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Theorem List for Metamath Proof Explorer - 13001-13100   *Has distinct variable group(s)
TypeLabelDescription
Statement
 
Theoremsqcld 13001 Closure of square. (Contributed by Mario Carneiro, 28-May-2016.)
(𝜑𝐴 ∈ ℂ)       (𝜑 → (𝐴↑2) ∈ ℂ)
 
Theoremsqeq0d 13002 A number is zero iff its square is zero. (Contributed by Mario Carneiro, 28-May-2016.)
(𝜑𝐴 ∈ ℂ)    &   (𝜑 → (𝐴↑2) = 0)       (𝜑𝐴 = 0)
 
Theoremexpcld 13003 Closure law for nonnegative integer exponentiation. (Contributed by Mario Carneiro, 28-May-2016.)
(𝜑𝐴 ∈ ℂ)    &   (𝜑𝑁 ∈ ℕ0)       (𝜑 → (𝐴𝑁) ∈ ℂ)
 
Theoremexpp1d 13004 Value of a complex number raised to a nonnegative integer power plus one. Part of Definition 10-4.1 of [Gleason] p. 134. (Contributed by Mario Carneiro, 28-May-2016.)
(𝜑𝐴 ∈ ℂ)    &   (𝜑𝑁 ∈ ℕ0)       (𝜑 → (𝐴↑(𝑁 + 1)) = ((𝐴𝑁) · 𝐴))
 
Theoremexpaddd 13005 Sum of exponents law for nonnegative integer exponentiation. Proposition 10-4.2(a) of [Gleason] p. 135. (Contributed by Mario Carneiro, 28-May-2016.)
(𝜑𝐴 ∈ ℂ)    &   (𝜑𝑁 ∈ ℕ0)    &   (𝜑𝑀 ∈ ℕ0)       (𝜑 → (𝐴↑(𝑀 + 𝑁)) = ((𝐴𝑀) · (𝐴𝑁)))
 
Theoremexpmuld 13006 Product of exponents law for positive integer exponentiation. Proposition 10-4.2(b) of [Gleason] p. 135, restricted to nonnegative integer exponents. (Contributed by Mario Carneiro, 28-May-2016.)
(𝜑𝐴 ∈ ℂ)    &   (𝜑𝑁 ∈ ℕ0)    &   (𝜑𝑀 ∈ ℕ0)       (𝜑 → (𝐴↑(𝑀 · 𝑁)) = ((𝐴𝑀)↑𝑁))
 
Theoremsqrecd 13007 Square of reciprocal. (Contributed by Mario Carneiro, 28-May-2016.)
(𝜑𝐴 ∈ ℂ)    &   (𝜑𝐴 ≠ 0)       (𝜑 → ((1 / 𝐴)↑2) = (1 / (𝐴↑2)))
 
Theoremexpclzd 13008 Closure law for integer exponentiation. (Contributed by Mario Carneiro, 28-May-2016.)
(𝜑𝐴 ∈ ℂ)    &   (𝜑𝐴 ≠ 0)    &   (𝜑𝑁 ∈ ℤ)       (𝜑 → (𝐴𝑁) ∈ ℂ)
 
Theoremexpne0d 13009 Nonnegative integer exponentiation is nonzero if its mantissa is nonzero. (Contributed by Mario Carneiro, 28-May-2016.)
(𝜑𝐴 ∈ ℂ)    &   (𝜑𝐴 ≠ 0)    &   (𝜑𝑁 ∈ ℤ)       (𝜑 → (𝐴𝑁) ≠ 0)
 
Theoremexpnegd 13010 Value of a complex number raised to a negative power. (Contributed by Mario Carneiro, 28-May-2016.)
(𝜑𝐴 ∈ ℂ)    &   (𝜑𝐴 ≠ 0)    &   (𝜑𝑁 ∈ ℤ)       (𝜑 → (𝐴↑-𝑁) = (1 / (𝐴𝑁)))
 
Theoremexprecd 13011 Nonnegative integer exponentiation of a reciprocal. (Contributed by Mario Carneiro, 28-May-2016.)
(𝜑𝐴 ∈ ℂ)    &   (𝜑𝐴 ≠ 0)    &   (𝜑𝑁 ∈ ℤ)       (𝜑 → ((1 / 𝐴)↑𝑁) = (1 / (𝐴𝑁)))
 
