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Theorem List for Metamath Proof Explorer - 20101-20200   *Has distinct variable group(s)
TypeLabelDescription
Statement

Theoremcxpcn 20101* Domain of continuity of the complex power function. (Contributed by Mario Carneiro, 1-May-2016.)
fld       t

Theoremcxpcn2 20102* Continuity of the complex power function, when the base is real. (Contributed by Mario Carneiro, 1-May-2016.)
fld       t

Theoremcxpcn3lem 20103* Lemma for cxpcn3 20104. (Contributed by Mario Carneiro, 2-May-2016.)
fld       t        t

Theoremcxpcn3 20104* Extend continuity of the complex power function to a base of zero, as long as the exponent has strictly positive real part. (Contributed by Mario Carneiro, 2-May-2016.)
fld       t        t

Theoremresqrcn 20105 Continuity of the real square root function. (Contributed by Mario Carneiro, 2-May-2016.)

Theoremsqrcn 20106 Continuity of the square root function. (Contributed by Mario Carneiro, 2-May-2016.)

Theoremcxpaddlelem 20107 Lemma for cxpaddle 20108. (Contributed by Mario Carneiro, 2-Aug-2014.)

Theoremcxpaddle 20108 Ordering property for complex exponentiation. (Contributed by Mario Carneiro, 8-Sep-2014.)

Theoremabscxpbnd 20109 Bound on the absolute value of a complex power. (Contributed by Mario Carneiro, 15-Sep-2014.)

Theoremroot1id 20110 Property of an -th root of unity. (Contributed by Mario Carneiro, 23-Apr-2015.)

Theoremroot1eq1 20111 The only powers of an -th root of unity that equal are the multiples of . In other words, has order in the multiplicative group of nonzero complex numbers. (In fact, these and their powers are the only elements of finite order in the complexes.) (Contributed by Mario Carneiro, 28-Apr-2016.)

Theoremroot1cj 20112 Within the -th roots of unity, the conjugate of the -th root is the -th root. (Contributed by Mario Carneiro, 23-Apr-2015.)

Theoremcxpeq 20113* Solve an equation involving an -th power. The expression is a way to write the primitive -th root of unity with the smallest positive argument. (Contributed by Mario Carneiro, 23-Apr-2015.)

Theoremloglesqr 20114 An upper bound on the logarithm. (Contributed by Mario Carneiro, 2-May-2016.)

13.3.5  Theorems of Pythagoras, isosceles triangles, and intersecting chords

Theoremangval 20115* Define the angle function, which takes two complex numbers, treated as vectors from the origin, and returns the angle between them, in the range . To convert from the geometry notation, , the measure of the angle with legs , where is more counterclockwise for positive angles, is represented by . (Contributed by Mario Carneiro, 23-Sep-2014.)

Theoremangcan 20116* Cancel a constant multiplier in the angle function. (Contributed by Mario Carneiro, 23-Sep-2014.)

Theoremangneg 20117* Cancel a negative sign in the angle function. (Contributed by Mario Carneiro, 23-Sep-2014.)

Theoremangvald 20118* The (signed) angle between two vectors is the argument of their quotient. Deduction form of angval 20115. (Contributed by David Moews, 28-Feb-2017.)

Theoremangcld 20119* The (signed) angle between two vectors is in . Deduction form. (Contributed by David Moews, 28-Feb-2017.)

Theoremangrteqvd 20120* Two vectors are at a right angle iff their quotient is purely imaginary. (Contributed by David Moews, 28-Feb-2017.)

Theoremcosangneg2d 20121* The cosine of the angle between and is the negative of that between and . If A, B and C are collinear points, this implies that the cosines of DBA and DBC sum to zero, i.e., that DBA and DBC are supplementary. (Contributed by David Moews, 28-Feb-2017.)

Theoremangrtmuld 20122* Perpendicularity of two vectors does not change under rescaling the second. (Contributed by David Moews, 28-Feb-2017.)

Theoremang180lem1 20123* Lemma for ang180 20128. Show that the "revolution number" is an integer, using efeq1 19907 to show that since the product of the three arguments is , the sum of the logarithms must be an integer multiple of away from . (Contributed by Mario Carneiro, 23-Sep-2014.)

