7$. {{begin-eqn}} {{eqn | l = 1^2       | o = \equiv       | r = 1       | rr= \pmod 7 }} {{eqn | l = 2^2       | o = \equiv       | r = 4       | rr= \pmod 7 }} {{eqn | l = 3^2       | o = \equiv       | r = 2       | rr= \pmod 7 }} {{eqn | l = 4^2       | o = \equiv       | r = 2       | rr= \pmod 7 }} {{eqn | l = 5^2       | o = \equiv       | r = 4       | rr= \pmod 7 }} {{eqn | l = 6^2       | o = \equiv       | r = 1       | rr= \pmod 7 }} {{end-eqn}} So the set of quadratic residues modulo $7$ is: :$\set {1, 2, 4}$ The set of quadratic non-residues of $7$ therefore consists of all the other non-zero least positive residues: :$\set {3, 5, 6}$  == Sources == * {{BookReference|Number Theory|1971|George E. Andrews|prev = Quadratic Residue/Examples/5|next = Quadratic Residue/Examples/17}}: $\text {3-5}$ The Use of Computers in Number Theory: Exercise $6$
Quadratic Residue/Examples/17
== Example of Quadratic Residues ==  The set of quadratic residues modulo $17$ is: :$\set {1, 2, 4, 8, 9, 13, 15, 16}$  {{OEIS|A010379}} == Proof == To list the quadratic residues of $17$ it is enough to work out the squares $1^2, 2^2, \dotsc, 16^2$ modulo $17$. {{begin-eqn}} {{eqn | l = 1^2       | o = \equiv       | r = 1       | rr= \pmod {17} }} {{eqn | l = 2^2       | o = \equiv       | r = 4       | rr= \pmod {17} }} {{eqn | l = 3^2       | o = \equiv       | r = 9       | rr= \pmod {17} }} {{eqn | l = 4^2       | o = \equiv       | r = 16       | rr= \pmod {17} }} {{eqn | l = 5^2       | o = \equiv       | r = 8       | rr= \pmod {17} }} {{eqn | l = 6^2       | o = \equiv       | r = 2       | rr= \pmod {17} }} {{eqn | l = 7^2       | o = \equiv       | r = 15       | rr= \pmod {17} }} {{eqn | l = 8^2       | o = \equiv       | r = 13       | rr= \pmod {17} }} {{eqn | l = 9^2       | o = \equiv       | r = 13       | rr= \pmod {17} }} {{eqn | l = 10^2       | o = \equiv       | r = 15       | rr= \pmod {17} }} {{eqn | l = 11^2       | o = \equiv       | r = 2       | rr= \pmod {17} }} {{eqn | l = 12^2       | o = \equiv       | r = 8       | rr= \pmod {17} }} {{eqn | l = 13^2       | o = \equiv       | r = 16       | rr= \pmod {17} }} {{eqn | l = 14^2       | o = \equiv       | r = 9       | rr= \pmod {17} }} {{eqn | l = 15^2       | o = \equiv       | r = 4       | rr= \pmod {17} }} {{eqn | l = 16^2       | o = \equiv       | r = 1       | rr= \pmod {17} }} {{end-eqn}} So the set of quadratic residues modulo $17$ is: :$\set {1, 2, 4, 8, 9, 13, 15, 16}$ The set of quadratic non-residues of $17$ therefore consists of all the other non-zero least positive residues: :$\set {3, 5, 6, 7, 10, 11, 12, 14}$ {{OEIS|A028730}} == Sources == * {{BookReference|Number Theory|1971|George E. Andrews|prev = Quadratic Residue/Examples/7|next = Quadratic Residue/Examples/29}}: $\text {3-5}$ The Use of Computers in Number Theory: Exercise $6$
Quadratic Residue/Examples/29