Theoremexpp1zd 13012 Value of a nonzero complex number raised to an integer power plus one. (Contributed by Mario Carneiro, 28-May-2016.)
(𝜑𝐴 ∈ ℂ)    &   (𝜑𝐴 ≠ 0)    &   (𝜑𝑁 ∈ ℤ)       (𝜑 → (𝐴↑(𝑁 + 1)) = ((𝐴𝑁) · 𝐴))
 
Theoremexpm1d 13013 Value of a complex number raised to an integer power minus one. (Contributed by Mario Carneiro, 28-May-2016.)
(𝜑𝐴 ∈ ℂ)    &   (𝜑𝐴 ≠ 0)    &   (𝜑𝑁 ∈ ℤ)       (𝜑 → (𝐴↑(𝑁 − 1)) = ((𝐴𝑁) / 𝐴))
 
Theoremexpsubd 13014 Exponent subtraction law for nonnegative integer exponentiation. (Contributed by Mario Carneiro, 28-May-2016.)
(𝜑𝐴 ∈ ℂ)    &   (𝜑𝐴 ≠ 0)    &   (𝜑𝑁 ∈ ℤ)    &   (𝜑𝑀 ∈ ℤ)       (𝜑 → (𝐴↑(𝑀𝑁)) = ((𝐴𝑀) / (𝐴𝑁)))
 
Theoremsqmuld 13015 Distribution of square over multiplication. (Contributed by Mario Carneiro, 28-May-2016.)
(𝜑𝐴 ∈ ℂ)    &   (𝜑𝐵 ∈ ℂ)       (𝜑 → ((𝐴 · 𝐵)↑2) = ((𝐴↑2) · (𝐵↑2)))
 
Theoremsqdivd 13016 Distribution of square over division. (Contributed by Mario Carneiro, 28-May-2016.)
(𝜑𝐴 ∈ ℂ)    &   (𝜑𝐵 ∈ ℂ)    &   (𝜑𝐵 ≠ 0)       (𝜑 → ((𝐴 / 𝐵)↑2) = ((𝐴↑2) / (𝐵↑2)))
 
Theoremexpdivd 13017 Nonnegative integer exponentiation of a quotient. (Contributed by Mario Carneiro, 28-May-2016.)
(𝜑𝐴 ∈ ℂ)    &   (𝜑𝐵 ∈ ℂ)    &   (𝜑𝐵 ≠ 0)    &   (𝜑𝑁 ∈ ℕ0)       (𝜑 → ((𝐴 / 𝐵)↑𝑁) = ((𝐴𝑁) / (𝐵𝑁)))
 
Theoremmulexpd 13018 Positive integer exponentiation of a product. Proposition 10-4.2(c) of [Gleason] p. 135, restricted to nonnegative integer exponents. (Contributed by Mario Carneiro, 28-May-2016.)
(𝜑𝐴 ∈ ℂ)    &   (𝜑𝐵 ∈ ℂ)    &   (𝜑𝑁 ∈ ℕ0)       (𝜑 → ((𝐴 · 𝐵)↑𝑁) = ((𝐴𝑁) · (𝐵𝑁)))
 
Theorem0expd 13019 Value of zero raised to a positive integer power. (Contributed by Mario Carneiro, 28-May-2016.)
(𝜑𝑁 ∈ ℕ)       (𝜑 → (0↑𝑁) = 0)
 
Theoremreexpcld 13020 Closure of exponentiation of reals. (Contributed by Mario Carneiro, 28-May-2016.)
(𝜑𝐴 ∈ ℝ)    &   (𝜑𝑁 ∈ ℕ0)       (𝜑 → (𝐴𝑁) ∈ ℝ)
 
Theoremexpge0d 13021 Nonnegative integer exponentiation with a nonnegative mantissa is nonnegative. (Contributed by Mario Carneiro, 28-May-2016.)
(𝜑𝐴 ∈ ℝ)    &   (𝜑𝑁 ∈ ℕ0)    &   (𝜑 → 0 ≤ 𝐴)       (𝜑 → 0 ≤ (𝐴𝑁))
 
Theoremexpge1d 13022 Nonnegative integer exponentiation with a nonnegative mantissa is nonnegative. (Contributed by Mario Carneiro, 28-May-2016.)
(𝜑𝐴 ∈ ℝ)    &   (𝜑𝑁 ∈ ℕ0)    &   (𝜑 → 1 ≤ 𝐴)       (𝜑 → 1 ≤ (𝐴𝑁))
 