Theoremang180lem2 20124* Lemma for ang180 20128. Show that the revolution number is strictly between and . Both bounds are established by iterating using the bounds on the imaginary part of the logarithm, logimcl 19943, but the resulting bound gives only for the upper bound. The case is not ruled out here, but it is in some sense an "edge case" that can only happen under very specific conditions; in particular we show that all the angle arguments must lie on the negative real axis, which is a contradiction because clearly if is negative then the other two are positive real. (Contributed by Mario Carneiro, 23-Sep-2014.)

Theoremang180lem3 20125* Lemma for ang180 20128. Since ang180lem1 20123 shows that is an integer and ang180lem2 20124 shows that is strictly between and , it follows that , and these two cases correspond to the two possible values for . (Contributed by Mario Carneiro, 23-Sep-2014.)

Theoremang180lem4 20126* Lemma for ang180 20128. Reduce the statement to one variable. (Contributed by Mario Carneiro, 23-Sep-2014.)

Theoremang180lem5 20127* Lemma for ang180 20128: Reduce the statement to two variables. (Contributed by Mario Carneiro, 23-Sep-2014.)

Theoremang180 20128* The sum of angles in a triangle adds up to either or , i.e. 180 degrees. (The sign is due to the two possible orientations of vertex arrangement and our signed notion of angle). (Contributed by Mario Carneiro, 23-Sep-2014.)

Theoremlawcoslem1 20129 Lemma for Law of Cosines lawcos 20130. Here we prove the law for a point at the origin and two distinct points U and V, using an expanded version of the signed angle expression on the complex plane. (Contributed by David A. Wheeler, 11-Jun-2015.)

Theoremlawcos 20130* Law of Cosines. Given three distinct points A, B, and C, prove a relationship between their segment lengths. This theorem is expressed using the complex number plane as a plane, where is the signed angle construct (as used in ang180 20128), is the distance of line segment BC, is the distance of line segment AC, is the distance of line segment AB, and is the distinguished (signed) angle m/_ BCA on the complex plane. We translate triangle ABC to move C to the origin (C-C), B to U=(B-C), and A to V=(A-C), then use lemma lawcoslem1 20129 to prove this algebraically simpler case. The metamath convention is to use a signed angle; in this case the sign doesn't matter because we use the cosine of the angle (see cosneg 12443). The Pythagorean Theorem pythag 20131 is a special case of the law of cosines. The theorem's expression and approach were suggested by Mario Carneiro. (Contributed by David A. Wheeler, 12-Jun-2015.)

Theorempythag 20131* Pythagorean Theorem. Given three distinct points A, B, and C that form a right triangle (with the right angle at C), prove a relationship between their segment lengths. This theorem is expressed using the complex number plane as a plane, where is the signed angle construct (as used in ang180 20128), is the distance of line segment BC, is the distance of line segment AC, is the distance of line segment AB (the hypotenuse), and is the distinguished (signed) right angle m/_ BCA. We use the law of cosines lawcos 20130 to prove this, along with simple trig facts like coshalfpi 19853 and cosneg 12443. (Contributed by David A. Wheeler, 13-Jun-2015.)

Theoremlogreclem 20132 Symmetry of the natural logarithm range by negation. Lemma for logrec 20133. (Contributed by Saveliy Skresanov, 27-Dec-2016.)

Theoremlogrec 20133 Logarithm of a reciprocal changes sign. (Contributed by Saveliy Skresanov, 28-Dec-2016.)

Theoremisosctrlem1 20134 Lemma for isosctr 20137. (Contributed by Saveliy Skresanov, 30-Dec-2016.)

Theoremisosctrlem2 20135 Lemma for isosctr 20137. Corresponds to the case where one vertex is at 0, another at 1 and the third lies on the unit circle. (Contributed by Saveliy Skresanov, 31-Dec-2016.)

Theoremisosctrlem3 20136* Lemma for isosctr 20137. Corresponds to the case where one vertex is at 0. (Contributed by Saveliy Skresanov, 1-Jan-2017.)

Theoremisosctr 20137* Isosceles triangle theorem. (Contributed by Saveliy Skresanov, 1-Jan-2017.)