== Example of Quadratic Residues ==  The set of quadratic residues modulo $29$ is: :$\set {1, 4, 5, 6, 7, 9, 13, 16, 20, 22, 23, 24, 25, 28}$  {{OEIS|A010391}} == Proof == From Square Modulo n Congruent to Square of Inverse Modulo n, to list the quadratic residues of $29$ it is sufficient to work out the squares $1^2, 2^2, \dotsc, \paren {\dfrac {28} 2}^2$ modulo $29$. So: {{begin-eqn}} {{eqn | l = 1^2       | o = \equiv       | r = 1       | rr= \pmod {29} }} {{eqn | l = 2^2       | o = \equiv       | r = 4       | rr= \pmod {29} }} {{eqn | l = 3^2       | o = \equiv       | r = 9       | rr= \pmod {29} }} {{eqn | l = 4^2       | o = \equiv       | r = 16       | rr= \pmod {29} }} {{eqn | l = 5^2       | o = \equiv       | r = 25       | rr= \pmod {29} }} {{eqn | l = 6^2       | o = \equiv       | r = 7       | rr= \pmod {29} }} {{eqn | l = 7^2       | o = \equiv       | r = 20       | rr= \pmod {29} }} {{eqn | l = 8^2       | o = \equiv       | r = 6       | rr= \pmod {29} }} {{eqn | l = 9^2       | o = \equiv       | r = 23       | rr= \pmod {29} }} {{eqn | l = 10^2       | o = \equiv       | r = 13       | rr= \pmod {29} }} {{eqn | l = 11^2       | o = \equiv       | r = 5       | rr= \pmod {29} }} {{eqn | l = 12^2       | o = \equiv       | r = 28       | rr= \pmod {29} }} {{eqn | l = 13^2       | o = \equiv       | r = 24       | rr= \pmod {29} }} {{eqn | l = 14^2       | o = \equiv       | r = 22       | rr= \pmod {29} }} {{end-eqn}} So the set of quadratic residues modulo $29$ is: :$\set {1, 4, 5, 6, 7, 9, 13, 16, 20, 22, 23, 24, 25, 28}$ The set of quadratic non-residues of $29$ therefore consists of all the other non-zero least positive residues: :$\set {2, 3, 8, 10, 11, 12, 14, 15, 17, 18, 19, 21, 26, 27}$ {{OEIS|A028742}} == Sources == * {{BookReference|Number Theory|1971|George E. Andrews|prev = Quadratic Residue/Examples/17|next = Quadratic Residue/Examples/61}}: $\text {3-5}$ The Use of Computers in Number Theory: Exercise $6$
Reduced Residue System/Examples/Modulo 18/Least Positive Residues
== Examples of Reduced Residue Systems ==  The least positive reduced residue system of $18$ is the set of positive integers: :$\set {1, 5, 7, 11, 13, 17}$  == Proof == The least positive residues of $18$ are: :$S := \set {0, 1, 2, \dotsc, 17}$ We have that: :$18 = 2 \times 3^2$ so the least positive reduced residue system of $18$ is the set of elements of $S$ which have neither $2$ or $3$ as a prime factor. That is: :$S \setminus \set {0, 2, 3, 4, 6, 8, 9, 10, 12, 14, 15, 16}$ Hence the result.  == Sources == * {{BookReference|Number Theory|1971|George E. Andrews|prev = Reduced Residue System/Examples/Modulo 18/Square Numbers|next = Congruence Modulo Power of p as Linear Combination of Congruences Modulo p}}: $\text {4-2}$ Residue Systems: Exercise $2 \ \text {(d)}$
Binomial Theorem/Examples/Cube of Difference
== Example of Use of Binomial Theorem ==  :$\paren {x - y}^3 = x^3 - 3 x^2 y + 3 x y^2 - y^3$  == Proof == Follows directly from the Binomial Theorem: :$\displaystyle \forall n \in \Z_{\ge 0}: \paren {x + \paren {-y} }^n = \sum_{k \mathop = 0}^n \binom n k x^{n - k} \paren {-y}^k$ putting $n = 3$.  == Sources == * {{BookReference|Mathematical Handbook of Formulas and Tables|1968|Murray R. Spiegel|prev = Cube of Sum|next = Fourth Power of Sum}}: $\S 2$: Special Products and Factors: $2.4$ * {{BookReference|Mathematical Handbook of Formulas and Tables|1968|Murray R. Spiegel|prev = Square of Sum/Algebraic Proof 2|next = Fourth Power of Sum}}: $\S 20$: Binomial Series: $20.6$
