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Type | Label | Description |
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Statement | ||
Theorem | logbrec 26701 | Logarithm of a reciprocal changes sign. See logrec 26682. Particular case of Property 3 of [Cohen4] p. 361. (Contributed by Thierry Arnoux, 27-Sep-2017.) |
β’ ((π΅ β (β€β₯β2) β§ π΄ β β+) β (π΅ logb (1 / π΄)) = -(π΅ logb π΄)) | ||
Theorem | logbleb 26702 | The general logarithm function is monotone/increasing. See logleb 26524. (Contributed by Stefan O'Rear, 19-Oct-2014.) (Revised by AV, 31-May-2020.) |
β’ ((π΅ β (β€β₯β2) β§ π β β+ β§ π β β+) β (π β€ π β (π΅ logb π) β€ (π΅ logb π))) | ||
Theorem | logblt 26703 | The general logarithm function is strictly monotone/increasing. Property 2 of [Cohen4] p. 377. See logltb 26521. (Contributed by Stefan O'Rear, 19-Oct-2014.) (Revised by Thierry Arnoux, 27-Sep-2017.) |
β’ ((π΅ β (β€β₯β2) β§ π β β+ β§ π β β+) β (π < π β (π΅ logb π) < (π΅ logb π))) | ||
Theorem | relogbcxp 26704 | Identity law for the general logarithm for real numbers. (Contributed by AV, 22-May-2020.) |
β’ ((π΅ β (β+ β {1}) β§ π β β) β (π΅ logb (π΅βππ)) = π) | ||
Theorem | cxplogb 26705 | Identity law for the general logarithm. (Contributed by AV, 22-May-2020.) |
β’ ((π΅ β (β β {0, 1}) β§ π β (β β {0})) β (π΅βπ(π΅ logb π)) = π) | ||
Theorem | relogbcxpb 26706 | The logarithm is the inverse of the exponentiation. Observation in [Cohen4] p. 348. (Contributed by AV, 11-Jun-2020.) |
β’ (((π΅ β β+ β§ π΅ β 1) β§ π β β+ β§ π β β) β ((π΅ logb π) = π β (π΅βππ) = π)) | ||
Theorem | logbmpt 26707* | The general logarithm to a fixed base regarded as mapping. (Contributed by AV, 11-Jun-2020.) |
β’ ((π΅ β β β§ π΅ β 0 β§ π΅ β 1) β (curry logb βπ΅) = (π¦ β (β β {0}) β¦ ((logβπ¦) / (logβπ΅)))) | ||
Theorem | logbf 26708 | The general logarithm to a fixed base regarded as function. (Contributed by AV, 11-Jun-2020.) |
β’ ((π΅ β β β§ π΅ β 0 β§ π΅ β 1) β (curry logb βπ΅):(β β {0})βΆβ) | ||
Theorem | logbfval 26709 | The general logarithm of a complex number to a fixed base. (Contributed by AV, 11-Jun-2020.) |
β’ (((π΅ β β β§ π΅ β 0 β§ π΅ β 1) β§ π β (β β {0})) β ((curry logb βπ΅)βπ) = (π΅ logb π)) | ||
Theorem | relogbf 26710 | The general logarithm to a real base greater than 1 regarded as function restricted to the positive integers. Property in [Cohen4] p. 349. (Contributed by AV, 12-Jun-2020.) |
β’ ((π΅ β β+ β§ 1 < π΅) β ((curry logb βπ΅) βΎ β+):β+βΆβ) | ||
Theorem | logblog 26711 | The general logarithm to the base being Euler's constant regarded as function is the natural logarithm. (Contributed by AV, 12-Jun-2020.) |
β’ (curry logb βe) = log | ||
Theorem | logbgt0b 26712 | The logarithm of a positive real number to a real base greater than 1 is positive iff the number is greater than 1. (Contributed by AV, 29-Dec-2022.) |
β’ ((π΄ β β+ β§ (π΅ β β+ β§ 1 < π΅)) β (0 < (π΅ logb π΄) β 1 < π΄)) | ||
Theorem | logbgcd1irr 26713 | The logarithm of an integer greater than 1 to an integer base greater than 1 is an irrational number if the argument and the base are relatively prime. For example, (2 logb 9) β (β β β) (see 2logb9irr 26714). (Contributed by AV, 29-Dec-2022.) |
β’ ((π β (β€β₯β2) β§ π΅ β (β€β₯β2) β§ (π gcd π΅) = 1) β (π΅ logb π) β (β β β)) | ||
Theorem | 2logb9irr 26714 | Example for logbgcd1irr 26713. The logarithm of nine to base two is irrational. (Contributed by AV, 29-Dec-2022.) |
β’ (2 logb 9) β (β β β) | ||
Theorem | logbprmirr 26715 | The logarithm of a prime to a different prime base is an irrational number. For example, (2 logb 3) β (β β β) (see 2logb3irr 26716). (Contributed by AV, 31-Dec-2022.) |
β’ ((π β β β§ π΅ β β β§ π β π΅) β (π΅ logb π) β (β β β)) | ||
Theorem | 2logb3irr 26716 | Example for logbprmirr 26715. The logarithm of three to base two is irrational. (Contributed by AV, 31-Dec-2022.) |
β’ (2 logb 3) β (β β β) | ||
Theorem | 2logb9irrALT 26717 | Alternate proof of 2logb9irr 26714: The logarithm of nine to base two is irrational. (Contributed by AV, 31-Dec-2022.) (Proof modification is discouraged.) (New usage is discouraged.) |
β’ (2 logb 9) β (β β β) | ||
Theorem | sqrt2cxp2logb9e3 26718 | The square root of two to the power of the logarithm of nine to base two is three. (ββ2) and (2 logb 9) are irrational numbers (see sqrt2irr0 16219 resp. 2logb9irr 26714), satisfying the statement in 2irrexpqALT 26719. (Contributed by AV, 29-Dec-2022.) |
β’ ((ββ2)βπ(2 logb 9)) = 3 | ||
Theorem | 2irrexpqALT 26719* | Alternate proof of 2irrexpq 26652: There exist irrational numbers π and π such that (πβπ) is rational. Statement in the Metamath book, section 1.1.5, footnote 27 on page 17, and the "constructive proof" for theorem 1.2 of [Bauer], p. 483. In contrast to 2irrexpq 26652, this is a constructive proof because it is based on two explicitly named irrational numbers (ββ2) and (2 logb 9), see sqrt2irr0 16219, 2logb9irr 26714 and sqrt2cxp2logb9e3 26718. Therefore, this proof is also acceptable/usable in intuitionistic logic. (Contributed by AV, 23-Dec-2022.) (New usage is discouraged.) (Proof modification is discouraged.) |