Theoremsqoddm1div8 13023 A squared odd number minus 1 divided by 8 is the odd number multiplied with its successor divided by 2. (Contributed by AV, 19-Jul-2021.)
((𝑁 ∈ ℤ ∧ 𝑀 = ((2 · 𝑁) + 1)) → (((𝑀↑2) − 1) / 8) = ((𝑁 · (𝑁 + 1)) / 2))
 
Theoremnnsqcld 13024 The naturals are closed under squaring. (Contributed by Mario Carneiro, 28-May-2016.)
(𝜑𝐴 ∈ ℕ)       (𝜑 → (𝐴↑2) ∈ ℕ)
 
Theoremnnexpcld 13025 Closure of exponentiation of nonnegative integers. (Contributed by Mario Carneiro, 28-May-2016.)
(𝜑𝐴 ∈ ℕ)    &   (𝜑𝑁 ∈ ℕ0)       (𝜑 → (𝐴𝑁) ∈ ℕ)
 
Theoremnn0expcld 13026 Closure of exponentiation of nonnegative integers. (Contributed by Mario Carneiro, 28-May-2016.)
(𝜑𝐴 ∈ ℕ0)    &   (𝜑𝑁 ∈ ℕ0)       (𝜑 → (𝐴𝑁) ∈ ℕ0)
 
Theoremrpexpcld 13027 Closure law for exponentiation of positive reals. (Contributed by Mario Carneiro, 28-May-2016.)
(𝜑𝐴 ∈ ℝ+)    &   (𝜑𝑁 ∈ ℤ)       (𝜑 → (𝐴𝑁) ∈ ℝ+)
 
Theoremltexp2rd 13028 The power of a positive number smaller than 1 decreases as its exponent increases. (Contributed by Mario Carneiro, 28-May-2016.)
(𝜑𝐴 ∈ ℝ+)    &   (𝜑𝑁 ∈ ℤ)    &   (𝜑𝑀 ∈ ℤ)    &   (𝜑𝐴 < 1)       (𝜑 → (𝑀 < 𝑁 ↔ (𝐴𝑁) < (𝐴𝑀)))
 
Theoremreexpclzd 13029 Closure of exponentiation of reals. (Contributed by Mario Carneiro, 28-May-2016.)
(𝜑𝐴 ∈ ℝ)    &   (𝜑𝐴 ≠ 0)    &   (𝜑𝑁 ∈ ℤ)       (𝜑 → (𝐴𝑁) ∈ ℝ)
 
Theoremresqcld 13030 Closure of square in reals. (Contributed by Mario Carneiro, 28-May-2016.)
(𝜑𝐴 ∈ ℝ)       (𝜑 → (𝐴↑2) ∈ ℝ)
 
Theoremsqge0d 13031 A square of a real is nonnegative. (Contributed by Mario Carneiro, 28-May-2016.)
(𝜑𝐴 ∈ ℝ)       (𝜑 → 0 ≤ (𝐴↑2))
 
Theoremsqgt0d 13032 The square of a nonzero real is positive. (Contributed by Mario Carneiro, 28-May-2016.)
(𝜑𝐴 ∈ ℝ)    &   (𝜑𝐴 ≠ 0)       (𝜑 → 0 < (𝐴↑2))
 
Theoremltexp2d 13033 Ordering relationship for exponentiation. (Contributed by Mario Carneiro, 28-May-2016.)
(𝜑𝐴 ∈ ℝ)    &   (𝜑𝑀 ∈ ℤ)    &   (𝜑𝑁 ∈ ℤ)    &   (𝜑 → 1 < 𝐴)       (𝜑 → (𝑀 < 𝑁 ↔ (𝐴𝑀) < (𝐴𝑁)))
 
Theoremleexp2d 13034 Ordering law for exponentiation. (Contributed by Mario Carneiro, 28-May-2016.)
(𝜑𝐴 ∈ ℝ)    &   (𝜑𝑀 ∈ ℤ)    &   (𝜑𝑁 ∈ ℤ)    &   (𝜑 → 1 < 𝐴)       (𝜑 → (𝑀𝑁 ↔ (𝐴𝑀) ≤ (𝐴𝑁)))
 