Binomial Theorem/Examples/4th Power of Difference
Binomial Theorem/Examples/4th Power of Difference 0 74937 393380 393379 2019-02-23T11:05:01Z Prime.mover 59 Prime.mover moved page Fourth Power of Difference to Binomial Theorem/Examples/4th Power of Difference wikitext text/x-wiki == Example of Use of Binomial Theorem ==  :$\paren {x - y}^4 = x^4 - 4 x^3 y + 6 x^2 y^2 - 4 x y^3 + y^4$  == Proof == Follows directly from the Binomial Theorem: :$\displaystyle \forall n \in \Z_{\ge 0}: \paren {x + \paren {-y} }^n = \sum_{k \mathop = 0}^n \binom n k x^{n - k} \paren {-y}^k$ putting $n = 4$.  == Sources == * {{BookReference|Mathematical Handbook of Formulas and Tables|1968|Murray R. Spiegel|prev = Fourth Power of Sum|next = Fifth Power of Sum}}: $\S 2$: Special Products and Factors: $2.6$
Square Root of Complex Number in Cartesian Form/Examples/-15-8i
Square Root of Complex Number in Cartesian Form/Examples/-15-8i 0 75200 394768 2019-03-05T20:37:56Z Prime.mover 59 Created page with "== Example of Square Root of Complex Number in Cartesian Form ==  :$\sqrt {-15 - 8 i} = \pm \paren {1 - 4 i}$   == Square Root of Complex Numb..." wikitext text/x-wiki == Example of Square Root of Complex Number in Cartesian Form ==  :$\sqrt {-15 - 8 i} = \pm \paren {1 - 4 i}$  == Proof 1 == {{:Square Root of Complex Number in Cartesian Form/Examples/-15-8i/Proof 1}} == Proof 2 == {{:Square Root of Complex Number in Cartesian Form/Examples/-15-8i/Proof 2}}
Complex Dot Product/Examples/3-4i dot -4+3i
== Examples of Complex Dot Product ==  Let: :$z_1 = 3 - 4 i$ :$z_2 = -4 + 3 i$ Then: :$z_1 \circ z_2 = -24$ where $\circ$ denotes (complex) dot product.  == Proof 1 == {{:Complex Dot Product/Examples/3-4i dot -4+3i/Proof 1}} == Proof 2 == {{:Complex Dot Product/Examples/3-4i dot -4+3i/Proof 2}}
Condition for Points in Complex Plane to form Isosceles Triangle
Condition for Points in Complex Plane to form Isosceles Triangle 0 75372 395335 2019-03-11T07:44:51Z Prime.mover 59 Created page with "== Theorem ==  Let $A = z_1 = x_1 + i y_1$, $B = z_2 = x_2 + i y_2$ and $C = z_3 = x_3 + i y_3$ represent on the complex plane the ..." wikitext text/x-wiki == Theorem ==  Let $A = z_1 = x_1 + i y_1$, $B = z_2 = x_2 + i y_2$ and $C = z_3 = x_3 + i y_3$ represent on the complex plane the vertices of a triangle. Then $\triangle ABC$ is isosceles, where $A$ is the apex, : :${x_2}^2 + {y_2}^2 - 2 \paren {x_1 x_2 + y_1 y_2} = {x_3}^2 + {y_3}^2 - 2 \paren {x_1 x_3 + y_1 y_3}$  == Proof == By definition of isosceles triangle: :$\triangle ABC$ is isosceles, where $A$ is the apex,  $AB = AC$. Hence: {{begin-eqn}} {{eqn | l = \cmod {z_1 - z_2}       | r = \cmod {z_1 - z_3}       | c =  }} {{eqn | ll= \leadstoandfrom       | l = \cmod {x_1 + i y_1 - x_2 + i y_2}^2       | r = \cmod {x_1 + i y_1 - x_3 + i y_3}^2       | c =  }} {{eqn | ll= \leadstoandfrom       | l = \paren {x_1 - x_2}^2 + \paren {y_1 - y_2}^2       | r = \paren {x_1 - x_3}^2 + \paren {y_1 - y_3}^2       | c = {{Defof|Complex Modulus}} }} {{eqn | ll= \leadstoandfrom       | l = {x_1}^2 - 2 x_1 x_2 + {x_2}^2 + {y_1}^2 - 2 y_1 y_2 + {y_2}^2       | r = {x_1}^2 - 2 x_1 x_3 + {x_3}^2 + {y_1}^2 - 2 y_1 y_3 + {y_3}^2       | c =  }} {{eqn | ll= \leadstoandfrom       | l = {x_2}^2 + {y_2}^2 - 2 \paren {x_1 x_2 + y_1 y_2}       | r = {x_3}^2 + {y_3}^2 - 2 \paren {x_1 x_3 + y_1 y_3}       | c =  }} {{end-eqn}}  == Examples == {{:Condition for Points in Complex Plane to form Isosceles Triangle/Examples}}