Theoremssscongptld 20138* If two triangles have equal sides, one angle in one triangle has the same cosine as the corresponding angle in the other triangle. This is a partial form of the SSS congruence theorem.

This theorem is proven by using lawcos 20130 on both triangles to express one side in terms of the other two, and then equating these expressions and reducing this algebraically to get an equality of cosines of angles. (Contributed by David Moews, 28-Feb-2017.)

Theoremaffineequiv 20139 Equivalence between two ways of expressing as an affine combination of and . (Contributed by David Moews, 28-Feb-2017.)

Theoremaffineequiv2 20140 Equivalence between two ways of expressing as an affine combination of and . (Contributed by David Moews, 28-Feb-2017.)

Theoremangpieqvdlem 20141 Equivalence used in the proof of angpieqvd 20144. (Contributed by David Moews, 28-Feb-2017.)

Theoremangpieqvdlem2 20142* Equivalence used in angpieqvd 20144. (Contributed by David Moews, 28-Feb-2017.)

Theoremangpined 20143* If the angle at ABC is , then A is not equal to C. (Contributed by David Moews, 28-Feb-2017.)

Theoremangpieqvd 20144* The angle ABC is iff B is a nontrivial convex combination of A and C, i.e., iff B is in the interior of the segment AC. (Contributed by David Moews, 28-Feb-2017.)

Theoremchordthmlem 20145* If M is the midpoint of AB and AQ = BQ, then QMB is a right angle. The proof uses ssscongptld 20138 to observe that, since AMQ and BMQ have equal sides, the angles QMB and QMA must be equal. Since they are supplementary, both must be right. (Contributed by David Moews, 28-Feb-2017.)

Theoremchordthmlem2 20146* If M is the midpoint of AB, AQ = BQ, and P is on the line AB, then QMP is a right angle. This is proven by reduction to the special case chordthmlem 20145, where P = B, and using angrtmuld 20122 to observe that QMP is right iff QMB is. (Contributed by David Moews, 28-Feb-2017.)

Theoremchordthmlem3 20147 If M is the midpoint of AB, AQ = BQ, and P is on the line AB, then PQ 2 = QM 2 PM 2 . This follows from chordthmlem2 20146 and the Pythagorean theorem (pythag 20131) in the case where P and Q are unequal to M. If either P or Q equals M, the result is trivial. (Contributed by David Moews, 28-Feb-2017.)

Theoremchordthmlem4 20148 If P is on the segment AB and M is the midpoint of AB, then PA PB = BM 2 PM 2 . If all lengths are reexpressed as fractions of AB, this reduces to the identity 2 2 . (Contributed by David Moews, 28-Feb-2017.)

Theoremchordthmlem5 20149 If P is on the segment AB and AQ = BQ, then PA PB = BQ 2 PQ 2 . This follows from two uses of chordthmlem3 20147 to show that PQ 2 = QM 2 PM 2 and BQ 2 = QM 2 BM 2 , so BQ 2 PQ 2 = (QM 2 BM 2 ) (QM 2 PM 2 ) = BM 2 PM 2 , which equals PA PB by chordthmlem4 20148. (Contributed by David Moews, 28-Feb-2017.)

Theoremchordthm 20150* The intersecting chords theorem. If points A, B, C, and D lie on a circle (with center Q, say), and the point P is on the interior of the segments AB and CD, then the two products of lengths PA PB and PC PD are equal. The Euclidean plane is identified with the complex plane, and the fact that P is on AB and on CD is expressed by the hypothesis that the angles APB and CPD are equal to . The result is proven by using chordthmlem5 20149 twice to show that PA PB and PC PD both equal BQ 2 PQ 2 . This is similar to the proof of the theorem given in Euclid's Elements, where it is Proposition III.35. (Contributed by David Moews, 28-Feb-2017.)

13.3.6  Solutions of quadratic, cubic, and quartic equations

Theoremquad2 20151 The quadratic equation, without specifying the particular branch to the square root. (Contributed by Mario Carneiro, 23-Apr-2015.)

Theoremquad 20152 The quadratic equation. (Contributed by Mario Carneiro, 23-Apr-2015.)

