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Theorem List for Metamath Proof Explorer - 25301-25400   *Has distinct variable group(s)
TypeLabelDescription
Statement

Theoremcxplt3d 25301 Ordering property for complex exponentiation. (Contributed by Mario Carneiro, 30-May-2016.)
(𝜑𝐴 ∈ ℝ+)    &   (𝜑𝐵 ∈ ℝ)    &   (𝜑𝐴 < 1)    &   (𝜑𝐶 ∈ ℝ)       (𝜑 → (𝐵 < 𝐶 ↔ (𝐴𝑐𝐶) < (𝐴𝑐𝐵)))

Theoremcxple3d 25302 Ordering property for complex exponentiation. (Contributed by Mario Carneiro, 30-May-2016.)
(𝜑𝐴 ∈ ℝ+)    &   (𝜑𝐵 ∈ ℝ)    &   (𝜑𝐴 < 1)    &   (𝜑𝐶 ∈ ℝ)       (𝜑 → (𝐵𝐶 ↔ (𝐴𝑐𝐶) ≤ (𝐴𝑐𝐵)))

Theoremcxpmuld 25303 Product of exponents law for complex exponentiation. Proposition 10-4.2(b) of [Gleason] p. 135. (Contributed by Mario Carneiro, 30-May-2016.)
(𝜑𝐴 ∈ ℝ+)    &   (𝜑𝐵 ∈ ℝ)    &   (𝜑𝐶 ∈ ℂ)       (𝜑 → (𝐴𝑐(𝐵 · 𝐶)) = ((𝐴𝑐𝐵)↑𝑐𝐶))

Theoremcxpcom 25304 Commutative law for real exponentiation. (Contributed by AV, 29-Dec-2022.)
((𝐴 ∈ ℝ+𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ) → ((𝐴𝑐𝐵)↑𝑐𝐶) = ((𝐴𝑐𝐶)↑𝑐𝐵))

Theoremdvcxp1 25305* The derivative of a complex power with respect to the first argument. (Contributed by Mario Carneiro, 24-Feb-2015.)
(𝐴 ∈ ℂ → (ℝ D (𝑥 ∈ ℝ+ ↦ (𝑥𝑐𝐴))) = (𝑥 ∈ ℝ+ ↦ (𝐴 · (𝑥𝑐(𝐴 − 1)))))

Theoremdvcxp2 25306* The derivative of a complex power with respect to the second argument. (Contributed by Mario Carneiro, 24-Feb-2015.)
(𝐴 ∈ ℝ+ → (ℂ D (𝑥 ∈ ℂ ↦ (𝐴𝑐𝑥))) = (𝑥 ∈ ℂ ↦ ((log‘𝐴) · (𝐴𝑐𝑥))))

Theoremdvsqrt 25307 The derivative of the real square root function. (Contributed by Mario Carneiro, 1-May-2016.)
(ℝ D (𝑥 ∈ ℝ+ ↦ (√‘𝑥))) = (𝑥 ∈ ℝ+ ↦ (1 / (2 · (√‘𝑥))))

Theoremdvcncxp1 25308* Derivative of complex power with respect to first argument on the complex plane. (Contributed by Brendan Leahy, 18-Dec-2018.)
𝐷 = (ℂ ∖ (-∞(,]0))       (𝐴 ∈ ℂ → (ℂ D (𝑥𝐷 ↦ (𝑥𝑐𝐴))) = (𝑥𝐷 ↦ (𝐴 · (𝑥𝑐(𝐴 − 1)))))

Theoremdvcnsqrt 25309* Derivative of square root function. (Contributed by Brendan Leahy, 18-Dec-2018.)
𝐷 = (ℂ ∖ (-∞(,]0))       (ℂ D (𝑥𝐷 ↦ (√‘𝑥))) = (𝑥𝐷 ↦ (1 / (2 · (√‘𝑥))))

Theoremcxpcn 25310* Domain of continuity of the complex power function. (Contributed by Mario Carneiro, 1-May-2016.)
𝐷 = (ℂ ∖ (-∞(,]0))    &   𝐽 = (TopOpen‘ℂfld)    &   𝐾 = (𝐽t 𝐷)       (𝑥𝐷, 𝑦 ∈ ℂ ↦ (𝑥𝑐𝑦)) ∈ ((𝐾 ×t 𝐽) Cn 𝐽)

Theoremcxpcn2 25311* Continuity of the complex power function, when the base is real. (Contributed by Mario Carneiro, 1-May-2016.)
𝐽 = (TopOpen‘ℂfld)    &   𝐾 = (𝐽t+)       (𝑥 ∈ ℝ+, 𝑦 ∈ ℂ ↦ (𝑥𝑐𝑦)) ∈ ((𝐾 ×t 𝐽) Cn 𝐽)

Theoremcxpcn3lem 25312* Lemma for cxpcn3 25313. (Contributed by Mario Carneiro, 2-May-2016.)
𝐷 = (ℜ “ ℝ+)    &   𝐽 = (TopOpen‘ℂfld)    &   𝐾 = (𝐽t (0[,)+∞))    &   𝐿 = (𝐽t 𝐷)    &   𝑈 = (if((ℜ‘𝐴) ≤ 1, (ℜ‘𝐴), 1) / 2)    &   𝑇 = if(𝑈 ≤ (𝐸𝑐(1 / 𝑈)), 𝑈, (𝐸𝑐(1 / 𝑈)))       ((𝐴𝐷𝐸 ∈ ℝ+) → ∃𝑑 ∈ ℝ+𝑎 ∈ (0[,)+∞)∀𝑏𝐷 (((abs‘𝑎) < 𝑑 ∧ (abs‘(𝐴𝑏)) < 𝑑) → (abs‘(𝑎𝑐𝑏)) < 𝐸))

Theoremcxpcn3 25313* 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.)
𝐷 = (ℜ “ ℝ+)    &   𝐽 = (TopOpen‘ℂfld)    &   𝐾 = (𝐽t (0[,)+∞))    &   𝐿 = (𝐽t 𝐷)       (𝑥 ∈ (0[,)+∞), 𝑦𝐷 ↦ (𝑥𝑐𝑦)) ∈ ((𝐾 ×t 𝐿) Cn 𝐽)

Theoremresqrtcn 25314 Continuity of the real square root function. (Contributed by Mario Carneiro, 2-May-2016.)
(√ ↾ (0[,)+∞)) ∈ ((0[,)+∞)–cn→ℝ)

Theoremsqrtcn 25315 Continuity of the square root function. (Contributed by Mario Carneiro, 2-May-2016.)
𝐷 = (ℂ ∖ (-∞(,]0))       (√ ↾ 𝐷) ∈ (𝐷cn→ℂ)