β’ βπ β (β β β)βπ β (β β β)(πβππ) β β | ||
Theorem | angval 26720* | Define the angle function, which takes two complex numbers, treated as vectors from the origin, and returns the angle between them, in the range ( β Ο, Ο]. To convert from the geometry notation, ππ΄π΅πΆ, the measure of the angle with legs π΄π΅, πΆπ΅ where πΆ is more counterclockwise for positive angles, is represented by ((πΆ β π΅)πΉ(π΄ β π΅)). (Contributed by Mario Carneiro, 23-Sep-2014.) |
β’ πΉ = (π₯ β (β β {0}), π¦ β (β β {0}) β¦ (ββ(logβ(π¦ / π₯)))) β β’ (((π΄ β β β§ π΄ β 0) β§ (π΅ β β β§ π΅ β 0)) β (π΄πΉπ΅) = (ββ(logβ(π΅ / π΄)))) | ||
Theorem | angcan 26721* | Cancel a constant multiplier in the angle function. (Contributed by Mario Carneiro, 23-Sep-2014.) |
β’ πΉ = (π₯ β (β β {0}), π¦ β (β β {0}) β¦ (ββ(logβ(π¦ / π₯)))) β β’ (((π΄ β β β§ π΄ β 0) β§ (π΅ β β β§ π΅ β 0) β§ (πΆ β β β§ πΆ β 0)) β ((πΆ Β· π΄)πΉ(πΆ Β· π΅)) = (π΄πΉπ΅)) | ||
Theorem | angneg 26722* | Cancel a negative sign in the angle function. (Contributed by Mario Carneiro, 23-Sep-2014.) |
β’ πΉ = (π₯ β (β β {0}), π¦ β (β β {0}) β¦ (ββ(logβ(π¦ / π₯)))) β β’ (((π΄ β β β§ π΄ β 0) β§ (π΅ β β β§ π΅ β 0)) β (-π΄πΉ-π΅) = (π΄πΉπ΅)) | ||
Theorem | angvald 26723* | The (signed) angle between two vectors is the argument of their quotient. Deduction form of angval 26720. (Contributed by David Moews, 28-Feb-2017.) |
β’ πΉ = (π₯ β (β β {0}), π¦ β (β β {0}) β¦ (ββ(logβ(π¦ / π₯)))) & β’ (π β π β β) & β’ (π β π β 0) & β’ (π β π β β) & β’ (π β π β 0) β β’ (π β (ππΉπ) = (ββ(logβ(π / π)))) | ||
Theorem | angcld 26724* | The (signed) angle between two vectors is in (-Ο(,]Ο). Deduction form. (Contributed by David Moews, 28-Feb-2017.) |
β’ πΉ = (π₯ β (β β {0}), π¦ β (β β {0}) β¦ (ββ(logβ(π¦ / π₯)))) & β’ (π β π β β) & β’ (π β π β 0) & β’ (π β π β β) & β’ (π β π β 0) β β’ (π β (ππΉπ) β (-Ο(,]Ο)) | ||
Theorem | angrteqvd 26725* | Two vectors are at a right angle iff their quotient is purely imaginary. (Contributed by David Moews, 28-Feb-2017.) |
β’ πΉ = (π₯ β (β β {0}), π¦ β (β β {0}) β¦ (ββ(logβ(π¦ / π₯)))) & β’ (π β π β β) & β’ (π β π β 0) & β’ (π β π β β) & β’ (π β π β 0) β β’ (π β ((ππΉπ) β {(Ο / 2), -(Ο / 2)} β (ββ(π / π)) = 0)) | ||
Theorem | cosangneg2d 26726* | The cosine of the angle between π and -π is the negative of that between π and π. 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.) |
β’ πΉ = (π₯ β (β β {0}), π¦ β (β β {0}) β¦ (ββ(logβ(π¦ / π₯)))) & β’ (π β π β β) & β’ (π β π β 0) & β’ (π β π β β) & β’ (π β π β 0) β β’ (π β (cosβ(ππΉ-π)) = -(cosβ(ππΉπ))) | ||
Theorem | angrtmuld 26727* | Perpendicularity of two vectors does not change under rescaling the second. (Contributed by David Moews, 28-Feb-2017.) |
β’ πΉ = (π₯ β (β β {0}), π¦ β (β β {0}) β¦ (ββ(logβ(π¦ / π₯)))) & β’ (π β π β β) & β’ (π β π β β) & β’ (π β π β β) & β’ (π β π β 0) & β’ (π β π β 0) & β’ (π β π β 0) & β’ (π β (π / π) β β) β β’ (π β ((ππΉπ) β {(Ο / 2), -(Ο / 2)} β (ππΉπ) β {(Ο / 2), -(Ο / 2)})) | ||
Theorem | ang180lem1 26728* | Lemma for ang180 26733. Show that the "revolution number" π is an integer, using efeq1 26449 to show that since the product of the three arguments π΄, 1 / (1 β π΄), (π΄ β 1) / π΄ is -1, the sum of the logarithms must be an integer multiple of 2Οi away from Οi = log(-1). (Contributed by Mario Carneiro, 23-Sep-2014.) |
β’ πΉ = (π₯ β (β β {0}), π¦ β (β β {0}) β¦ (ββ(logβ(π¦ / π₯)))) & β’ π = (((logβ(1 / (1 β π΄))) + (logβ((π΄ β 1) / π΄))) + (logβπ΄)) & β’ π = (((π / i) / (2 Β· Ο)) β (1 / 2)) β β’ ((π΄ β β β§ π΄ β 0 β§ π΄ β 1) β (π β β€ β§ (π / i) β β)) | ||
Theorem | ang180lem2 26729* | Lemma for ang180 26733. Show that the revolution number π is strictly between -2 and 1. Both bounds are established by iterating using the bounds on the imaginary part of the logarithm, logimcl 26490, but the resulting bound gives only π β€ 1 for the upper bound. The case π = 1 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 π΄, 1 / (1 β π΄), (π΄ β 1) / π΄ must lie on the negative real axis, which is a contradiction because clearly if π΄ is negative then the other two are positive real. (Contributed by Mario Carneiro, 23-Sep-2014.) |
β’ πΉ = (π₯ β (β β {0}), π¦ β (β β {0}) β¦ (ββ(logβ(π¦ / π₯)))) & β’ π = (((logβ(1 / (1 β π΄))) + (logβ((π΄ β 1) / π΄))) + (logβπ΄)) & β’ π = (((π / i) / (2 Β· Ο)) β (1 / 2)) β β’ ((π΄ β β β§ π΄ β 0 β§ π΄ β 1) β (-2 < π β§ π < 1)) | ||
Theorem | ang180lem3 26730* | Lemma for ang180 26733. Since ang180lem1 26728 shows that π is an integer and ang180lem2 26729 shows that π is strictly between -2 and 1, it follows that π β {-1, 0}, and these two cases correspond to the two possible values for π. (Contributed by Mario Carneiro, 23-Sep-2014.) |
β’ πΉ = (π₯ β (β β {0}), π¦ β (β β {0}) β¦ (ββ(logβ(π¦ / π₯)))) & β’ π = (((logβ(1 / (1 β π΄))) + (logβ((π΄ β 1) / π΄))) + (logβπ΄)) & β’ π = (((π / i) / (2 Β· Ο)) β (1 / 2)) β β’ ((π΄ β β β§ π΄ β 0 β§ π΄ β 1) β π β {-(i Β· Ο), (i Β· Ο)}) | ||