Theoremexpcand 13035 Ordering relationship for exponentiation. (Contributed by Mario Carneiro, 28-May-2016.)
(𝜑𝐴 ∈ ℝ)    &   (𝜑𝑀 ∈ ℤ)    &   (𝜑𝑁 ∈ ℤ)    &   (𝜑 → 1 < 𝐴)    &   (𝜑 → (𝐴𝑀) = (𝐴𝑁))       (𝜑𝑀 = 𝑁)
 
Theoremleexp2ad 13036 Ordering relationship for exponentiation. (Contributed by Mario Carneiro, 28-May-2016.)
(𝜑𝐴 ∈ ℝ)    &   (𝜑 → 1 ≤ 𝐴)    &   (𝜑𝑁 ∈ (ℤ𝑀))       (𝜑 → (𝐴𝑀) ≤ (𝐴𝑁))
 
Theoremleexp2rd 13037 Ordering relationship for exponentiation. (Contributed by Mario Carneiro, 28-May-2016.)
(𝜑𝐴 ∈ ℝ)    &   (𝜑𝑀 ∈ ℕ0)    &   (𝜑𝑁 ∈ (ℤ𝑀))    &   (𝜑 → 0 ≤ 𝐴)    &   (𝜑𝐴 ≤ 1)       (𝜑 → (𝐴𝑁) ≤ (𝐴𝑀))
 
Theoremlt2sqd 13038 The square function on nonnegative reals is strictly monotonic. (Contributed by Mario Carneiro, 28-May-2016.)
(𝜑𝐴 ∈ ℝ)    &   (𝜑𝐵 ∈ ℝ)    &   (𝜑 → 0 ≤ 𝐴)    &   (𝜑 → 0 ≤ 𝐵)       (𝜑 → (𝐴 < 𝐵 ↔ (𝐴↑2) < (𝐵↑2)))
 
Theoremle2sqd 13039 The square function on nonnegative reals is monotonic. (Contributed by Mario Carneiro, 28-May-2016.)
(𝜑𝐴 ∈ ℝ)    &   (𝜑𝐵 ∈ ℝ)    &   (𝜑 → 0 ≤ 𝐴)    &   (𝜑 → 0 ≤ 𝐵)       (𝜑 → (𝐴𝐵 ↔ (𝐴↑2) ≤ (𝐵↑2)))
 
Theoremsq11d 13040 The square function is one-to-one for nonnegative reals. (Contributed by Mario Carneiro, 28-May-2016.)
(𝜑𝐴 ∈ ℝ)    &   (𝜑𝐵 ∈ ℝ)    &   (𝜑 → 0 ≤ 𝐴)    &   (𝜑 → 0 ≤ 𝐵)    &   (𝜑 → (𝐴↑2) = (𝐵↑2))       (𝜑𝐴 = 𝐵)
 
Theoremmulsubdivbinom2 13041 The square of a binomial with factor minus a number divided by a nonzero number. (Contributed by AV, 19-Jul-2021.)
(((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐷 ∈ ℂ) ∧ (𝐶 ∈ ℂ ∧ 𝐶 ≠ 0)) → (((((𝐶 · 𝐴) + 𝐵)↑2) − 𝐷) / 𝐶) = (((𝐶 · (𝐴↑2)) + (2 · (𝐴 · 𝐵))) + (((𝐵↑2) − 𝐷) / 𝐶)))
 
Theoremmuldivbinom2 13042 The square of a binomial with factor divided by a nonzero number. (Contributed by AV, 19-Jul-2021.)
((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ (𝐶 ∈ ℂ ∧ 𝐶 ≠ 0)) → ((((𝐶 · 𝐴) + 𝐵)↑2) / 𝐶) = (((𝐶 · (𝐴↑2)) + (2 · (𝐴 · 𝐵))) + ((𝐵↑2) / 𝐶)))
 
Theoremsq10 13043 The square of 10 is 100. (Contributed by AV, 14-Jun-2021.) (Revised by AV, 1-Aug-2021.)
(10↑2) = 100
 
Theoremsq10e99m1 13044 The square of 10 is 99 plus 1. (Contributed by AV, 14-Jun-2021.) (Revised by AV, 1-Aug-2021.)
(10↑2) = (99 + 1)
 
Theorem3dec 13045 A "decimal constructor" which is used to build up "decimal integers" or "numeric terms" in base 10 with 3 "digits". (Contributed by AV, 14-Jun-2021.) (Revised by AV, 1-Aug-2021.)
𝐴 ∈ ℕ0    &   𝐵 ∈ ℕ0       𝐴𝐵𝐶 = ((((10↑2) · 𝐴) + (10 · 𝐵)) + 𝐶)
 