(𝜑𝐴 ∈ ℝ)    &   (𝜑 → 0 ≤ 𝐴)    &   (𝜑𝐴 ≤ 1)    &   (𝜑𝐵 ∈ ℝ+)    &   (𝜑𝐵 ≤ 1)       (𝜑𝐴 ≤ (𝐴𝑐𝐵))

Theoremcxpaddle 25317 Ordering property for complex exponentiation. (Contributed by Mario Carneiro, 8-Sep-2014.)
(𝜑𝐴 ∈ ℝ)    &   (𝜑 → 0 ≤ 𝐴)    &   (𝜑𝐵 ∈ ℝ)    &   (𝜑 → 0 ≤ 𝐵)    &   (𝜑𝐶 ∈ ℝ+)    &   (𝜑𝐶 ≤ 1)       (𝜑 → ((𝐴 + 𝐵)↑𝑐𝐶) ≤ ((𝐴𝑐𝐶) + (𝐵𝑐𝐶)))

Theoremabscxpbnd 25318 Bound on the absolute value of a complex power. (Contributed by Mario Carneiro, 15-Sep-2014.)
(𝜑𝐴 ∈ ℂ)    &   (𝜑𝐵 ∈ ℂ)    &   (𝜑 → 0 ≤ (ℜ‘𝐵))    &   (𝜑𝑀 ∈ ℝ)    &   (𝜑 → (abs‘𝐴) ≤ 𝑀)       (𝜑 → (abs‘(𝐴𝑐𝐵)) ≤ ((𝑀𝑐(ℜ‘𝐵)) · (exp‘((abs‘𝐵) · π))))

Theoremroot1id 25319 Property of an 𝑁-th root of unity. (Contributed by Mario Carneiro, 23-Apr-2015.)
(𝑁 ∈ ℕ → ((-1↑𝑐(2 / 𝑁))↑𝑁) = 1)

Theoremroot1eq1 25320 The only powers of an 𝑁-th root of unity that equal 1 are the multiples of 𝑁. In other words, -1↑𝑐(2 / 𝑁) has order 𝑁 in the multiplicative group of nonzero complex numbers. (In fact, these and their powers are the only elements of finite order in the complex numbers.) (Contributed by Mario Carneiro, 28-Apr-2016.)
((𝑁 ∈ ℕ ∧ 𝐾 ∈ ℤ) → (((-1↑𝑐(2 / 𝑁))↑𝐾) = 1 ↔ 𝑁𝐾))

Theoremroot1cj 25321 Within the 𝑁-th roots of unity, the conjugate of the 𝐾-th root is the 𝑁𝐾-th root. (Contributed by Mario Carneiro, 23-Apr-2015.)
((𝑁 ∈ ℕ ∧ 𝐾 ∈ ℤ) → (∗‘((-1↑𝑐(2 / 𝑁))↑𝐾)) = ((-1↑𝑐(2 / 𝑁))↑(𝑁𝐾)))

Theoremcxpeq 25322* Solve an equation involving an 𝑁-th power. The expression -1↑𝑐(2 / 𝑁) = exp(2πi / 𝑁) is a way to write the primitive 𝑁-th root of unity with the smallest positive argument. (Contributed by Mario Carneiro, 23-Apr-2015.)
((𝐴 ∈ ℂ ∧ 𝑁 ∈ ℕ ∧ 𝐵 ∈ ℂ) → ((𝐴𝑁) = 𝐵 ↔ ∃𝑛 ∈ (0...(𝑁 − 1))𝐴 = ((𝐵𝑐(1 / 𝑁)) · ((-1↑𝑐(2 / 𝑁))↑𝑛))))

Theoremloglesqrt 25323 An upper bound on the logarithm. (Contributed by Mario Carneiro, 2-May-2016.) (Proof shortened by AV, 2-Aug-2021.)
((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) → (log‘(𝐴 + 1)) ≤ (√‘𝐴))

Theoremlogreclem 25324 Symmetry of the natural logarithm range by negation. Lemma for logrec 25325. (Contributed by Saveliy Skresanov, 27-Dec-2016.)
((𝐴 ∈ ran log ∧ ¬ (ℑ‘𝐴) = π) → -𝐴 ∈ ran log)

Theoremlogrec 25325 Logarithm of a reciprocal changes sign. (Contributed by Saveliy Skresanov, 28-Dec-2016.)
((𝐴 ∈ ℂ ∧ 𝐴 ≠ 0 ∧ (ℑ‘(log‘𝐴)) ≠ π) → (log‘𝐴) = -(log‘(1 / 𝐴)))

14.3.5  Logarithms to an arbitrary base

Define "log using an arbitrary base" function and then prove some of its properties. Note that logb is generalized to an arbitrary base and arbitrary parameter in , but it doesn't accept infinities as arguments, unlike log.

Metamath doesn't care what letters are used to represent classes. Usually classes begin with the letter "A", but here we use "B" and "X" to more clearly distinguish between "base" and "other parameter of log".

There are different ways this could be defined in Metamath. The approach used here is intentionally similar to existing 2-parameter Metamath functions (operations): (𝐵 logb 𝑋) where 𝐵 is the base and 𝑋 is the argument of the logarithm function. An alternative would be to support the notational form (( logb𝐵)‘𝑋); that looks a little more like traditional notation. Such a function ( logb𝐵) for a fixed base can be obtained in Metamath (without the need for a new definition) by the curry function: (curry logb𝐵), see logbmpt 25350, logbf 25351 and logbfval 25352.

Syntaxclogb 25326 Extend class notation to include the logarithm generalized to an arbitrary base.
class logb

Definitiondf-logb 25327* Define the logb operator. This is the logarithm generalized to an arbitrary base. It can be used as (𝐵 logb 𝑋) for "log base B of X". In the most common traditional notation, base B is a subscript of "log". The definition is according to Wikipedia "Complex logarithm": https://en.wikipedia.org/wiki/Complex_logarithm#Logarithms_to_other_bases (10-Jun-2020). (Contributed by David A. Wheeler, 21-Jan-2017.)
logb = (𝑥 ∈ (ℂ ∖ {0, 1}), 𝑦 ∈ (ℂ ∖ {0}) ↦ ((log‘𝑦) / (log‘𝑥)))

Theoremlogbval 25328 Define the value of the logb function, the logarithm generalized to an arbitrary base, when used as infix. Most Metamath statements select variables in order of their use, but to make the order clearer we use "B" for base and "X" for the argument of the logarithm function here. (Contributed by David A. Wheeler, 21-Jan-2017.) (Revised by David A. Wheeler, 16-Jul-2017.)
((𝐵 ∈ (ℂ ∖ {0, 1}) ∧ 𝑋 ∈ (ℂ ∖ {0})) → (𝐵 logb 𝑋) = ((log‘𝑋) / (log‘𝐵)))