Theorem | ang180lem4 26731* | Lemma for ang180 26733. Reduce the statement to one variable. (Contributed by Mario Carneiro, 23-Sep-2014.) |
β’ πΉ = (π₯ β (β β {0}), π¦ β (β β {0}) β¦ (ββ(logβ(π¦ / π₯)))) β β’ ((π΄ β β β§ π΄ β 0 β§ π΄ β 1) β ((((1 β π΄)πΉ1) + (π΄πΉ(π΄ β 1))) + (1πΉπ΄)) β {-Ο, Ο}) | ||
Theorem | ang180lem5 26732* | Lemma for ang180 26733: Reduce the statement to two variables. (Contributed by Mario Carneiro, 23-Sep-2014.) |
β’ πΉ = (π₯ β (β β {0}), π¦ β (β β {0}) β¦ (ββ(logβ(π¦ / π₯)))) β β’ (((π΄ β β β§ π΄ β 0) β§ (π΅ β β β§ π΅ β 0) β§ π΄ β π΅) β ((((π΄ β π΅)πΉπ΄) + (π΅πΉ(π΅ β π΄))) + (π΄πΉπ΅)) β {-Ο, Ο}) | ||
Theorem | ang180 26733* | The sum of angles ππ΄π΅πΆ + ππ΅πΆπ΄ + ππΆπ΄π΅ in a triangle adds up to either Ο or -Ο, i.e. 180 degrees. (The sign is due to the two possible orientations of vertex arrangement and our signed notion of angle). This is Metamath 100 proof #27. (Contributed by Mario Carneiro, 23-Sep-2014.) |
β’ πΉ = (π₯ β (β β {0}), π¦ β (β β {0}) β¦ (ββ(logβ(π¦ / π₯)))) β β’ (((π΄ β β β§ π΅ β β β§ πΆ β β) β§ (π΄ β π΅ β§ π΅ β πΆ β§ π΄ β πΆ)) β ((((πΆ β π΅)πΉ(π΄ β π΅)) + ((π΄ β πΆ)πΉ(π΅ β πΆ))) + ((π΅ β π΄)πΉ(πΆ β π΄))) β {-Ο, Ο}) | ||
Theorem | lawcoslem1 26734 | Lemma for lawcos 26735. 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.) |
β’ (π β π β β) & β’ (π β π β β) & β’ (π β π β 0) & β’ (π β π β 0) β β’ (π β ((absβ(π β π))β2) = ((((absβπ)β2) + ((absβπ)β2)) β (2 Β· (((absβπ) Β· (absβπ)) Β· ((ββ(π / π)) / (absβ(π / π))))))) | ||
Theorem | lawcos 26735* | Law of cosines (also known as the Al-Kashi theorem or the generalized Pythagorean theorem, or the cosine formula or cosine rule). 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 πΉ is the signed angle construct (as used in ang180 26733), π is the distance of line segment BC, π is the distance of line segment AC, π is the distance of line segment AB, and π is the 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 26734 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 16115). The Pythagorean theorem pythag 26736 is a special case of the law of cosines. The theorem's expression and approach were suggested by Mario Carneiro. This is Metamath 100 proof #94. (Contributed by David A. Wheeler, 12-Jun-2015.) |
β’ πΉ = (π₯ β (β β {0}), π¦ β (β β {0}) β¦ (ββ(logβ(π¦ / π₯)))) & β’ π = (absβ(π΅ β πΆ)) & β’ π = (absβ(π΄ β πΆ)) & β’ π = (absβ(π΄ β π΅)) & β’ π = ((π΅ β πΆ)πΉ(π΄ β πΆ)) β β’ (((π΄ β β β§ π΅ β β β§ πΆ β β) β§ (π΄ β πΆ β§ π΅ β πΆ)) β (πβ2) = (((πβ2) + (πβ2)) β (2 Β· ((π Β· π) Β· (cosβπ))))) | ||
Theorem | pythag 26736* | 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 πΉ is the signed angle construct (as used in ang180 26733), π is the distance of line segment BC, π is the distance of line segment AC, π is the distance of line segment AB (the hypotenuse), and π is the signed right angle m/_ BCA. We use the law of cosines lawcos 26735 to prove this, along with simple trigonometry facts like coshalfpi 26391 and cosneg 16115. (Contributed by David A. Wheeler, 13-Jun-2015.) |
β’ πΉ = (π₯ β (β β {0}), π¦ β (β β {0}) β¦ (ββ(logβ(π¦ / π₯)))) & β’ π = (absβ(π΅ β πΆ)) & β’ π = (absβ(π΄ β πΆ)) & β’ π = (absβ(π΄ β π΅)) & β’ π = ((π΅ β πΆ)πΉ(π΄ β πΆ)) β β’ (((π΄ β β β§ π΅ β β β§ πΆ β β) β§ (π΄ β πΆ β§ π΅ β πΆ) β§ π β {(Ο / 2), -(Ο / 2)}) β (πβ2) = ((πβ2) + (πβ2))) | ||
Theorem | isosctrlem1 26737 | Lemma for isosctr 26740. (Contributed by Saveliy Skresanov, 30-Dec-2016.) |
β’ ((π΄ β β β§ (absβπ΄) = 1 β§ Β¬ 1 = π΄) β (ββ(logβ(1 β π΄))) β Ο) | ||
Theorem | isosctrlem2 26738 | Lemma for isosctr 26740. 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.) |
β’ ((π΄ β β β§ (absβπ΄) = 1 β§ Β¬ 1 = π΄) β (ββ(logβ(1 β π΄))) = (ββ(logβ(-π΄ / (1 β π΄))))) | ||
Theorem | isosctrlem3 26739* | Lemma for isosctr 26740. Corresponds to the case where one vertex is at 0. (Contributed by Saveliy Skresanov, 1-Jan-2017.) |
β’ πΉ = (π₯ β (β β {0}), π¦ β (β β {0}) β¦ (ββ(logβ(π¦ / π₯)))) β β’ (((π΄ β β β§ π΅ β β) β§ (π΄ β 0 β§ π΅ β 0 β§ π΄ β π΅) β§ (absβπ΄) = (absβπ΅)) β (-π΄πΉ(π΅ β π΄)) = ((π΄ β π΅)πΉ-π΅)) | ||
Theorem | isosctr 26740* | Isosceles triangle theorem. This is Metamath 100 proof #65. (Contributed by Saveliy Skresanov, 1-Jan-2017.) |
β’ πΉ = (π₯ β (β β {0}), π¦ β (β β {0}) β¦ (ββ(logβ(π¦ / π₯)))) β β’ (((π΄ β β β§ π΅ β β β§ πΆ β β) β§ (π΄ β πΆ β§ π΅ β πΆ β§ π΄ β π΅) β§ (absβ(π΄ β πΆ)) = (absβ(π΅ β πΆ))) β ((πΆ β π΄)πΉ(π΅ β π΄)) = ((π΄ β π΅)πΉ(πΆ β π΅))) | ||
Theorem | ssscongptld 26741* |
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 26735 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.) |
β’ πΉ = (π₯ β (β β {0}), π¦ β (β β {0}) β¦ (ββ(logβ(π¦ / π₯)))) & β’ (π β π΄ β β) & β’ (π β π΅ β β) & β’ (π β πΆ β β) & β’ (π β π· β β) & β’ (π β πΈ β β) & β’ (π β πΊ β β) & β’ (π β π΄ β π΅) & β’ (π β π΅ β πΆ) & β’ (π β π· β πΈ) & β’ (π β πΈ β πΊ) & β’ (π β (absβ(π΄ β π΅)) = (absβ(π· β πΈ))) & β’ (π β (absβ(π΅ β πΆ)) = (absβ(πΈ β πΊ))) & β’ (π β (absβ(πΆ β π΄)) = (absβ(πΊ β π·))) β β’ (π β (cosβ((π΄ β π΅)πΉ(πΆ β π΅))) = (cosβ((π· β πΈ)πΉ(πΊ β πΈ)))) | ||