Theorem1cubrlem 20153 The cube roots of unity. (Contributed by Mario Carneiro, 23-Apr-2015.)

Theorem1cubr 20154 The cube roots of unity. (Contributed by Mario Carneiro, 23-Apr-2015.)

Theoremdcubic1lem 20155 Lemma for dcubic1 20157 and dcubic2 20156: simplify the cubic equation under the substitution . (Contributed by Mario Carneiro, 26-Apr-2015.)

Theoremdcubic2 20156* Reverse direction of dcubic 20158. Given a solution to the "substitution" quadratic equation , show that is in the desired form. (Contributed by Mario Carneiro, 25-Apr-2015.)

Theoremdcubic1 20157 Forward direction of dcubic 20158: the claimed formula produces solutions to the cubic equation. (Contributed by Mario Carneiro, 25-Apr-2015.)

Theoremdcubic 20158* Solutions to the depressed cubic, a special case of cubic 20161. (The definitions of here differ from mcubic 20159 by scale factors of , , and respectively, to simplify the algebra and presentation.) (Contributed by Mario Carneiro, 26-Apr-2015.)

Theoremmcubic 20159* Solutions to a monic cubic equation, a special case of cubic 20161. (Contributed by Mario Carneiro, 24-Apr-2015.)
;

Theoremcubic2 20160* The solution to the general cubic equation, for arbitrary choices and of the square and cube roots. (Contributed by Mario Carneiro, 23-Apr-2015.)
;

Theoremcubic 20161* The cubic equation, which gives the roots of an arbitrary (nondegenerate) cubic function. Use rextp 3702 to convert the existential quantifier to a triple disjunction. (Contributed by Mario Carneiro, 26-Apr-2015.)
;

Theorembinom4 20162 Work out a quartic binomial. (You would think that by this point it would be faster to use binom 12304, but it turns out to be just as much work to put it into this form after clearing all the sums and calculating binomial coefficients.) (Contributed by Mario Carneiro, 6-May-2015.)

Theoremdquartlem1 20163 Lemma for dquart 20165. (Contributed by Mario Carneiro, 6-May-2015.)

Theoremdquartlem2 20164 Lemma for dquart 20165. (Contributed by Mario Carneiro, 6-May-2015.)

Theoremdquart 20165 Solve a depressed quartic equation. To eliminate , which is the square root of a solution to the resolvent cubic equation, apply cubic 20161 or one of its variants. (Contributed by Mario Carneiro, 6-May-2015.)

Theoremquart1cl 20166 Closure lemmas for quart 20173. (Contributed by Mario Carneiro, 7-May-2015.)
; ;;

Theoremquart1lem 20167 Lemma for quart1 20168. (Contributed by Mario Carneiro, 6-May-2015.)
; ;;                      ;;

Theoremquart1 20168 Depress a quartic equation. (Contributed by Mario Carneiro, 6-May-2015.)
; ;;

Theoremquartlem1 20169 Lemma for quart 20173. (Contributed by Mario Carneiro, 6-May-2015.)
;        ; ;        ;

Theoremquartlem2 20170 Closure lemmas for quart 20173. (Contributed by Mario Carneiro, 7-May-2015.)
; ;;        ;        ; ;

Theoremquartlem3 20171 Closure lemmas for quart 20173. (Contributed by Mario Carneiro, 7-May-2015.)
; ;;        ;        ; ;

Theoremquartlem4 20172 Closure lemmas for quart 20173. (Contributed by Mario Carneiro, 7-May-2015.)
; ;;        ;        ; ;

Theoremquart 20173 The quartic equation, writing out all roots using square and cube root functions so that only direct substitutions remain, and we can actually claim to have a "quartic equation". Naturally, this theorem is ridiculously long (see quartfull 23701) if all the substitutions are performed. (Contributed by Mario Carneiro, 6-May-2015.)
; ;;        ;        ; ;