Theoremlogbcl 25329 General logarithm closure. (Contributed by David A. Wheeler, 17-Jul-2017.)
((𝐵 ∈ (ℂ ∖ {0, 1}) ∧ 𝑋 ∈ (ℂ ∖ {0})) → (𝐵 logb 𝑋) ∈ ℂ)

Theoremlogbid1 25330 General logarithm is 1 when base and arg match. Property 1(a) of [Cohen4] p. 361. (Contributed by Stefan O'Rear, 19-Sep-2014.) (Revised by David A. Wheeler, 22-Jul-2017.)
((𝐴 ∈ ℂ ∧ 𝐴 ≠ 0 ∧ 𝐴 ≠ 1) → (𝐴 logb 𝐴) = 1)

Theoremlogb1 25331 The logarithm of 1 to an arbitrary base 𝐵 is 0. Property 1(b) of [Cohen4] p. 361. See log1 25153. (Contributed by Stefan O'Rear, 19-Sep-2014.) (Revised by Thierry Arnoux, 27-Sep-2017.)
((𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ∧ 𝐵 ≠ 1) → (𝐵 logb 1) = 0)

Theoremelogb 25332 The general logarithm of a number to the base being Euler's constant is the natural logarithm of the number. Put another way, using e as the base in logb is the same as log. Definition in [Cohen4] p. 352. (Contributed by David A. Wheeler, 17-Oct-2017.) (Revised by David A. Wheeler and AV, 16-Jun-2020.)
(𝐴 ∈ (ℂ ∖ {0}) → (e logb 𝐴) = (log‘𝐴))

Theoremlogbchbase 25333 Change of base for logarithms. Property in [Cohen4] p. 367. (Contributed by AV, 11-Jun-2020.)
(((𝐴 ∈ ℂ ∧ 𝐴 ≠ 0 ∧ 𝐴 ≠ 1) ∧ (𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ∧ 𝐵 ≠ 1) ∧ 𝑋 ∈ (ℂ ∖ {0})) → (𝐴 logb 𝑋) = ((𝐵 logb 𝑋) / (𝐵 logb 𝐴)))

Theoremrelogbval 25334 Value of the general logarithm with integer base. (Contributed by Thierry Arnoux, 27-Sep-2017.)
((𝐵 ∈ (ℤ‘2) ∧ 𝑋 ∈ ℝ+) → (𝐵 logb 𝑋) = ((log‘𝑋) / (log‘𝐵)))

Theoremrelogbcl 25335 Closure of the general logarithm with a positive real base on positive reals. (Contributed by Stefan O'Rear, 19-Sep-2014.) (Revised by Thierry Arnoux, 27-Sep-2017.)
((𝐵 ∈ ℝ+𝑋 ∈ ℝ+𝐵 ≠ 1) → (𝐵 logb 𝑋) ∈ ℝ)

Theoremrelogbzcl 25336 Closure of the general logarithm with integer base on positive reals. (Contributed by Thierry Arnoux, 27-Sep-2017.) (Proof shortened by AV, 9-Jun-2020.)
((𝐵 ∈ (ℤ‘2) ∧ 𝑋 ∈ ℝ+) → (𝐵 logb 𝑋) ∈ ℝ)

Theoremrelogbreexp 25337 Power law for the general logarithm for real powers: The logarithm of a positive real number to the power of a real number is equal to the product of the exponent and the logarithm of the base of the power. Property 4 of [Cohen4] p. 361. (Contributed by AV, 9-Jun-2020.)
((𝐵 ∈ (ℂ ∖ {0, 1}) ∧ 𝐶 ∈ ℝ+𝐸 ∈ ℝ) → (𝐵 logb (𝐶𝑐𝐸)) = (𝐸 · (𝐵 logb 𝐶)))

Theoremrelogbzexp 25338 Power law for the general logarithm for integer powers: The logarithm of a positive real number to the power of an integer is equal to the product of the exponent and the logarithm of the base of the power. (Contributed by Stefan O'Rear, 19-Sep-2014.) (Revised by AV, 9-Jun-2020.)
((𝐵 ∈ (ℂ ∖ {0, 1}) ∧ 𝐶 ∈ ℝ+𝑁 ∈ ℤ) → (𝐵 logb (𝐶𝑁)) = (𝑁 · (𝐵 logb 𝐶)))

Theoremrelogbmul 25339 The logarithm of the product of two positive real numbers is the sum of logarithms. Property 2 of [Cohen4] p. 361. (Contributed by Stefan O'Rear, 19-Sep-2014.) (Revised by AV, 29-May-2020.)
((𝐵 ∈ (ℂ ∖ {0, 1}) ∧ (𝐴 ∈ ℝ+𝐶 ∈ ℝ+)) → (𝐵 logb (𝐴 · 𝐶)) = ((𝐵 logb 𝐴) + (𝐵 logb 𝐶)))

Theoremrelogbmulexp 25340 The logarithm of the product of a positive real and a positive real number to the power of a real number is the sum of the logarithm of the first real number and the scaled logarithm of the second real number. (Contributed by AV, 29-May-2020.)
((𝐵 ∈ (ℂ ∖ {0, 1}) ∧ (𝐴 ∈ ℝ+𝐶 ∈ ℝ+𝐸 ∈ ℝ)) → (𝐵 logb (𝐴 · (𝐶𝑐𝐸))) = ((𝐵 logb 𝐴) + (𝐸 · (𝐵 logb 𝐶))))

Theoremrelogbdiv 25341 The logarithm of the quotient of two positive real numbers is the difference of logarithms. Property 3 of [Cohen4] p. 361. (Contributed by AV, 29-May-2020.)
((𝐵 ∈ (ℂ ∖ {0, 1}) ∧ (𝐴 ∈ ℝ+𝐶 ∈ ℝ+)) → (𝐵 logb (𝐴 / 𝐶)) = ((𝐵 logb 𝐴) − (𝐵 logb 𝐶)))

Theoremrelogbexp 25342 Identity law for general logarithm: the logarithm of a power to the base is the exponent. Property 6 of [Cohen4] p. 361. (Contributed by Stefan O'Rear, 19-Sep-2014.) (Revised by AV, 9-Jun-2020.)
((𝐵 ∈ ℝ+𝐵 ≠ 1 ∧ 𝑀 ∈ ℤ) → (𝐵 logb (𝐵𝑀)) = 𝑀)