Theorem | affineequiv 26742 | Equivalence between two ways of expressing π΅ as an affine combination of π΄ and πΆ. (Contributed by David Moews, 28-Feb-2017.) |
β’ (π β π΄ β β) & β’ (π β π΅ β β) & β’ (π β πΆ β β) & β’ (π β π· β β) β β’ (π β (π΅ = ((π· Β· π΄) + ((1 β π·) Β· πΆ)) β (πΆ β π΅) = (π· Β· (πΆ β π΄)))) | ||
Theorem | affineequiv2 26743 | Equivalence between two ways of expressing π΅ as an affine combination of π΄ and πΆ. (Contributed by David Moews, 28-Feb-2017.) |
β’ (π β π΄ β β) & β’ (π β π΅ β β) & β’ (π β πΆ β β) & β’ (π β π· β β) β β’ (π β (π΅ = ((π· Β· π΄) + ((1 β π·) Β· πΆ)) β (π΅ β π΄) = ((1 β π·) Β· (πΆ β π΄)))) | ||
Theorem | affineequiv3 26744 | Equivalence between two ways of expressing π΄ as an affine combination of π΅ and πΆ. (Contributed by AV, 22-Jan-2023.) |
β’ (π β π΄ β β) & β’ (π β π΅ β β) & β’ (π β πΆ β β) & β’ (π β π· β β) β β’ (π β (π΄ = (((1 β π·) Β· π΅) + (π· Β· πΆ)) β (π΄ β π΅) = (π· Β· (πΆ β π΅)))) | ||
Theorem | affineequiv4 26745 | Equivalence between two ways of expressing π΄ as an affine combination of π΅ and πΆ. (Contributed by AV, 22-Jan-2023.) |
β’ (π β π΄ β β) & β’ (π β π΅ β β) & β’ (π β πΆ β β) & β’ (π β π· β β) β β’ (π β (π΄ = (((1 β π·) Β· π΅) + (π· Β· πΆ)) β π΄ = ((π· Β· (πΆ β π΅)) + π΅))) | ||
Theorem | affineequivne 26746 | Equivalence between two ways of expressing π΄ as an affine combination of π΅ and πΆ if π΅ and πΆ are not equal. (Contributed by AV, 22-Jan-2023.) |
β’ (π β π΄ β β) & β’ (π β π΅ β β) & β’ (π β πΆ β β) & β’ (π β π· β β) & β’ (π β π΅ β πΆ) β β’ (π β (π΄ = (((1 β π·) Β· π΅) + (π· Β· πΆ)) β π· = ((π΄ β π΅) / (πΆ β π΅)))) | ||
Theorem | angpieqvdlem 26747 | Equivalence used in the proof of angpieqvd 26750. (Contributed by David Moews, 28-Feb-2017.) |
β’ (π β π΄ β β) & β’ (π β π΅ β β) & β’ (π β πΆ β β) & β’ (π β π΄ β π΅) & β’ (π β π΄ β πΆ) β β’ (π β (-((πΆ β π΅) / (π΄ β π΅)) β β+ β ((πΆ β π΅) / (πΆ β π΄)) β (0(,)1))) | ||
Theorem | angpieqvdlem2 26748* | Equivalence used in angpieqvd 26750. (Contributed by David Moews, 28-Feb-2017.) |
β’ πΉ = (π₯ β (β β {0}), π¦ β (β β {0}) β¦ (ββ(logβ(π¦ / π₯)))) & β’ (π β π΄ β β) & β’ (π β π΅ β β) & β’ (π β πΆ β β) & β’ (π β π΄ β π΅) & β’ (π β π΅ β πΆ) β β’ (π β (-((πΆ β π΅) / (π΄ β π΅)) β β+ β ((π΄ β π΅)πΉ(πΆ β π΅)) = Ο)) | ||
Theorem | angpined 26749* | If the angle at ABC is Ο, then π΄ is not equal to πΆ. (Contributed by David Moews, 28-Feb-2017.) |
β’ πΉ = (π₯ β (β β {0}), π¦ β (β β {0}) β¦ (ββ(logβ(π¦ / π₯)))) & β’ (π β π΄ β β) & β’ (π β π΅ β β) & β’ (π β πΆ β β) & β’ (π β π΄ β π΅) & β’ (π β π΅ β πΆ) β β’ (π β (((π΄ β π΅)πΉ(πΆ β π΅)) = Ο β π΄ β πΆ)) | ||
Theorem | angpieqvd 26750* | The angle ABC is Ο iff π΅ is a nontrivial convex combination of π΄ and πΆ, i.e., iff π΅ is in the interior of the segment AC. (Contributed by David Moews, 28-Feb-2017.) |
β’ πΉ = (π₯ β (β β {0}), π¦ β (β β {0}) β¦ (ββ(logβ(π¦ / π₯)))) & β’ (π β π΄ β β) & β’ (π β π΅ β β) & β’ (π β πΆ β β) & β’ (π β π΄ β π΅) & β’ (π β π΅ β πΆ) β β’ (π β (((π΄ β π΅)πΉ(πΆ β π΅)) = Ο β βπ€ β (0(,)1)π΅ = ((π€ Β· π΄) + ((1 β π€) Β· πΆ)))) | ||
Theorem | chordthmlem 26751* | If π is the midpoint of AB and AQ = BQ, then QMB is a right angle. The proof uses ssscongptld 26741 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 angles. (Contributed by David Moews, 28-Feb-2017.) |
β’ πΉ = (π₯ β (β β {0}), π¦ β (β β {0}) β¦ (ββ(logβ(π¦ / π₯)))) & β’ (π β π΄ β β) & β’ (π β π΅ β β) & β’ (π β π β β) & β’ (π β π = ((π΄ + π΅) / 2)) & β’ (π β (absβ(π΄ β π)) = (absβ(π΅ β π))) & β’ (π β π΄ β π΅) & β’ (π β π β π) β β’ (π β ((π β π)πΉ(π΅ β π)) β {(Ο / 2), -(Ο / 2)}) | ||
Theorem | chordthmlem2 26752* | 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 26751, where P = B, and using angrtmuld 26727 to observe that QMP is right iff QMB is. (Contributed by David Moews, 28-Feb-2017.) |
β’ πΉ = (π₯ β (β β {0}), π¦ β (β β {0}) β¦ (ββ(logβ(π¦ / π₯)))) & β’ (π β π΄ β β) & β’ (π β π΅ β β) & β’ (π β π β β) & β’ (π β π β β) & β’ (π β π = ((π΄ + π΅) / 2)) & β’ (π β π = ((π Β· π΄) + ((1 β π) Β· π΅))) & β’ (π β (absβ(π΄ β π)) = (absβ(π΅ β π))) & β’ (π β π β π) & β’ (π β π β π) β β’ (π β ((π β π)πΉ(π β π)) β {(Ο / 2), -(Ο / 2)}) | ||
Theorem | chordthmlem3 26753 | 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 26752 and the Pythagorean theorem (pythag 26736) 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.) |
β’ (π β π΄ β β) & β’ (π β π΅ β β) & β’ (π β π β β) & β’ (π β π β β) & β’ (π β π = ((π΄ + π΅) / 2)) & β’ (π β π = ((π Β· π΄) + ((1 β π) Β· π΅))) & β’ (π β (absβ(π΄ β π)) = (absβ(π΅ β π))) β β’ (π β ((absβ(π β π))β2) = (((absβ(π β π))β2) + ((absβ(π β π))β2))) | ||
Theorem | chordthmlem4 26754 | If P is on the segment AB and M is the midpoint of AB, then PA Β· PB = BM 2 β PM 2 . If all lengths are reexpressed as fractions of AB, this reduces to the identity π Β· (1 β π) = (1 / 2) 2 β ((1 / 2) β π) 2 . (Contributed by David Moews, 28-Feb-2017.) |