Equation of Hyperbola in Complex Plane
Equation of Hyperbola in Complex Plane 0 75404 395426 2019-03-12T06:57:38Z Prime.mover 59 Created page with "== Theorem ==  Let $\C$ be the complex plane.  Let $H$ be a hyperbola in $\C$ whose Definition:Major Axis..." wikitext text/x-wiki == Theorem ==  Let $\C$ be the complex plane. Let $H$ be a hyperbola in $\C$ whose major axis is $d \in \R_{>0}$ and whose foci are at $\alpha, \beta \in \C$. Then $C$ may be written as: :$\cmod {z - \alpha} - \cmod {z - \beta} = d$ where $\cmod {\, \cdot \,}$ denotes complex modulus.  == Proof == By definition of complex modulus: :$\cmod {z - \alpha}$ is the distance from $z$ to $\alpha$ :$\cmod {z - \beta}$ is the distance from $z$ to $\beta$. Thus $\cmod {z - \alpha} - \cmod {z - \beta}$ is the difference of the distance from $z$ to $\alpha$ and from $z$ to $\beta$. This is precisely the equidistance property of the hyperbola. From Equidistance of Hyperbola equals Transverse Axis, the constant distance $d$ is equal to the transverse axis of $H$.  == Examples == {{:Equation of Hyperbola in Complex Plane/Examples}}
Complex Dot Product/Examples/2+5i dot 3-i
== Example of Complex Dot Product ==  Let: :$z_1 = 2 + 5 i$ :$z_2 = 3 - i$ Then: :$z_1 \circ z_2 = 1$ where $\circ$ denotes (complex) dot product.  == Proof == {{begin-eqn}} {{eqn | l = z_1 \circ z_2       | r = \paren {2 + 5 i} \circ \paren {3 - i}       | c =  }} {{eqn | r = 2 \times 3 + 5 \times \paren {-1}       | c = {{Defof|Dot Product|subdef = Complex|index = 1}} }} {{eqn | r = 6 - 5       | c =  }} {{eqn | r = 1       | c =  }} {{end-eqn}}  == Sources == * {{BookReference|Theory and Problems of Complex Variables|1981|Murray R. Spiegel|ed = SI|edpage = SI (Metric) Edition|prev = Complex Algebra/Examples/(1+z)^5 = (1-z)^5|next = Complex Cross Product/Examples/2+5i cross 3-i}}: $1$: Complex Numbers: Supplementary Problems: The Dot and Cross Product: $110 \ \text {(a)}$
Complex Cross Product/Examples/2+5i cross 3-i
Complex Cross Product/Examples/2+5i cross 3-i 0 75562 395984 2019-03-17T11:32:37Z Prime.mover 59 Created page with "== Example of Complex Cross Product ==  Let: :$z_1 = 2 + 5 i$ :$z_2 = 3 - i$ Then: :$z_1 \times z_2 = -17$ where $\times$ den..." wikitext text/x-wiki == Example of Complex Cross Product ==  Let: :$z_1 = 2 + 5 i$ :$z_2 = 3 - i$ Then: :$z_1 \times z_2 = -17$ where $\times$ denotes (complex) cross product.  == Proof == {{begin-eqn}} {{eqn | l = z_1 \circ z_2       | r = \paren {2 + 5 i} \times \paren {3 - i}       | c =  }} {{eqn | r = 2 \times \paren {-1} - 3 \times 5       | c = {{Defof|Vector Cross Product|subdef = Complex|index = 1|Complex Cross Product}} }} {{eqn | r = -2 - 15       | c =  }} {{eqn | r = -17       | c =  }} {{end-eqn}}  == Sources == * {{BookReference|Theory and Problems of Complex Variables|1981|Murray R. Spiegel|ed = SI|edpage = SI (Metric) Edition|prev = Complex Dot Product/Examples/2+5i dot 3-i|next = Complex Dot Product/Examples/3-i dot 2+5i}}: $1$: Complex Numbers: Supplementary Problems: The Dot and Cross Product: $110 \ \text {(b)}$