Theoremnnlogbexp 25343 Identity law for general logarithm with integer base. (Contributed by Stefan O'Rear, 19-Sep-2014.) (Revised by Thierry Arnoux, 27-Sep-2017.)
((𝐵 ∈ (ℤ‘2) ∧ 𝑀 ∈ ℤ) → (𝐵 logb (𝐵𝑀)) = 𝑀)

Theoremlogbrec 25344 Logarithm of a reciprocal changes sign. See logrec 25325. Particular case of Property 3 of [Cohen4] p. 361. (Contributed by Thierry Arnoux, 27-Sep-2017.)
((𝐵 ∈ (ℤ‘2) ∧ 𝐴 ∈ ℝ+) → (𝐵 logb (1 / 𝐴)) = -(𝐵 logb 𝐴))

Theoremlogbleb 25345 The general logarithm function is monotone/increasing. See logleb 25170. (Contributed by Stefan O'Rear, 19-Oct-2014.) (Revised by AV, 31-May-2020.)
((𝐵 ∈ (ℤ‘2) ∧ 𝑋 ∈ ℝ+𝑌 ∈ ℝ+) → (𝑋𝑌 ↔ (𝐵 logb 𝑋) ≤ (𝐵 logb 𝑌)))

Theoremlogblt 25346 The general logarithm function is strictly monotone/increasing. Property 2 of [Cohen4] p. 377. See logltb 25167. (Contributed by Stefan O'Rear, 19-Oct-2014.) (Revised by Thierry Arnoux, 27-Sep-2017.)
((𝐵 ∈ (ℤ‘2) ∧ 𝑋 ∈ ℝ+𝑌 ∈ ℝ+) → (𝑋 < 𝑌 ↔ (𝐵 logb 𝑋) < (𝐵 logb 𝑌)))

Theoremrelogbcxp 25347 Identity law for the general logarithm for real numbers. (Contributed by AV, 22-May-2020.)
((𝐵 ∈ (ℝ+ ∖ {1}) ∧ 𝑋 ∈ ℝ) → (𝐵 logb (𝐵𝑐𝑋)) = 𝑋)

Theoremcxplogb 25348 Identity law for the general logarithm. (Contributed by AV, 22-May-2020.)
((𝐵 ∈ (ℂ ∖ {0, 1}) ∧ 𝑋 ∈ (ℂ ∖ {0})) → (𝐵𝑐(𝐵 logb 𝑋)) = 𝑋)

Theoremrelogbcxpb 25349 The logarithm is the inverse of the exponentiation. Observation in [Cohen4] p. 348. (Contributed by AV, 11-Jun-2020.)
(((𝐵 ∈ ℝ+𝐵 ≠ 1) ∧ 𝑋 ∈ ℝ+𝑌 ∈ ℝ) → ((𝐵 logb 𝑋) = 𝑌 ↔ (𝐵𝑐𝑌) = 𝑋))

Theoremlogbmpt 25350* The general logarithm to a fixed base regarded as mapping. (Contributed by AV, 11-Jun-2020.)
((𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ∧ 𝐵 ≠ 1) → (curry logb𝐵) = (𝑦 ∈ (ℂ ∖ {0}) ↦ ((log‘𝑦) / (log‘𝐵))))

Theoremlogbf 25351 The general logarithm to a fixed base regarded as function. (Contributed by AV, 11-Jun-2020.)
((𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ∧ 𝐵 ≠ 1) → (curry logb𝐵):(ℂ ∖ {0})⟶ℂ)

Theoremlogbfval 25352 The general logarithm of a complex number to a fixed base. (Contributed by AV, 11-Jun-2020.)
(((𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ∧ 𝐵 ≠ 1) ∧ 𝑋 ∈ (ℂ ∖ {0})) → ((curry logb𝐵)‘𝑋) = (𝐵 logb 𝑋))

Theoremrelogbf 25353 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.)
((𝐵 ∈ ℝ+ ∧ 1 < 𝐵) → ((curry logb𝐵) ↾ ℝ+):ℝ+⟶ℝ)

Theoremlogblog 25354 The general logarithm to the base being Euler's constant regarded as function is the natural logarithm. (Contributed by AV, 12-Jun-2020.)
(curry logb ‘e) = log

Theoremlogbgt0b 25355 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.)
((𝐴 ∈ ℝ+ ∧ (𝐵 ∈ ℝ+ ∧ 1 < 𝐵)) → (0 < (𝐵 logb 𝐴) ↔ 1 < 𝐴))

Theoremlogbgcd1irr 25356 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) ∈ (ℝ ∖ ℚ) (see 2logb9irr 25357). (Contributed by AV, 29-Dec-2022.)
((𝑋 ∈ (ℤ‘2) ∧ 𝐵 ∈ (ℤ‘2) ∧ (𝑋 gcd 𝐵) = 1) → (𝐵 logb 𝑋) ∈ (ℝ ∖ ℚ))

Theorem2logb9irr 25357 Example for logbgcd1irr 25356. The logarithm of nine to base two is irrational. (Contributed by AV, 29-Dec-2022.)
(2 logb 9) ∈ (ℝ ∖ ℚ)

Theoremlogbprmirr 25358 The logarithm of a prime to a different prime base is an irrational number. For example, (2 logb 3) ∈ (ℝ ∖ ℚ) (see 2logb3irr 25359). (Contributed by AV, 31-Dec-2022.)
((𝑋 ∈ ℙ ∧ 𝐵 ∈ ℙ ∧ 𝑋𝐵) → (𝐵 logb 𝑋) ∈ (ℝ ∖ ℚ))

Theorem2logb3irr 25359 Example for logbprmirr 25358. The logarithm of three to base two is irrational. (Contributed by AV, 31-Dec-2022.)
(2 logb 3) ∈ (ℝ ∖ ℚ)

Theorem2logb9irrALT 25360 Alternate proof of 2logb9irr 25357: The logarithm of nine to base two is irrational. (Contributed by AV, 31-Dec-2022.) (Proof modification is discouraged.) (New usage is discouraged.)
(2 logb 9) ∈ (ℝ ∖ ℚ)

Theoremsqrt2cxp2logb9e3 25361 The square root of two to the power of the logarithm of nine to base two is three. (√‘2) and (2 logb 9) are irrational numbers (see sqrt2irr0 15582 resp. 2logb9irr 25357), satisfying the statement in 2irrexpqALT 25362. (Contributed by AV, 29-Dec-2022.)
((√‘2)↑𝑐(2 logb 9)) = 3