β’ (π β π΄ β β) & β’ (π β π΅ β β) & β’ (π β π β (0[,]1)) & β’ (π β π = ((π΄ + π΅) / 2)) & β’ (π β π = ((π Β· π΄) + ((1 β π) Β· π΅))) β β’ (π β ((absβ(π β π΄)) Β· (absβ(π β π΅))) = (((absβ(π΅ β π))β2) β ((absβ(π β π))β2))) | ||
Theorem | chordthmlem5 26755 | If P is on the segment AB and AQ = BQ, then PA Β· PB = BQ 2 β PQ 2 . This follows from two uses of chordthmlem3 26753 to show that PQ 2 = QM 2 + PM 2 and BQ 2 = QM 2 + BM 2 , so BQ 2 β PQ 2 = (QM 2 + BM 2 ) β (QM 2 + PM 2 ) = BM 2 β PM 2 , which equals PA Β· PB by chordthmlem4 26754. (Contributed by David Moews, 28-Feb-2017.) |
β’ (π β π΄ β β) & β’ (π β π΅ β β) & β’ (π β π β β) & β’ (π β π β (0[,]1)) & β’ (π β π = ((π Β· π΄) + ((1 β π) Β· π΅))) & β’ (π β (absβ(π΄ β π)) = (absβ(π΅ β π))) β β’ (π β ((absβ(π β π΄)) Β· (absβ(π β π΅))) = (((absβ(π΅ β π))β2) β ((absβ(π β π))β2))) | ||
Theorem | chordthm 26756* | 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 Β· PB and PC Β· 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 Ο. The result is proven by using chordthmlem5 26755 twice to show that PA Β· PB and PC Β· PD both equal BQ 2 β PQ 2 . This is similar to the proof of the theorem given in Euclid's Elements, where it is Proposition III.35. This is Metamath 100 proof #55. (Contributed by David Moews, 28-Feb-2017.) |
β’ πΉ = (π₯ β (β β {0}), π¦ β (β β {0}) β¦ (ββ(logβ(π¦ / π₯)))) & β’ (π β π΄ β β) & β’ (π β π΅ β β) & β’ (π β πΆ β β) & β’ (π β π· β β) & β’ (π β π β β) & β’ (π β π΄ β π) & β’ (π β π΅ β π) & β’ (π β πΆ β π) & β’ (π β π· β π) & β’ (π β ((π΄ β π)πΉ(π΅ β π)) = Ο) & β’ (π β ((πΆ β π)πΉ(π· β π)) = Ο) & β’ (π β π β β) & β’ (π β (absβ(π΄ β π)) = (absβ(π΅ β π))) & β’ (π β (absβ(π΄ β π)) = (absβ(πΆ β π))) & β’ (π β (absβ(π΄ β π)) = (absβ(π· β π))) β β’ (π β ((absβ(π β π΄)) Β· (absβ(π β π΅))) = ((absβ(π β πΆ)) Β· (absβ(π β π·)))) | ||
Theorem | heron 26757* | Heron's formula gives the area of a triangle given only the side lengths. If points A, B, C form a triangle, then the area of the triangle, represented here as (1 / 2) Β· π Β· π Β· abs(sinπ), is equal to the square root of π Β· (π β π) Β· (π β π) Β· (π β π), where π = (π + π + π) / 2 is half the perimeter of the triangle. Based on work by Jon Pennant. This is Metamath 100 proof #57. (Contributed by Mario Carneiro, 10-Mar-2019.) |
β’ πΉ = (π₯ β (β β {0}), π¦ β (β β {0}) β¦ (ββ(logβ(π¦ / π₯)))) & β’ π = (absβ(π΅ β πΆ)) & β’ π = (absβ(π΄ β πΆ)) & β’ π = (absβ(π΄ β π΅)) & β’ π = ((π΅ β πΆ)πΉ(π΄ β πΆ)) & β’ π = (((π + π) + π) / 2) & β’ (π β π΄ β β) & β’ (π β π΅ β β) & β’ (π β πΆ β β) & β’ (π β π΄ β πΆ) & β’ (π β π΅ β πΆ) β β’ (π β (((1 / 2) Β· (π Β· π)) Β· (absβ(sinβπ))) = (ββ((π Β· (π β π)) Β· ((π β π) Β· (π β π))))) | ||
Theorem | quad2 26758 | The quadratic equation, without specifying the particular branch π· to the square root. (Contributed by Mario Carneiro, 23-Apr-2015.) |
β’ (π β π΄ β β) & β’ (π β π΄ β 0) & β’ (π β π΅ β β) & β’ (π β πΆ β β) & β’ (π β π β β) & β’ (π β π· β β) & β’ (π β (π·β2) = ((π΅β2) β (4 Β· (π΄ Β· πΆ)))) β β’ (π β (((π΄ Β· (πβ2)) + ((π΅ Β· π) + πΆ)) = 0 β (π = ((-π΅ + π·) / (2 Β· π΄)) β¨ π = ((-π΅ β π·) / (2 Β· π΄))))) | ||
Theorem | quad 26759 | The quadratic equation. (Contributed by Mario Carneiro, 23-Apr-2015.) |
β’ (π β π΄ β β) & β’ (π β π΄ β 0) & β’ (π β π΅ β β) & β’ (π β πΆ β β) & β’ (π β π β β) & β’ (π β π· = ((π΅β2) β (4 Β· (π΄ Β· πΆ)))) β β’ (π β (((π΄ Β· (πβ2)) + ((π΅ Β· π) + πΆ)) = 0 β (π = ((-π΅ + (ββπ·)) / (2 Β· π΄)) β¨ π = ((-π΅ β (ββπ·)) / (2 Β· π΄))))) | ||
Theorem | 1cubrlem 26760 | The cube roots of unity. (Contributed by Mario Carneiro, 23-Apr-2015.) |
β’ ((-1βπ(2 / 3)) = ((-1 + (i Β· (ββ3))) / 2) β§ ((-1βπ(2 / 3))β2) = ((-1 β (i Β· (ββ3))) / 2)) | ||
Theorem | 1cubr 26761 | The cube roots of unity. (Contributed by Mario Carneiro, 23-Apr-2015.) |
β’ π = {1, ((-1 + (i Β· (ββ3))) / 2), ((-1 β (i Β· (ββ3))) / 2)} β β’ (π΄ β π β (π΄ β β β§ (π΄β3) = 1)) | ||
Theorem | dcubic1lem 26762 | Lemma for dcubic1 26764 and dcubic2 26763: simplify the cubic equation under the substitution π = π β π / π. (Contributed by Mario Carneiro, 26-Apr-2015.) |
β’ (π β π β β) & β’ (π β π β β) & β’ (π β π β β) & β’ (π β π β β) & β’ (π β (πβ3) = (πΊ β π)) & β’ (π β πΊ β β) & β’ (π β (πΊβ2) = ((πβ2) + (πβ3))) & β’ (π β π = (π / 3)) & β’ (π β π = (π / 2)) & β’ (π β π β 0) & β’ (π β π β β) & β’ (π β π β 0) & β’ (π β π = (π β (π / π))) β β’ (π β (((πβ3) + ((π Β· π) + π)) = 0 β (((πβ3)β2) + ((π Β· (πβ3)) β (πβ3))) = 0)) | ||
Theorem | dcubic2 26763* | Reverse direction of dcubic 26765. Given a solution π to the "substitution" quadratic equation π = π β π / π, show that π is in the desired form. (Contributed by Mario Carneiro, 25-Apr-2015.) |
β’ (π β π β β) & β’ (π β π β β) & β’ (π β π β β) & β’ (π β π β β) & β’ (π β (πβ3) = (πΊ β π)) & β’ (π β πΊ β β) & β’ (π β (πΊβ2) = ((πβ2) + (πβ3))) & β’ (π β π = (π / 3)) & β’ (π β π = (π / 2)) & β’ (π β π β 0) & β’ (π β π β β) & β’ (π β π β 0) & β’ (π β π = (π β (π / π))) & β’ (π β ((πβ3) + ((π Β· π) + π)) = 0) β β’ (π β βπ β β ((πβ3) = 1 β§ π = ((π Β· π) β (π / (π Β· π))))) | ||