Theoremsq10OLD 13046 Old version of sq10 13043. Obsolete as of 1-Aug-2021. (Contributed by AV, 14-Jun-2021.) (Proof modification is discouraged.) (New usage is discouraged.)
(10↑2) = 100
 
Theoremsq10e99m1OLD 13047 Old version of sq10e99m1 13044. Obsolete as of 1-Aug-2021. (Contributed by AV, 14-Jun-2021.) (Proof modification is discouraged.) (New usage is discouraged.)
(10↑2) = (99 + 1)
 
Theorem3decOLD 13048 Old version of 3dec 13045. Obsolete as of 1-Aug-2021. (Contributed by AV, 14-Jun-2021.) (Proof modification is discouraged.) (New usage is discouraged.)
𝐴 ∈ ℕ0    &   𝐵 ∈ ℕ0       𝐴𝐵𝐶 = ((((10↑2) · 𝐴) + (10 · 𝐵)) + 𝐶)
 
5.6.8  Ordered pair theorem for nonnegative integers
 
Theoremnn0le2msqi 13049 The square function on nonnegative integers is monotonic. (Contributed by Raph Levien, 10-Dec-2002.)
𝐴 ∈ ℕ0    &   𝐵 ∈ ℕ0       (𝐴𝐵 ↔ (𝐴 · 𝐴) ≤ (𝐵 · 𝐵))
 
Theoremnn0opthlem1 13050 A rather pretty lemma for nn0opthi 13052. (Contributed by Raph Levien, 10-Dec-2002.)
𝐴 ∈ ℕ0    &   𝐶 ∈ ℕ0       (𝐴 < 𝐶 ↔ ((𝐴 · 𝐴) + (2 · 𝐴)) < (𝐶 · 𝐶))
 
Theoremnn0opthlem2 13051 Lemma for nn0opthi 13052. (Contributed by Raph Levien, 10-Dec-2002.) (Revised by Scott Fenton, 8-Sep-2010.)
𝐴 ∈ ℕ0    &   𝐵 ∈ ℕ0    &   𝐶 ∈ ℕ0    &   𝐷 ∈ ℕ0       ((𝐴 + 𝐵) < 𝐶 → ((𝐶 · 𝐶) + 𝐷) ≠ (((𝐴 + 𝐵) · (𝐴 + 𝐵)) + 𝐵))
 
Theoremnn0opthi 13052 An ordered pair theorem for nonnegative integers. Theorem 17.3 of [Quine] p. 124. We can represent an ordered pair of nonnegative integers 𝐴 and 𝐵 by (((𝐴 + 𝐵) · (𝐴 + 𝐵)) + 𝐵). If two such ordered pairs are equal, their first elements are equal and their second elements are equal. Contrast this ordered pair representation with the standard one df-op 4182 that works for any set. (Contributed by Raph Levien, 10-Dec-2002.) (Proof shortened by Scott Fenton, 8-Sep-2010.)
𝐴 ∈ ℕ0    &   𝐵 ∈ ℕ0    &   𝐶 ∈ ℕ0    &   𝐷 ∈ ℕ0       ((((𝐴 + 𝐵) · (𝐴 + 𝐵)) + 𝐵) = (((𝐶 + 𝐷) · (𝐶 + 𝐷)) + 𝐷) ↔ (𝐴 = 𝐶𝐵 = 𝐷))
 
Theoremnn0opth2i 13053 An ordered pair theorem for nonnegative integers. Theorem 17.3 of [Quine] p. 124. See comments for nn0opthi 13052. (Contributed by NM, 22-Jul-2004.)
𝐴 ∈ ℕ0    &   𝐵 ∈ ℕ0    &   𝐶 ∈ ℕ0    &   𝐷 ∈ ℕ0       ((((𝐴 + 𝐵)↑2) + 𝐵) = (((𝐶 + 𝐷)↑2) + 𝐷) ↔ (𝐴 = 𝐶𝐵 = 𝐷))
 
Theoremnn0opth2 13054 An ordered pair theorem for nonnegative integers. Theorem 17.3 of [Quine] p. 124. See nn0opthi 13052. (Contributed by NM, 22-Jul-2004.)
(((𝐴 ∈ ℕ0𝐵 ∈ ℕ0) ∧ (𝐶 ∈ ℕ0𝐷 ∈ ℕ0)) → ((((𝐴 + 𝐵)↑2) + 𝐵) = (((𝐶 + 𝐷)↑2) + 𝐷) ↔ (𝐴 = 𝐶𝐵 = 𝐷)))
 
5.6.9  Factorial function
 
Syntaxcfa 13055 Extend class notation to include the factorial of nonnegative integers.
class !
 