Theorem2irrexpqALT 25362* Alternate proof of 2irrexpq 25297: There exist irrational numbers 𝑎 and 𝑏 such that (𝑎𝑏) 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 25297, this is a constructive proof because it is based on two explicitly named irrational numbers (√‘2) and (2 logb 9), see sqrt2irr0 15582, 2logb9irr 25357 and sqrt2cxp2logb9e3 25361. Therefore, this proof is also acceptable/usable in intuitionistic logic. (Contributed by AV, 23-Dec-2022.) (New usage is discouraged.) (Proof modification is discouraged.)
𝑎 ∈ (ℝ ∖ ℚ)∃𝑏 ∈ (ℝ ∖ ℚ)(𝑎𝑐𝑏) ∈ ℚ

14.3.6  Theorems of Pythagoras, isosceles triangles, and intersecting chords

Theoremangval 25363* Define the angle function, which takes two complex numbers, treated as vectors from the origin, and returns the angle between them, in the range ( − π, π]. To convert from the geometry notation, 𝑚𝐴𝐵𝐶, the measure of the angle with legs 𝐴𝐵, 𝐶𝐵 where 𝐶 is more counterclockwise for positive angles, is represented by ((𝐶𝐵)𝐹(𝐴𝐵)). (Contributed by Mario Carneiro, 23-Sep-2014.)
𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥))))       (((𝐴 ∈ ℂ ∧ 𝐴 ≠ 0) ∧ (𝐵 ∈ ℂ ∧ 𝐵 ≠ 0)) → (𝐴𝐹𝐵) = (ℑ‘(log‘(𝐵 / 𝐴))))

Theoremangcan 25364* Cancel a constant multiplier in the angle function. (Contributed by Mario Carneiro, 23-Sep-2014.)
𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥))))       (((𝐴 ∈ ℂ ∧ 𝐴 ≠ 0) ∧ (𝐵 ∈ ℂ ∧ 𝐵 ≠ 0) ∧ (𝐶 ∈ ℂ ∧ 𝐶 ≠ 0)) → ((𝐶 · 𝐴)𝐹(𝐶 · 𝐵)) = (𝐴𝐹𝐵))

Theoremangneg 25365* Cancel a negative sign in the angle function. (Contributed by Mario Carneiro, 23-Sep-2014.)
𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥))))       (((𝐴 ∈ ℂ ∧ 𝐴 ≠ 0) ∧ (𝐵 ∈ ℂ ∧ 𝐵 ≠ 0)) → (-𝐴𝐹-𝐵) = (𝐴𝐹𝐵))

Theoremangvald 25366* The (signed) angle between two vectors is the argument of their quotient. Deduction form of angval 25363. (Contributed by David Moews, 28-Feb-2017.)
𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥))))    &   (𝜑𝑋 ∈ ℂ)    &   (𝜑𝑋 ≠ 0)    &   (𝜑𝑌 ∈ ℂ)    &   (𝜑𝑌 ≠ 0)       (𝜑 → (𝑋𝐹𝑌) = (ℑ‘(log‘(𝑌 / 𝑋))))

Theoremangcld 25367* The (signed) angle between two vectors is in (-π(,]π). Deduction form. (Contributed by David Moews, 28-Feb-2017.)
𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥))))    &   (𝜑𝑋 ∈ ℂ)    &   (𝜑𝑋 ≠ 0)    &   (𝜑𝑌 ∈ ℂ)    &   (𝜑𝑌 ≠ 0)       (𝜑 → (𝑋𝐹𝑌) ∈ (-π(,]π))

Theoremangrteqvd 25368* Two vectors are at a right angle iff their quotient is purely imaginary. (Contributed by David Moews, 28-Feb-2017.)
𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥))))    &   (𝜑𝑋 ∈ ℂ)    &   (𝜑𝑋 ≠ 0)    &   (𝜑𝑌 ∈ ℂ)    &   (𝜑𝑌 ≠ 0)       (𝜑 → ((𝑋𝐹𝑌) ∈ {(π / 2), -(π / 2)} ↔ (ℜ‘(𝑌 / 𝑋)) = 0))

Theoremcosangneg2d 25369* The cosine of the angle between 𝑋 and -𝑌 is the negative of that between 𝑋 and 𝑌. 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.)
𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥))))    &   (𝜑𝑋 ∈ ℂ)    &   (𝜑𝑋 ≠ 0)    &   (𝜑𝑌 ∈ ℂ)    &   (𝜑𝑌 ≠ 0)       (𝜑 → (cos‘(𝑋𝐹-𝑌)) = -(cos‘(𝑋𝐹𝑌)))

Theoremangrtmuld 25370* Perpendicularity of two vectors does not change under rescaling the second. (Contributed by David Moews, 28-Feb-2017.)
𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥))))    &   (𝜑𝑋 ∈ ℂ)    &   (𝜑𝑌 ∈ ℂ)    &   (𝜑𝑍 ∈ ℂ)    &   (𝜑𝑋 ≠ 0)    &   (𝜑𝑌 ≠ 0)    &   (𝜑𝑍 ≠ 0)    &   (𝜑 → (𝑍 / 𝑌) ∈ ℝ)       (𝜑 → ((𝑋𝐹𝑌) ∈ {(π / 2), -(π / 2)} ↔ (𝑋𝐹𝑍) ∈ {(π / 2), -(π / 2)}))

Theoremang180lem1 25371* Lemma for ang180 25376. Show that the "revolution number" 𝑁 is an integer, using efeq1 25096 to show that since the product of the three arguments 𝐴, 1 / (1 − 𝐴), (𝐴 − 1) / 𝐴 is -1, the sum of the logarithms must be an integer multiple of 2πi away from πi = log(-1). (Contributed by Mario Carneiro, 23-Sep-2014.)
𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥))))    &   𝑇 = (((log‘(1 / (1 − 𝐴))) + (log‘((𝐴 − 1) / 𝐴))) + (log‘𝐴))    &   𝑁 = (((𝑇 / i) / (2 · π)) − (1 / 2))       ((𝐴 ∈ ℂ ∧ 𝐴 ≠ 0 ∧ 𝐴 ≠ 1) → (𝑁 ∈ ℤ ∧ (𝑇 / i) ∈ ℝ))

Theoremang180lem2 25372* Lemma for ang180 25376. Show that the revolution number 𝑁 is strictly between -2 and 1. Both bounds are established by iterating using the bounds on the imaginary part of the logarithm, logimcl 25137, but the resulting bound gives only 𝑁 ≤ 1 for the upper bound. The case 𝑁 = 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 𝐴, 1 / (1 − 𝐴), (𝐴 − 1) / 𝐴 must lie on the negative real axis, which is a contradiction because clearly if 𝐴 is negative then the other two are positive real. (Contributed by Mario Carneiro, 23-Sep-2014.)
𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥))))    &   𝑇 = (((log‘(1 / (1 − 𝐴))) + (log‘((𝐴 − 1) / 𝐴))) + (log‘𝐴))    &   𝑁 = (((𝑇 / i) / (2 · π)) − (1 / 2))       ((𝐴 ∈ ℂ ∧ 𝐴 ≠ 0 ∧ 𝐴 ≠ 1) → (-2 < 𝑁𝑁 < 1))