Theorem | dcubic1 26764 | Forward direction of dcubic 26765: the claimed formula produces solutions to the cubic equation. (Contributed by Mario Carneiro, 25-Apr-2015.) |
β’ (π β π β β) & β’ (π β π β β) & β’ (π β π β β) & β’ (π β π β β) & β’ (π β (πβ3) = (πΊ β π)) & β’ (π β πΊ β β) & β’ (π β (πΊβ2) = ((πβ2) + (πβ3))) & β’ (π β π = (π / 3)) & β’ (π β π = (π / 2)) & β’ (π β π β 0) & β’ (π β π = (π β (π / π))) β β’ (π β ((πβ3) + ((π Β· π) + π)) = 0) | ||
Theorem | dcubic 26765* | Solutions to the depressed cubic, a special case of cubic 26768. (The definitions of π, π, πΊ, π here differ from mcubic 26766 by scale factors of -9, 54, 54 and -27 respectively, to simplify the algebra and presentation.) (Contributed by Mario Carneiro, 26-Apr-2015.) |
β’ (π β π β β) & β’ (π β π β β) & β’ (π β π β β) & β’ (π β π β β) & β’ (π β (πβ3) = (πΊ β π)) & β’ (π β πΊ β β) & β’ (π β (πΊβ2) = ((πβ2) + (πβ3))) & β’ (π β π = (π / 3)) & β’ (π β π = (π / 2)) & β’ (π β π β 0) β β’ (π β (((πβ3) + ((π Β· π) + π)) = 0 β βπ β β ((πβ3) = 1 β§ π = ((π Β· π) β (π / (π Β· π)))))) | ||
Theorem | mcubic 26766* | Solutions to a monic cubic equation, a special case of cubic 26768. (Contributed by Mario Carneiro, 24-Apr-2015.) |
β’ (π β π΅ β β) & β’ (π β πΆ β β) & β’ (π β π· β β) & β’ (π β π β β) & β’ (π β π β β) & β’ (π β (πβ3) = ((π + πΊ) / 2)) & β’ (π β πΊ β β) & β’ (π β (πΊβ2) = ((πβ2) β (4 Β· (πβ3)))) & β’ (π β π = ((π΅β2) β (3 Β· πΆ))) & β’ (π β π = (((2 Β· (π΅β3)) β (9 Β· (π΅ Β· πΆ))) + (;27 Β· π·))) & β’ (π β π β 0) β β’ (π β ((((πβ3) + (π΅ Β· (πβ2))) + ((πΆ Β· π) + π·)) = 0 β βπ β β ((πβ3) = 1 β§ π = -(((π΅ + (π Β· π)) + (π / (π Β· π))) / 3)))) | ||
Theorem | cubic2 26767* | The solution to the general cubic equation, for arbitrary choices πΊ and π of the square and cube roots. (Contributed by Mario Carneiro, 23-Apr-2015.) |
β’ (π β π΄ β β) & β’ (π β π΄ β 0) & β’ (π β π΅ β β) & β’ (π β πΆ β β) & β’ (π β π· β β) & β’ (π β π β β) & β’ (π β π β β) & β’ (π β (πβ3) = ((π + πΊ) / 2)) & β’ (π β πΊ β β) & β’ (π β (πΊβ2) = ((πβ2) β (4 Β· (πβ3)))) & β’ (π β π = ((π΅β2) β (3 Β· (π΄ Β· πΆ)))) & β’ (π β π = (((2 Β· (π΅β3)) β ((9 Β· π΄) Β· (π΅ Β· πΆ))) + (;27 Β· ((π΄β2) Β· π·)))) & β’ (π β π β 0) β β’ (π β ((((π΄ Β· (πβ3)) + (π΅ Β· (πβ2))) + ((πΆ Β· π) + π·)) = 0 β βπ β β ((πβ3) = 1 β§ π = -(((π΅ + (π Β· π)) + (π / (π Β· π))) / (3 Β· π΄))))) | ||
Theorem | cubic 26768* | The cubic equation, which gives the roots of an arbitrary (nondegenerate) cubic function. Use rextp 4706 to convert the existential quantifier to a triple disjunction. This is Metamath 100 proof #37. (Contributed by Mario Carneiro, 26-Apr-2015.) |
β’ π = {1, ((-1 + (i Β· (ββ3))) / 2), ((-1 β (i Β· (ββ3))) / 2)} & β’ (π β π΄ β β) & β’ (π β π΄ β 0) & β’ (π β π΅ β β) & β’ (π β πΆ β β) & β’ (π β π· β β) & β’ (π β π β β) & β’ (π β π = (((π + (ββπΊ)) / 2)βπ(1 / 3))) & β’ (π β πΊ = ((πβ2) β (4 Β· (πβ3)))) & β’ (π β π = ((π΅β2) β (3 Β· (π΄ Β· πΆ)))) & β’ (π β π = (((2 Β· (π΅β3)) β ((9 Β· π΄) Β· (π΅ Β· πΆ))) + (;27 Β· ((π΄β2) Β· π·)))) & β’ (π β π β 0) β β’ (π β ((((π΄ Β· (πβ3)) + (π΅ Β· (πβ2))) + ((πΆ Β· π) + π·)) = 0 β βπ β π π = -(((π΅ + (π Β· π)) + (π / (π Β· π))) / (3 Β· π΄)))) | ||
Theorem | binom4 26769 | Work out a quartic binomial. (You would think that by this point it would be faster to use binom 15800, 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.) |
β’ ((π΄ β β β§ π΅ β β) β ((π΄ + π΅)β4) = (((π΄β4) + (4 Β· ((π΄β3) Β· π΅))) + ((6 Β· ((π΄β2) Β· (π΅β2))) + ((4 Β· (π΄ Β· (π΅β3))) + (π΅β4))))) | ||
Theorem | dquartlem1 26770 | Lemma for dquart 26772. (Contributed by Mario Carneiro, 6-May-2015.) |
β’ (π β π΅ β β) & β’ (π β πΆ β β) & β’ (π β π β β) & β’ (π β π β β) & β’ (π β π = ((2 Β· π)β2)) & β’ (π β π β 0) & β’ (π β πΌ β β) & β’ (π β (πΌβ2) = ((-(πβ2) β (π΅ / 2)) + ((πΆ / 4) / π))) β β’ (π β ((((πβ2) + ((π + π΅) / 2)) + ((((π / 2) Β· π) β (πΆ / 4)) / π)) = 0 β (π = (-π + πΌ) β¨ π = (-π β πΌ)))) | ||
Theorem | dquartlem2 26771 | Lemma for dquart 26772. (Contributed by Mario Carneiro, 6-May-2015.) |
β’ (π β π΅ β β) & β’ (π β πΆ β β) & β’ (π β π β β) & β’ (π β π β β) & β’ (π β π = ((2 Β· π)β2)) & β’ (π β π β 0) & β’ (π β πΌ β β) & β’ (π β (πΌβ2) = ((-(πβ2) β (π΅ / 2)) + ((πΆ / 4) / π))) & β’ (π β π· β β) & β’ (π β (((πβ3) + ((2 Β· π΅) Β· (πβ2))) + ((((π΅β2) β (4 Β· π·)) Β· π) + -(πΆβ2))) = 0) β β’ (π β ((((π + π΅) / 2)β2) β (((πΆβ2) / 4) / π)) = π·) | ||
Theorem | dquart 26772 | Solve a depressed quartic equation. To eliminate π, which is the square root of a solution π to the resolvent cubic equation, apply cubic 26768 or one of its variants. (Contributed by Mario Carneiro, 6-May-2015.) |