Definitiondf-fac 13056 Define the factorial function on nonnegative integers. For example, (!‘5) = 120 because 1 · 2 · 3 · 4 · 5 = 120 (ex-fac 27292). In the literature, the factorial function is written as a postscript exclamation point. (Contributed by NM, 2-Dec-2004.)
! = ({⟨0, 1⟩} ∪ seq1( · , I ))
 
Theoremfacnn 13057 Value of the factorial function for positive integers. (Contributed by NM, 2-Dec-2004.) (Revised by Mario Carneiro, 13-Jul-2013.)
(𝑁 ∈ ℕ → (!‘𝑁) = (seq1( · , I )‘𝑁))
 
Theoremfac0 13058 The factorial of 0. (Contributed by NM, 2-Dec-2004.) (Revised by Mario Carneiro, 13-Jul-2013.)
(!‘0) = 1
 
Theoremfac1 13059 The factorial of 1. (Contributed by NM, 2-Dec-2004.) (Revised by Mario Carneiro, 13-Jul-2013.)
(!‘1) = 1
 
Theoremfacp1 13060 The factorial of a successor. (Contributed by NM, 2-Dec-2004.) (Revised by Mario Carneiro, 13-Jul-2013.)
(𝑁 ∈ ℕ0 → (!‘(𝑁 + 1)) = ((!‘𝑁) · (𝑁 + 1)))
 
Theoremfac2 13061 The factorial of 2. (Contributed by NM, 17-Mar-2005.)
(!‘2) = 2
 
Theoremfac3 13062 The factorial of 3. (Contributed by NM, 17-Mar-2005.)
(!‘3) = 6
 
Theoremfac4 13063 The factorial of 4. (Contributed by Mario Carneiro, 18-Jun-2015.)
(!‘4) = 24
 
Theoremfacnn2 13064 Value of the factorial function expressed recursively. (Contributed by NM, 2-Dec-2004.)
(𝑁 ∈ ℕ → (!‘𝑁) = ((!‘(𝑁 − 1)) · 𝑁))
 
Theoremfaccl 13065 Closure of the factorial function. (Contributed by NM, 2-Dec-2004.)
(𝑁 ∈ ℕ0 → (!‘𝑁) ∈ ℕ)
 
Theoremfaccld 13066 Closure of the factorial function, deduction version of faccl 13065. (Contributed by Glauco Siliprandi, 5-Apr-2020.)
(𝜑𝑁 ∈ ℕ0)       (𝜑 → (!‘𝑁) ∈ ℕ)
 
Theoremfacmapnn 13067 The factorial function restricted to positive integers is a mapping from the positive integers to the positive integers. (Contributed by AV, 8-Aug-2020.)
(𝑛 ∈ ℕ ↦ (!‘𝑛)) ∈ (ℕ ↑𝑚 ℕ)
 
Theoremfacne0 13068 The factorial function is nonzero. (Contributed by NM, 26-Apr-2005.)
(𝑁 ∈ ℕ0 → (!‘𝑁) ≠ 0)
 
Theoremfacdiv 13069 A positive integer divides the factorial of an equal or larger number. (Contributed by NM, 2-May-2005.)
((𝑀 ∈ ℕ0𝑁 ∈ ℕ ∧ 𝑁𝑀) → ((!‘𝑀) / 𝑁) ∈ ℕ)
 
Theoremfacndiv 13070 No positive integer (greater than one) divides the factorial plus one of an equal or larger number. (Contributed by NM, 3-May-2005.)
(((𝑀 ∈ ℕ0𝑁 ∈ ℕ) ∧ (1 < 𝑁𝑁𝑀)) → ¬ (((!‘𝑀) + 1) / 𝑁) ∈ ℤ)
 
Theoremfacwordi 13071 Ordering property of factorial. (Contributed by NM, 9-Dec-2005.)
((𝑀 ∈ ℕ0𝑁 ∈ ℕ0𝑀𝑁) → (!‘𝑀) ≤ (!‘𝑁))
 