Theoremang180lem3 25373* Lemma for ang180 25376. Since ang180lem1 25371 shows that 𝑁 is an integer and ang180lem2 25372 shows that 𝑁 is strictly between -2 and 1, it follows that 𝑁 ∈ {-1, 0}, and these two cases correspond to the two possible values for 𝑇. (Contributed by Mario Carneiro, 23-Sep-2014.)
𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥))))    &   𝑇 = (((log‘(1 / (1 − 𝐴))) + (log‘((𝐴 − 1) / 𝐴))) + (log‘𝐴))    &   𝑁 = (((𝑇 / i) / (2 · π)) − (1 / 2))       ((𝐴 ∈ ℂ ∧ 𝐴 ≠ 0 ∧ 𝐴 ≠ 1) → 𝑇 ∈ {-(i · π), (i · π)})

Theoremang180lem4 25374* Lemma for ang180 25376. Reduce the statement to one variable. (Contributed by Mario Carneiro, 23-Sep-2014.)
𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥))))       ((𝐴 ∈ ℂ ∧ 𝐴 ≠ 0 ∧ 𝐴 ≠ 1) → ((((1 − 𝐴)𝐹1) + (𝐴𝐹(𝐴 − 1))) + (1𝐹𝐴)) ∈ {-π, π})

Theoremang180lem5 25375* Lemma for ang180 25376: Reduce the statement to two variables. (Contributed by Mario Carneiro, 23-Sep-2014.)
𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥))))       (((𝐴 ∈ ℂ ∧ 𝐴 ≠ 0) ∧ (𝐵 ∈ ℂ ∧ 𝐵 ≠ 0) ∧ 𝐴𝐵) → ((((𝐴𝐵)𝐹𝐴) + (𝐵𝐹(𝐵𝐴))) + (𝐴𝐹𝐵)) ∈ {-π, π})

Theoremang180 25376* The sum of angles 𝑚𝐴𝐵𝐶 + 𝑚𝐵𝐶𝐴 + 𝑚𝐶𝐴𝐵 in a triangle adds up to either π or , 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.)
𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥))))       (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝐴𝐵𝐵𝐶𝐴𝐶)) → ((((𝐶𝐵)𝐹(𝐴𝐵)) + ((𝐴𝐶)𝐹(𝐵𝐶))) + ((𝐵𝐴)𝐹(𝐶𝐴))) ∈ {-π, π})

Theoremlawcoslem1 25377 Lemma for lawcos 25378. 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.)
(𝜑𝑈 ∈ ℂ)    &   (𝜑𝑉 ∈ ℂ)    &   (𝜑𝑈 ≠ 0)    &   (𝜑𝑉 ≠ 0)       (𝜑 → ((abs‘(𝑈𝑉))↑2) = ((((abs‘𝑈)↑2) + ((abs‘𝑉)↑2)) − (2 · (((abs‘𝑈) · (abs‘𝑉)) · ((ℜ‘(𝑈 / 𝑉)) / (abs‘(𝑈 / 𝑉)))))))

Theoremlawcos 25378* 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 𝐹 is the signed angle construct (as used in ang180 25376), 𝑋 is the distance of line segment BC, 𝑌 is the distance of line segment AC, 𝑍 is the distance of line segment AB, and 𝑂 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 25377 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 15478). The Pythagorean theorem pythag 25379 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.)
𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥))))    &   𝑋 = (abs‘(𝐵𝐶))    &   𝑌 = (abs‘(𝐴𝐶))    &   𝑍 = (abs‘(𝐴𝐵))    &   𝑂 = ((𝐵𝐶)𝐹(𝐴𝐶))       (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝐴𝐶𝐵𝐶)) → (𝑍↑2) = (((𝑋↑2) + (𝑌↑2)) − (2 · ((𝑋 · 𝑌) · (cos‘𝑂)))))

Theorempythag 25379* 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 𝐹 is the signed angle construct (as used in ang180 25376), 𝑋 is the distance of line segment BC, 𝑌 is the distance of line segment AC, 𝑍 is the distance of line segment AB (the hypotenuse), and 𝑂 is the signed right angle m/_ BCA. We use the law of cosines lawcos 25378 to prove this, along with simple trigonometry facts like coshalfpi 25038 and cosneg 15478. (Contributed by David A. Wheeler, 13-Jun-2015.)
𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥))))    &   𝑋 = (abs‘(𝐵𝐶))    &   𝑌 = (abs‘(𝐴𝐶))    &   𝑍 = (abs‘(𝐴𝐵))    &   𝑂 = ((𝐵𝐶)𝐹(𝐴𝐶))       (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝐴𝐶𝐵𝐶) ∧ 𝑂 ∈ {(π / 2), -(π / 2)}) → (𝑍↑2) = ((𝑋↑2) + (𝑌↑2)))

Theoremisosctrlem1 25380 Lemma for isosctr 25383. (Contributed by Saveliy Skresanov, 30-Dec-2016.)
((𝐴 ∈ ℂ ∧ (abs‘𝐴) = 1 ∧ ¬ 1 = 𝐴) → (ℑ‘(log‘(1 − 𝐴))) ≠ π)

Theoremisosctrlem2 25381 Lemma for isosctr 25383. 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.)
((𝐴 ∈ ℂ ∧ (abs‘𝐴) = 1 ∧ ¬ 1 = 𝐴) → (ℑ‘(log‘(1 − 𝐴))) = (ℑ‘(log‘(-𝐴 / (1 − 𝐴)))))

Theoremisosctrlem3 25382* Lemma for isosctr 25383. Corresponds to the case where one vertex is at 0. (Contributed by Saveliy Skresanov, 1-Jan-2017.)
𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥))))       (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) ∧ (𝐴 ≠ 0 ∧ 𝐵 ≠ 0 ∧ 𝐴𝐵) ∧ (abs‘𝐴) = (abs‘𝐵)) → (-𝐴𝐹(𝐵𝐴)) = ((𝐴𝐵)𝐹-𝐵))

Theoremisosctr 25383* Isosceles triangle theorem. This is Metamath 100 proof #65. (Contributed by Saveliy Skresanov, 1-Jan-2017.)
𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥))))       (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝐴𝐶𝐵𝐶𝐴𝐵) ∧ (abs‘(𝐴𝐶)) = (abs‘(𝐵𝐶))) → ((𝐶𝐴)𝐹(𝐵𝐴)) = ((𝐴𝐵)𝐹(𝐶𝐵)))

Theoremssscongptld 25384* 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 25378 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.)

𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥))))    &   (𝜑𝐴 ∈ ℂ)    &   (𝜑𝐵 ∈ ℂ)    &   (𝜑𝐶 ∈ ℂ)    &   (𝜑𝐷 ∈ ℂ)    &   (𝜑𝐸 ∈ ℂ)    &   (𝜑𝐺 ∈ ℂ)    &   (𝜑𝐴𝐵)    &   (𝜑𝐵𝐶)    &   (𝜑𝐷𝐸)    &   (𝜑𝐸𝐺)    &   (𝜑 → (abs‘(𝐴𝐵)) = (abs‘(𝐷𝐸)))    &   (𝜑 → (abs‘(𝐵𝐶)) = (abs‘(𝐸𝐺)))    &   (𝜑 → (abs‘(𝐶𝐴)) = (abs‘(𝐺𝐷)))       (𝜑 → (cos‘((𝐴𝐵)𝐹(𝐶𝐵))) = (cos‘((𝐷𝐸)𝐹(𝐺𝐸))))

Theoremaffineequiv 25385 Equivalence between two ways of expressing 𝐵 as an affine combination of 𝐴 and 𝐶. (Contributed by David Moews, 28-Feb-2017.)
(𝜑𝐴 ∈ ℂ)    &   (𝜑𝐵 ∈ ℂ)    &   (𝜑𝐶 ∈ ℂ)    &   (𝜑𝐷 ∈ ℂ)       (𝜑 → (𝐵 = ((𝐷 · 𝐴) + ((1 − 𝐷) · 𝐶)) ↔ (𝐶𝐵) = (𝐷 · (𝐶𝐴))))

Theoremaffineequiv2 25386 Equivalence between two ways of expressing 𝐵 as an affine combination of 𝐴 and 𝐶. (Contributed by David Moews, 28-Feb-2017.)
(𝜑𝐴 ∈ ℂ)    &   (𝜑𝐵 ∈ ℂ)    &   (𝜑𝐶 ∈ ℂ)    &   (𝜑𝐷 ∈ ℂ)       (𝜑 → (𝐵 = ((𝐷 · 𝐴) + ((1 − 𝐷) · 𝐶)) ↔ (𝐵𝐴) = ((1 − 𝐷) · (𝐶𝐴))))

Theoremaffineequiv3 25387 Equivalence between two ways of expressing 𝐴 as an affine combination of 𝐵 and 𝐶. (Contributed by AV, 22-Jan-2023.)
(𝜑𝐴 ∈ ℂ)    &   (𝜑𝐵 ∈ ℂ)    &   (𝜑𝐶 ∈ ℂ)    &   (𝜑𝐷 ∈ ℂ)       (𝜑 → (𝐴 = (((1 − 𝐷) · 𝐵) + (𝐷 · 𝐶)) ↔ (𝐴𝐵) = (𝐷 · (𝐶𝐵))))

Theoremaffineequiv4 25388 Equivalence between two ways of expressing 𝐴 as an affine combination of 𝐵 and 𝐶. (Contributed by AV, 22-Jan-2023.)
(𝜑𝐴 ∈ ℂ)    &   (𝜑𝐵 ∈ ℂ)    &   (𝜑𝐶 ∈ ℂ)    &   (𝜑𝐷 ∈ ℂ)       (𝜑 → (𝐴 = (((1 − 𝐷) · 𝐵) + (𝐷 · 𝐶)) ↔ 𝐴 = ((𝐷 · (𝐶𝐵)) + 𝐵)))

Theoremaffineequivne 25389 Equivalence between two ways of expressing 𝐴 as an affine combination of 𝐵 and 𝐶 if 𝐵 and 𝐶 are not equal. (Contributed by AV, 22-Jan-2023.)
(𝜑𝐴 ∈ ℂ)    &   (𝜑𝐵 ∈ ℂ)    &   (𝜑𝐶 ∈ ℂ)    &   (𝜑𝐷 ∈ ℂ)    &   (𝜑𝐵𝐶)       (𝜑 → (𝐴 = (((1 − 𝐷) · 𝐵) + (𝐷 · 𝐶)) ↔ 𝐷 = ((𝐴𝐵) / (𝐶𝐵))))

Theoremangpieqvdlem 25390 Equivalence used in the proof of angpieqvd 25393. (Contributed by David Moews, 28-Feb-2017.)
(𝜑𝐴 ∈ ℂ)    &   (𝜑𝐵 ∈ ℂ)    &   (𝜑𝐶 ∈ ℂ)    &   (𝜑𝐴𝐵)    &   (𝜑𝐴𝐶)       (𝜑 → (-((𝐶𝐵) / (𝐴𝐵)) ∈ ℝ+ ↔ ((𝐶𝐵) / (𝐶𝐴)) ∈ (0(,)1)))

Theoremangpieqvdlem2 25391* Equivalence used in angpieqvd 25393. (Contributed by David Moews, 28-Feb-2017.)
𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥))))    &   (𝜑𝐴 ∈ ℂ)    &   (𝜑𝐵 ∈ ℂ)    &   (𝜑𝐶 ∈ ℂ)    &   (𝜑𝐴𝐵)    &   (𝜑𝐵𝐶)       (𝜑 → (-((𝐶𝐵) / (𝐴𝐵)) ∈ ℝ+ ↔ ((𝐴𝐵)𝐹(𝐶𝐵)) = π))

Theoremangpined 25392* If the angle at ABC is π, then A is not equal to C. (Contributed by David Moews, 28-Feb-2017.)
𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥))))    &   (𝜑𝐴 ∈ ℂ)    &   (𝜑𝐵 ∈ ℂ)    &   (𝜑𝐶 ∈ ℂ)    &   (𝜑𝐴𝐵)    &   (𝜑𝐵𝐶)       (𝜑 → (((𝐴𝐵)𝐹(𝐶𝐵)) = π → 𝐴𝐶))

Theoremangpieqvd 25393* The angle ABC is π 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.)
𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥))))    &   (𝜑𝐴 ∈ ℂ)    &   (𝜑𝐵 ∈ ℂ)    &   (𝜑𝐶 ∈ ℂ)    &   (𝜑𝐴𝐵)    &   (𝜑𝐵𝐶)       (𝜑 → (((𝐴𝐵)𝐹(𝐶𝐵)) = π ↔ ∃𝑤 ∈ (0(,)1)𝐵 = ((𝑤 · 𝐴) + ((1 − 𝑤) · 𝐶))))