β’ (π β π΅ β β) & β’ (π β πΆ β β) & β’ (π β π β β) & β’ (π β π β β) & β’ (π β π = ((2 Β· π)β2)) & β’ (π β π β 0) & β’ (π β πΌ β β) & β’ (π β (πΌβ2) = ((-(πβ2) β (π΅ / 2)) + ((πΆ / 4) / π))) & β’ (π β π· β β) & β’ (π β (((πβ3) + ((2 Β· π΅) Β· (πβ2))) + ((((π΅β2) β (4 Β· π·)) Β· π) + -(πΆβ2))) = 0) & β’ (π β π½ β β) & β’ (π β (π½β2) = ((-(πβ2) β (π΅ / 2)) β ((πΆ / 4) / π))) β β’ (π β ((((πβ4) + (π΅ Β· (πβ2))) + ((πΆ Β· π) + π·)) = 0 β ((π = (-π + πΌ) β¨ π = (-π β πΌ)) β¨ (π = (π + π½) β¨ π = (π β π½))))) | ||
Theorem | quart1cl 26773 | Closure lemmas for quart 26780. (Contributed by Mario Carneiro, 7-May-2015.) |
β’ (π β π΄ β β) & β’ (π β π΅ β β) & β’ (π β πΆ β β) & β’ (π β π· β β) & β’ (π β π = (π΅ β ((3 / 8) Β· (π΄β2)))) & β’ (π β π = ((πΆ β ((π΄ Β· π΅) / 2)) + ((π΄β3) / 8))) & β’ (π β π = ((π· β ((πΆ Β· π΄) / 4)) + ((((π΄β2) Β· π΅) / ;16) β ((3 / ;;256) Β· (π΄β4))))) β β’ (π β (π β β β§ π β β β§ π β β)) | ||
Theorem | quart1lem 26774 | Lemma for quart1 26775. (Contributed by Mario Carneiro, 6-May-2015.) |
β’ (π β π΄ β β) & β’ (π β π΅ β β) & β’ (π β πΆ β β) & β’ (π β π· β β) & β’ (π β π = (π΅ β ((3 / 8) Β· (π΄β2)))) & β’ (π β π = ((πΆ β ((π΄ Β· π΅) / 2)) + ((π΄β3) / 8))) & β’ (π β π = ((π· β ((πΆ Β· π΄) / 4)) + ((((π΄β2) Β· π΅) / ;16) β ((3 / ;;256) Β· (π΄β4))))) & β’ (π β π β β) & β’ (π β π = (π + (π΄ / 4))) β β’ (π β π· = ((((π΄β4) / ;;256) + (π Β· ((π΄ / 4)β2))) + ((π Β· (π΄ / 4)) + π ))) | ||
Theorem | quart1 26775 | Depress a quartic equation. (Contributed by Mario Carneiro, 6-May-2015.) |
β’ (π β π΄ β β) & β’ (π β π΅ β β) & β’ (π β πΆ β β) & β’ (π β π· β β) & β’ (π β π = (π΅ β ((3 / 8) Β· (π΄β2)))) & β’ (π β π = ((πΆ β ((π΄ Β· π΅) / 2)) + ((π΄β3) / 8))) & β’ (π β π = ((π· β ((πΆ Β· π΄) / 4)) + ((((π΄β2) Β· π΅) / ;16) β ((3 / ;;256) Β· (π΄β4))))) & β’ (π β π β β) & β’ (π β π = (π + (π΄ / 4))) β β’ (π β (((πβ4) + (π΄ Β· (πβ3))) + ((π΅ Β· (πβ2)) + ((πΆ Β· π) + π·))) = (((πβ4) + (π Β· (πβ2))) + ((π Β· π) + π ))) | ||
Theorem | quartlem1 26776 | Lemma for quart 26780. (Contributed by Mario Carneiro, 6-May-2015.) |
β’ (π β π β β) & β’ (π β π β β) & β’ (π β π β β) & β’ (π β π = ((πβ2) + (;12 Β· π ))) & β’ (π β π = ((-(2 Β· (πβ3)) β (;27 Β· (πβ2))) + (;72 Β· (π Β· π )))) β β’ (π β (π = (((2 Β· π)β2) β (3 Β· ((πβ2) β (4 Β· π )))) β§ π = (((2 Β· ((2 Β· π)β3)) β (9 Β· ((2 Β· π) Β· ((πβ2) β (4 Β· π ))))) + (;27 Β· -(πβ2))))) | ||
Theorem | quartlem2 26777 | Closure lemmas for quart 26780. (Contributed by Mario Carneiro, 7-May-2015.) |
β’ (π β π΄ β β) & β’ (π β π΅ β β) & β’ (π β πΆ β β) & β’ (π β π· β β) & β’ (π β π β β) & β’ (π β πΈ = -(π΄ / 4)) & β’ (π β π = (π΅ β ((3 / 8) Β· (π΄β2)))) & β’ (π β π = ((πΆ β ((π΄ Β· π΅) / 2)) + ((π΄β3) / 8))) & β’ (π β π = ((π· β ((πΆ Β· π΄) / 4)) + ((((π΄β2) Β· π΅) / ;16) β ((3 / ;;256) Β· (π΄β4))))) & β’ (π β π = ((πβ2) + (;12 Β· π ))) & β’ (π β π = ((-(2 Β· (πβ3)) β (;27 Β· (πβ2))) + (;72 Β· (π Β· π )))) & β’ (π β π = (ββ((πβ2) β (4 Β· (πβ3))))) β β’ (π β (π β β β§ π β β β§ π β β)) | ||
Theorem | quartlem3 26778 | Closure lemmas for quart 26780. (Contributed by Mario Carneiro, 7-May-2015.) |
β’ (π β π΄ β β) & β’ (π β π΅ β β) & β’ (π β πΆ β β) & β’ (π β π· β β) & β’ (π β π β β) & β’ (π β πΈ = -(π΄ / 4)) & β’ (π β π = (π΅ β ((3 / 8) Β· (π΄β2)))) & β’ (π β π = ((πΆ β ((π΄ Β· π΅) / 2)) + ((π΄β3) / 8))) & β’ (π β π = ((π· β ((πΆ Β· π΄) / 4)) + ((((π΄β2) Β· π΅) / ;16) β ((3 / ;;256) Β· (π΄β4))))) & β’ (π β π = ((πβ2) + (;12 Β· π ))) & β’ (π β π = ((-(2 Β· (πβ3)) β (;27 Β· (πβ2))) + (;72 Β· (π Β· π )))) & β’ (π β π = (ββ((πβ2) β (4 Β· (πβ3))))) & β’ (π β π = ((ββπ) / 2)) & β’ (π β π = -((((2 Β· π) + π) + (π / π)) / 3)) & β’ (π β π = (((π + π) / 2)βπ(1 / 3))) & β’ (π β π β 0) β β’ (π β (π β β β§ π β β β§ π β β)) | ||
Theorem | quartlem4 26779 | Closure lemmas for quart 26780. (Contributed by Mario Carneiro, 7-May-2015.) |
β’ (π β π΄ β β) & β’ (π β π΅ β β) & β’ (π β πΆ β β) & β’ (π β π· β β) & β’ (π β π β β) & β’ (π β πΈ = -(π΄ / 4)) & β’ (π β π = (π΅ β ((3 / 8) Β· (π΄β2)))) & β’ (π β π = ((πΆ β ((π΄ Β· π΅) / 2)) + ((π΄β3) / 8))) & β’ (π β π = ((π· β ((πΆ Β· π΄) / 4)) + ((((π΄β2) Β· π΅) / ;16) β ((3 / ;;256) Β· (π΄β4))))) & β’ (π β π = ((πβ2) + (;12 Β· π ))) & β’ (π β π = ((-(2 Β· (πβ3)) β (;27 Β· (πβ2))) + (;72 Β· (π Β· π )))) & β’ (π β π = (ββ((πβ2) β (4 Β· (πβ3))))) & β’ (π β π = ((ββπ) / 2)) & β’ (π β π = -((((2 Β· π) + π) + (π / π)) / 3)) & β’ (π β π = (((π + π) / 2)βπ(1 / 3))) & β’ (π β π β 0) & β’ (π β π β 0) & β’ (π β πΌ = (ββ((-(πβ2) β (π / 2)) + ((π / 4) / π)))) & β’ (π β π½ = (ββ((-(πβ2) β (π / 2)) β ((π / 4) / π)))) β β’ (π β (π β 0 β§ πΌ β β β§ π½ β β)) | ||
Theorem | quart 26780 | 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 34711) if all the substitutions are performed. This is Metamath 100 proof #46. (Contributed by Mario Carneiro, 6-May-2015.) |