13.3.7  Inverse trigonometric functions

Syntaxcasin 20174 The arcsine function.
arcsin

Syntaxcacos 20175 The arccosine function.
arccos

Syntaxcatan 20176 The arctangent function.
arctan

Definitiondf-asin 20177 Define the arcsine function. Because is not a one-to-one function, the literal inverse is not a function. Rather than attempt to find the right domain on which to restrict in order to get a total function, we just define it in terms of , which we already know is total (except at ). There are branch points at and (at which the function is defined), and branch cuts along the real line not between and , which is to say . (Contributed by Mario Carneiro, 31-Mar-2015.)
arcsin

Definitiondf-acos 20178 Define the arccosine function. See also remarks for df-asin 20177. Since we define arccos in terms of arcsin, it shares the same branch points and cuts, namely . (Contributed by Mario Carneiro, 31-Mar-2015.)
arccos arcsin

Definitiondf-atan 20179 Define the arctangent function. See also remarks for df-asin 20177. Unlike arcsin and arccos, this function is not defined everywhere, because for all . For all other , there is a formula for arctan in terms of , and we take that as the definition. Branch points are at ; branch cuts are on the pure imaginary axis not between and , which is to say . (Contributed by Mario Carneiro, 31-Mar-2015.)
arctan

Theoremasinlem 20180 The argument to the logarithm in df-asin 20177 is always nonzero. (Contributed by Mario Carneiro, 31-Mar-2015.)

Theoremasinlem2 20181 The argument to the logarithm in df-asin 20177 has the property that replacing with in the expression gives the reciprocal. (Contributed by Mario Carneiro, 1-Apr-2015.)

Theoremasinlem3a 20182 Lemma for asinlem3 20183. (Contributed by Mario Carneiro, 1-Apr-2015.)

Theoremasinlem3 20183 The argument to the logarithm in df-asin 20177 has nonnegative real part. (Contributed by Mario Carneiro, 1-Apr-2015.)

Theoremasinf 20184 Domain and range of the arcsin function. (Contributed by Mario Carneiro, 31-Mar-2015.)
arcsin

Theoremasincl 20185 Closure for the arcsin function. (Contributed by Mario Carneiro, 31-Mar-2015.)
arcsin

Theoremacosf 20186 Domain and range of the arccos function. (Contributed by Mario Carneiro, 31-Mar-2015.)
arccos

Theoremacoscl 20187 Closure for the arccos function. (Contributed by Mario Carneiro, 31-Mar-2015.)
arccos

Theorematandm 20188 Since the property is a little lengthy, we abbreviate as arctan. This is the necessary precondition for the definition of arctan to make sense. (Contributed by Mario Carneiro, 31-Mar-2015.)
arctan

Theorematandm2 20189 This form of atandm 20188 is a bit more useful for showing that the logarithms in df-atan 20179 are well-defined. (Contributed by Mario Carneiro, 31-Mar-2015.)
arctan

Theorematandm3 20190 A compact form of atandm 20188. (Contributed by Mario Carneiro, 31-Mar-2015.)
arctan

Theorematandm4 20191 A compact form of atandm 20188. (Contributed by Mario Carneiro, 3-Apr-2015.)
arctan

Theorematanf 20192 Domain and range of the arctan function. (Contributed by Mario Carneiro, 31-Mar-2015.)
arctan

Theorematancl 20193 Closure for the arctan function. (Contributed by Mario Carneiro, 31-Mar-2015.)
arctan arctan

Theoremasinval 20194 Value of the arcsin function. (Contributed by Mario Carneiro, 31-Mar-2015.)
arcsin

Theoremacosval 20195 Value of the arccos function. (Contributed by Mario Carneiro, 31-Mar-2015.)
arccos arcsin

Theorematanval 20196 Value of the arctan function. (Contributed by Mario Carneiro, 31-Mar-2015.)
arctan arctan

Theorematanre 20197 A real number is in the domain of the arctangent function. (Contributed by Mario Carneiro, 31-Mar-2015.)
arctan

Theoremasinneg 20198 The arcsine function is odd. (Contributed by Mario Carneiro, 1-Apr-2015.)
arcsin arcsin

Theoremacosneg 20199 The negative symmetry relation of the arccosine. (Contributed by Mario Carneiro, 2-Apr-2015.)
arccos arccos

Theoremefiasin 20200 The exponential of the arcsine function. (Contributed by Mario Carneiro, 31-Mar-2015.)
arcsin

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