Theoremfaclbnd 13072 A lower bound for the factorial function. (Contributed by NM, 17-Dec-2005.)
((𝑀 ∈ ℕ0𝑁 ∈ ℕ0) → (𝑀↑(𝑁 + 1)) ≤ ((𝑀𝑀) · (!‘𝑁)))
 
Theoremfaclbnd2 13073 A lower bound for the factorial function. (Contributed by NM, 17-Dec-2005.)
(𝑁 ∈ ℕ0 → ((2↑𝑁) / 2) ≤ (!‘𝑁))
 
Theoremfaclbnd3 13074 A lower bound for the factorial function. (Contributed by NM, 19-Dec-2005.)
((𝑀 ∈ ℕ0𝑁 ∈ ℕ0) → (𝑀𝑁) ≤ ((𝑀𝑀) · (!‘𝑁)))
 
Theoremfaclbnd4lem1 13075 Lemma for faclbnd4 13079. Prepare the induction step. (Contributed by NM, 20-Dec-2005.)
𝑁 ∈ ℕ    &   𝐾 ∈ ℕ0    &   𝑀 ∈ ℕ0       ((((𝑁 − 1)↑𝐾) · (𝑀↑(𝑁 − 1))) ≤ (((2↑(𝐾↑2)) · (𝑀↑(𝑀 + 𝐾))) · (!‘(𝑁 − 1))) → ((𝑁↑(𝐾 + 1)) · (𝑀𝑁)) ≤ (((2↑((𝐾 + 1)↑2)) · (𝑀↑(𝑀 + (𝐾 + 1)))) · (!‘𝑁)))
 
Theoremfaclbnd4lem2 13076 Lemma for faclbnd4 13079. Use the weak deduction theorem to convert the hypotheses of faclbnd4lem1 13075 to antecedents. (Contributed by NM, 23-Dec-2005.)
((𝑀 ∈ ℕ0𝐾 ∈ ℕ0𝑁 ∈ ℕ) → ((((𝑁 − 1)↑𝐾) · (𝑀↑(𝑁 − 1))) ≤ (((2↑(𝐾↑2)) · (𝑀↑(𝑀 + 𝐾))) · (!‘(𝑁 − 1))) → ((𝑁↑(𝐾 + 1)) · (𝑀𝑁)) ≤ (((2↑((𝐾 + 1)↑2)) · (𝑀↑(𝑀 + (𝐾 + 1)))) · (!‘𝑁))))
 
Theoremfaclbnd4lem3 13077 Lemma for faclbnd4 13079. The 𝑁 = 0 case. (Contributed by NM, 23-Dec-2005.)
(((𝑀 ∈ ℕ0𝐾 ∈ ℕ0) ∧ 𝑁 = 0) → ((𝑁𝐾) · (𝑀𝑁)) ≤ (((2↑(𝐾↑2)) · (𝑀↑(𝑀 + 𝐾))) · (!‘𝑁)))
 
Theoremfaclbnd4lem4 13078 Lemma for faclbnd4 13079. Prove the 0 < 𝑁 case by induction on 𝐾. (Contributed by NM, 19-Dec-2005.)
((𝑁 ∈ ℕ ∧ 𝐾 ∈ ℕ0𝑀 ∈ ℕ0) → ((𝑁𝐾) · (𝑀𝑁)) ≤ (((2↑(𝐾↑2)) · (𝑀↑(𝑀 + 𝐾))) · (!‘𝑁)))
 
Theoremfaclbnd4 13079 Variant of faclbnd5 13080 providing a non-strict lower bound. (Contributed by NM, 23-Dec-2005.)
((𝑁 ∈ ℕ0𝐾 ∈ ℕ0𝑀 ∈ ℕ0) → ((𝑁𝐾) · (𝑀𝑁)) ≤ (((2↑(𝐾↑2)) · (𝑀↑(𝑀 + 𝐾))) · (!‘𝑁)))
 
Theoremfaclbnd5 13080 The factorial function grows faster than powers and exponentiations. If we consider 𝐾 and 𝑀 to be constants, the right-hand side of the inequality is a constant times 𝑁-factorial. (Contributed by NM, 24-Dec-2005.)
((𝑁 ∈ ℕ0𝐾 ∈ ℕ0𝑀 ∈ ℕ) → ((𝑁𝐾) · (𝑀𝑁)) < ((2 · ((2↑(𝐾↑2)) · (𝑀↑(𝑀 + 𝐾)))) · (!‘𝑁)))
 