β’ (π β π΄ β β) & β’ (π β π΅ β β) & β’ (π β πΆ β β) & β’ (π β π· β β) & β’ (π β π β β) & β’ (π β πΈ = -(π΄ / 4)) & β’ (π β π = (π΅ β ((3 / 8) Β· (π΄β2)))) & β’ (π β π = ((πΆ β ((π΄ Β· π΅) / 2)) + ((π΄β3) / 8))) & β’ (π β π = ((π· β ((πΆ Β· π΄) / 4)) + ((((π΄β2) Β· π΅) / ;16) β ((3 / ;;256) Β· (π΄β4))))) & β’ (π β π = ((πβ2) + (;12 Β· π ))) & β’ (π β π = ((-(2 Β· (πβ3)) β (;27 Β· (πβ2))) + (;72 Β· (π Β· π )))) & β’ (π β π = (ββ((πβ2) β (4 Β· (πβ3))))) & β’ (π β π = ((ββπ) / 2)) & β’ (π β π = -((((2 Β· π) + π) + (π / π)) / 3)) & β’ (π β π = (((π + π) / 2)βπ(1 / 3))) & β’ (π β π β 0) & β’ (π β π β 0) & β’ (π β πΌ = (ββ((-(πβ2) β (π / 2)) + ((π / 4) / π)))) & β’ (π β π½ = (ββ((-(πβ2) β (π / 2)) β ((π / 4) / π)))) β β’ (π β ((((πβ4) + (π΄ Β· (πβ3))) + ((π΅ Β· (πβ2)) + ((πΆ Β· π) + π·))) = 0 β ((π = ((πΈ β π) + πΌ) β¨ π = ((πΈ β π) β πΌ)) β¨ (π = ((πΈ + π) + π½) β¨ π = ((πΈ + π) β π½))))) | ||
Syntax | casin 26781 | The arcsine function. |
class arcsin | ||
Syntax | cacos 26782 | The arccosine function. |
class arccos | ||
Syntax | catan 26783 | The arctangent function. |
class arctan | ||
Definition | df-asin 26784 | Define the arcsine function. Because sin is not a one-to-one function, the literal inverse β‘sin is not a function. Rather than attempt to find the right domain on which to restrict sin in order to get a total function, we just define it in terms of log, which we already know is total (except at 0). There are branch points at -1 and 1 (at which the function is defined), and branch cuts along the real line not between -1 and 1, which is to say (-β, -1) βͺ (1, +β). (Contributed by Mario Carneiro, 31-Mar-2015.) |
β’ arcsin = (π₯ β β β¦ (-i Β· (logβ((i Β· π₯) + (ββ(1 β (π₯β2))))))) | ||
Definition | df-acos 26785 | Define the arccosine function. See also remarks for df-asin 26784. Since we define arccos in terms of arcsin, it shares the same branch points and cuts, namely (-β, -1) βͺ (1, +β). (Contributed by Mario Carneiro, 31-Mar-2015.) |
β’ arccos = (π₯ β β β¦ ((Ο / 2) β (arcsinβπ₯))) | ||
Definition | df-atan 26786 | Define the arctangent function. See also remarks for df-asin 26784. Unlike arcsin and arccos, this function is not defined everywhere, because tan(π§) β Β±i for all π§ β β. For all other π§, there is a formula for arctan(π§) in terms of log, and we take that as the definition. Branch points are at Β±i; branch cuts are on the pure imaginary axis not between -i and i, which is to say {π§ β β β£ (i Β· π§) β (-β, -1) βͺ (1, +β)}. (Contributed by Mario Carneiro, 31-Mar-2015.) |
β’ arctan = (π₯ β (β β {-i, i}) β¦ ((i / 2) Β· ((logβ(1 β (i Β· π₯))) β (logβ(1 + (i Β· π₯)))))) | ||
Theorem | asinlem 26787 | The argument to the logarithm in df-asin 26784 is always nonzero. (Contributed by Mario Carneiro, 31-Mar-2015.) |
β’ (π΄ β β β ((i Β· π΄) + (ββ(1 β (π΄β2)))) β 0) | ||
Theorem | asinlem2 26788 | The argument to the logarithm in df-asin 26784 has the property that replacing π΄ with -π΄ in the expression gives the reciprocal. (Contributed by Mario Carneiro, 1-Apr-2015.) |
β’ (π΄ β β β (((i Β· π΄) + (ββ(1 β (π΄β2)))) Β· ((i Β· -π΄) + (ββ(1 β (-π΄β2))))) = 1) | ||
Theorem | asinlem3a 26789 | Lemma for asinlem3 26790. (Contributed by Mario Carneiro, 1-Apr-2015.) |
β’ ((π΄ β β β§ (ββπ΄) β€ 0) β 0 β€ (ββ((i Β· π΄) + (ββ(1 β (π΄β2)))))) | ||
Theorem | asinlem3 26790 | The argument to the logarithm in df-asin 26784 has nonnegative real part. (Contributed by Mario Carneiro, 1-Apr-2015.) |
β’ (π΄ β β β 0 β€ (ββ((i Β· π΄) + (ββ(1 β (π΄β2)))))) | ||
Theorem | asinf 26791 | Domain and codomain of the arcsin function. (Contributed by Mario Carneiro, 31-Mar-2015.) |
β’ arcsin:ββΆβ | ||
Theorem | asincl 26792 | Closure for the arcsin function. (Contributed by Mario Carneiro, 31-Mar-2015.) |
β’ (π΄ β β β (arcsinβπ΄) β β) | ||
Theorem | acosf 26793 | Domain and codoamin of the arccos function. (Contributed by Mario Carneiro, 31-Mar-2015.) |
β’ arccos:ββΆβ | ||
Theorem | acoscl 26794 | Closure for the arccos function. (Contributed by Mario Carneiro, 31-Mar-2015.) |
β’ (π΄ β β β (arccosβπ΄) β β) | ||
Theorem | atandm 26795 | Since the property is a little lengthy, we abbreviate π΄ β β β§ π΄ β -i β§ π΄ β i as π΄ β dom arctan. This is the necessary precondition for the definition of arctan to make sense. (Contributed by Mario Carneiro, 31-Mar-2015.) |
β’ (π΄ β dom arctan β (π΄ β β β§ π΄ β -i β§ π΄ β i)) | ||
Theorem | atandm2 26796 | This form of atandm 26795 is a bit more useful for showing that the logarithms in df-atan 26786 are well-defined. (Contributed by Mario Carneiro, 31-Mar-2015.) |
β’ (π΄ β dom arctan β (π΄ β β β§ (1 β (i Β· π΄)) β 0 β§ (1 + (i Β· π΄)) β 0)) | ||
Theorem | atandm3 26797 | A compact form of atandm 26795. (Contributed by Mario Carneiro, 31-Mar-2015.) |
β’ (π΄ β dom arctan β (π΄ β β β§ (π΄β2) β -1)) | ||
Theorem | atandm4 26798 | A compact form of atandm 26795. (Contributed by Mario Carneiro, 3-Apr-2015.) |
β’ (π΄ β dom arctan β (π΄ β β β§ (1 + (π΄β2)) β 0)) | ||
Theorem | atanf 26799 | Domain and codoamin of the arctan function. (Contributed by Mario Carneiro, 31-Mar-2015.) |
β’ arctan:(β β {-i, i})βΆβ | ||
Theorem | atancl 26800 | Closure for the arctan function. (Contributed by Mario Carneiro, 31-Mar-2015.) |
β’ (π΄ β dom arctan β (arctanβπ΄) β β) |
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