Theoremchordthmlem 25394* If M is the midpoint of AB and AQ = BQ, then QMB is a right angle. The proof uses ssscongptld 25384 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.)
𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥))))    &   (𝜑𝐴 ∈ ℂ)    &   (𝜑𝐵 ∈ ℂ)    &   (𝜑𝑄 ∈ ℂ)    &   (𝜑𝑀 = ((𝐴 + 𝐵) / 2))    &   (𝜑 → (abs‘(𝐴𝑄)) = (abs‘(𝐵𝑄)))    &   (𝜑𝐴𝐵)    &   (𝜑𝑄𝑀)       (𝜑 → ((𝑄𝑀)𝐹(𝐵𝑀)) ∈ {(π / 2), -(π / 2)})

Theoremchordthmlem2 25395* 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 25394, where P = B, and using angrtmuld 25370 to observe that QMP is right iff QMB is. (Contributed by David Moews, 28-Feb-2017.)
𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥))))    &   (𝜑𝐴 ∈ ℂ)    &   (𝜑𝐵 ∈ ℂ)    &   (𝜑𝑄 ∈ ℂ)    &   (𝜑𝑋 ∈ ℝ)    &   (𝜑𝑀 = ((𝐴 + 𝐵) / 2))    &   (𝜑𝑃 = ((𝑋 · 𝐴) + ((1 − 𝑋) · 𝐵)))    &   (𝜑 → (abs‘(𝐴𝑄)) = (abs‘(𝐵𝑄)))    &   (𝜑𝑃𝑀)    &   (𝜑𝑄𝑀)       (𝜑 → ((𝑄𝑀)𝐹(𝑃𝑀)) ∈ {(π / 2), -(π / 2)})

Theoremchordthmlem3 25396 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 25395 and the Pythagorean theorem (pythag 25379) 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.)
(𝜑𝐴 ∈ ℂ)    &   (𝜑𝐵 ∈ ℂ)    &   (𝜑𝑄 ∈ ℂ)    &   (𝜑𝑋 ∈ ℝ)    &   (𝜑𝑀 = ((𝐴 + 𝐵) / 2))    &   (𝜑𝑃 = ((𝑋 · 𝐴) + ((1 − 𝑋) · 𝐵)))    &   (𝜑 → (abs‘(𝐴𝑄)) = (abs‘(𝐵𝑄)))       (𝜑 → ((abs‘(𝑃𝑄))↑2) = (((abs‘(𝑄𝑀))↑2) + ((abs‘(𝑃𝑀))↑2)))

Theoremchordthmlem4 25397 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 𝑋 · (1 − 𝑋) = (1 / 2) 2 − ((1 / 2) − 𝑋) 2 . (Contributed by David Moews, 28-Feb-2017.)
(𝜑𝐴 ∈ ℂ)    &   (𝜑𝐵 ∈ ℂ)    &   (𝜑𝑋 ∈ (0[,]1))    &   (𝜑𝑀 = ((𝐴 + 𝐵) / 2))    &   (𝜑𝑃 = ((𝑋 · 𝐴) + ((1 − 𝑋) · 𝐵)))       (𝜑 → ((abs‘(𝑃𝐴)) · (abs‘(𝑃𝐵))) = (((abs‘(𝐵𝑀))↑2) − ((abs‘(𝑃𝑀))↑2)))

Theoremchordthmlem5 25398 If P is on the segment AB and AQ = BQ, then PA · PB = BQ 2 PQ 2 . This follows from two uses of chordthmlem3 25396 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 25397. (Contributed by David Moews, 28-Feb-2017.)
(𝜑𝐴 ∈ ℂ)    &   (𝜑𝐵 ∈ ℂ)    &   (𝜑𝑄 ∈ ℂ)    &   (𝜑𝑋 ∈ (0[,]1))    &   (𝜑𝑃 = ((𝑋 · 𝐴) + ((1 − 𝑋) · 𝐵)))    &   (𝜑 → (abs‘(𝐴𝑄)) = (abs‘(𝐵𝑄)))       (𝜑 → ((abs‘(𝑃𝐴)) · (abs‘(𝑃𝐵))) = (((abs‘(𝐵𝑄))↑2) − ((abs‘(𝑃𝑄))↑2)))

Theoremchordthm 25399* 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 π. The result is proven by using chordthmlem5 25398 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. This is Metamath 100 proof #55. (Contributed by David Moews, 28-Feb-2017.)
𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥))))    &   (𝜑𝐴 ∈ ℂ)    &   (𝜑𝐵 ∈ ℂ)    &   (𝜑𝐶 ∈ ℂ)    &   (𝜑𝐷 ∈ ℂ)    &   (𝜑𝑃 ∈ ℂ)    &   (𝜑𝐴𝑃)    &   (𝜑𝐵𝑃)    &   (𝜑𝐶𝑃)    &   (𝜑𝐷𝑃)    &   (𝜑 → ((𝐴𝑃)𝐹(𝐵𝑃)) = π)    &   (𝜑 → ((𝐶𝑃)𝐹(𝐷𝑃)) = π)    &   (𝜑𝑄 ∈ ℂ)    &   (𝜑 → (abs‘(𝐴𝑄)) = (abs‘(𝐵𝑄)))    &   (𝜑 → (abs‘(𝐴𝑄)) = (abs‘(𝐶𝑄)))    &   (𝜑 → (abs‘(𝐴𝑄)) = (abs‘(𝐷𝑄)))       (𝜑 → ((abs‘(𝑃𝐴)) · (abs‘(𝑃𝐵))) = ((abs‘(𝑃𝐶)) · (abs‘(𝑃𝐷))))

Theoremheron 25400* 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) · 𝑋 · 𝑌 · abs(sin𝑂), is equal to the square root of 𝑆 · (𝑆𝑋) · (𝑆𝑌) · (𝑆𝑍), where 𝑆 = (𝑋 + 𝑌 + 𝑍) / 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.)
𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥))))    &   𝑋 = (abs‘(𝐵𝐶))    &   𝑌 = (abs‘(𝐴𝐶))    &   𝑍 = (abs‘(𝐴𝐵))    &   𝑂 = ((𝐵𝐶)𝐹(𝐴𝐶))    &   𝑆 = (((𝑋 + 𝑌) + 𝑍) / 2)    &   (𝜑𝐴 ∈ ℂ)    &   (𝜑𝐵 ∈ ℂ)    &   (𝜑𝐶 ∈ ℂ)    &   (𝜑𝐴𝐶)    &   (𝜑𝐵𝐶)       (𝜑 → (((1 / 2) · (𝑋 · 𝑌)) · (abs‘(sin‘𝑂))) = (√‘((𝑆 · (𝑆𝑋)) · ((𝑆𝑌) · (𝑆𝑍)))))

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