Complex Dot Product/Examples/3-i dot 2+5i
Complex Dot Product/Examples/3-i dot 2+5i 0 75563 395985 2019-03-17T11:33:50Z Prime.mover 59 Created page with "== Examples of Complex Dot Product ==  Let: :$z_1 = 3 - i$ :$z_1 = 2 + 5 i$ Then: :$z_1 \circ z_2 = 1$ where $\circ$ denotes ..." wikitext text/x-wiki == Examples of Complex Dot Product ==  Let: :$z_1 = 3 - i$ :$z_1 = 2 + 5 i$ Then: :$z_1 \circ z_2 = 1$ where $\circ$ denotes (complex) dot product.  == Proof == {{begin-eqn}} {{eqn | l = z_1 \circ z_2       | r = \paren {3 - i} \circ \paren {2 + 5 i}       | c =  }} {{eqn | r = 3 \times 2 + \paren {-1} \times 5       | c = {{Defof|Dot Product|subdef = Complex|index = 1}} }} {{eqn | r = 6 - 5       | c =  }} {{eqn | r = 1       | c =  }} {{end-eqn}}  == Sources == * {{BookReference|Theory and Problems of Complex Variables|1981|Murray R. Spiegel|ed = SI|edpage = SI (Metric) Edition|prev = Complex Cross Product/Examples/2+5i cross 3-i|next = Complex Cross Product/Examples/3-i cross 2+5i}}: $1$: Complex Numbers: Supplementary Problems: The Dot and Cross Product: $110 \ \text {(c)}$
Complex Cross Product/Examples/3-i cross 2+5i
== Example of Complex Cross Product ==  Let: :$z_1 = 3 - i$ :$z_2 = 2 + 5 i$ Then: :$z_1 \times z_2 = 17$ where $\times$ denotes (complex) cross product.  == Proof == {{begin-eqn}} {{eqn | l = z_1 \circ z_2       | r = \paren {3 - i} \times \paren {2 + 5 i}       | c =  }} {{eqn | r = 3 \times \paren 5 - \paren {-1} \times 2       | c = {{Defof|Vector Cross Product|subdef = Complex|index = 1|Complex Cross Product}} }} {{eqn | r = 15 - \paren {-2}       | c =  }} {{eqn | r = 17       | c =  }} {{end-eqn}}  == Sources == * {{BookReference|Theory and Problems of Complex Variables|1981|Murray R. Spiegel|ed = SI|edpage = SI (Metric) Edition|prev = Complex Dot Product/Examples/3-i dot 2+5i|next = Complex Dot Product/Examples/Size of 2+5i dot 3-i}}: $1$: Complex Numbers: Supplementary Problems: The Dot and Cross Product: $110 \ \text {(d)}$
Property of Group Automorphism which Fixes Identity Only/Corollary 2
== Corollary to Property of Group Automorphism which Fixes Identity Only == Let $G$ be a finite group whose identity is $e$. Let $\phi: G \to G$ be a group automorphism. Let $\phi$ have the property that: :$\forall g \in G \setminus \set e: \map \phi t \ne t$ That is, the only fixed element of $\phi$ is $e$.  Let: :$\phi^2 = I_G$ where $I_G$ denotes the identity mapping on $G$. Then: :$\forall g \in G: \map \phi g = g^{-1}$  == Proof == Let $g \in G$. Then: {{begin-eqn}} {{eqn | lo= \exists x \in G:       | l = \map \phi g       | r = \map \phi {x^{-1} \, \map \phi x}       | c = Corollary 1 }} {{eqn | r = \paren {\map \phi x}^{-1} x       | c =  }} {{eqn | r = g^{-1}       | c =  }} {{end-eqn}}  == Sources == * {{BookReference|An Introduction to Abstract Algebra|1978|Thomas A. Whitelaw|prev = Property of Group Automorphism which Fixes Identity Only/Corollary 1|next = Property of Group Automorphism which Fixes Identity Only/Corollary 3}}: Chapter $8$: Homomorphisms, Normal Subgroups and Quotient Groups: Exercise $26$