Theoremfaclbnd6 13081 Geometric lower bound for the factorial function, where N is usually held constant. (Contributed by Paul Chapman, 28-Dec-2007.)
((𝑁 ∈ ℕ0𝑀 ∈ ℕ0) → ((!‘𝑁) · ((𝑁 + 1)↑𝑀)) ≤ (!‘(𝑁 + 𝑀)))
 
Theoremfacubnd 13082 An upper bound for the factorial function. (Contributed by Mario Carneiro, 15-Apr-2016.)
(𝑁 ∈ ℕ0 → (!‘𝑁) ≤ (𝑁𝑁))
 
Theoremfacavg 13083 The product of two factorials is greater than or equal to the factorial of (the floor of) their average. (Contributed by NM, 9-Dec-2005.)
((𝑀 ∈ ℕ0𝑁 ∈ ℕ0) → (!‘(⌊‘((𝑀 + 𝑁) / 2))) ≤ ((!‘𝑀) · (!‘𝑁)))
 
5.6.10  The binomial coefficient operation
 
Syntaxcbc 13084 Extend class notation to include the binomial coefficient operation (combinatorial choose operation).
class C
 
Definitiondf-bc 13085* Define the binomial coefficient operation. For example, (5C3) = 10 (ex-bc 27293).

In the literature, this function is often written as a column vector of the two arguments, or with the arguments as subscripts before and after the letter "C". (𝑁C𝐾) is read "𝑁 choose 𝐾." Definition of binomial coefficient in [Gleason] p. 295. As suggested by Gleason, we define it to be 0 when 0 ≤ 𝑘𝑛 does not hold. (Contributed by NM, 10-Jul-2005.)

C = (𝑛 ∈ ℕ0, 𝑘 ∈ ℤ ↦ if(𝑘 ∈ (0...𝑛), ((!‘𝑛) / ((!‘(𝑛𝑘)) · (!‘𝑘))), 0))
 
Theorembcval 13086 Value of the binomial coefficient, 𝑁 choose 𝐾. Definition of binomial coefficient in [Gleason] p. 295. As suggested by Gleason, we define it to be 0 when 0 ≤ 𝐾𝑁 does not hold. See bcval2 13087 for the value in the standard domain. (Contributed by NM, 10-Jul-2005.) (Revised by Mario Carneiro, 7-Nov-2013.)
((𝑁 ∈ ℕ0𝐾 ∈ ℤ) → (𝑁C𝐾) = if(𝐾 ∈ (0...𝑁), ((!‘𝑁) / ((!‘(𝑁𝐾)) · (!‘𝐾))), 0))
 
Theorembcval2 13087 Value of the binomial coefficient, 𝑁 choose 𝐾, in its standard domain. (Contributed by NM, 9-Jun-2005.) (Revised by Mario Carneiro, 7-Nov-2013.)
(𝐾 ∈ (0...𝑁) → (𝑁C𝐾) = ((!‘𝑁) / ((!‘(𝑁𝐾)) · (!‘𝐾))))
 
Theorembcval3 13088 Value of the binomial coefficient, 𝑁 choose 𝐾, outside of its standard domain. Remark in [Gleason] p. 295. (Contributed by NM, 14-Jul-2005.) (Revised by Mario Carneiro, 8-Nov-2013.)
((𝑁 ∈ ℕ0𝐾 ∈ ℤ ∧ ¬ 𝐾 ∈ (0...𝑁)) → (𝑁C𝐾) = 0)
 
Theorembcval4 13089 Value of the binomial coefficient, 𝑁 choose 𝐾, outside of its standard domain. Remark in [Gleason] p. 295. (Contributed by NM, 14-Jul-2005.) (Revised by Mario Carneiro, 7-Nov-2013.)
((𝑁 ∈ ℕ0𝐾 ∈ ℤ ∧ (𝐾 < 0 ∨ 𝑁 < 𝐾)) → (𝑁C𝐾) = 0)
 
Theorembcrpcl 13090 Closure of the binomial coefficient in the positive reals. (This is mostly a lemma before we have bccl2 13105.) (Contributed by Mario Carneiro, 10-Mar-2014.)
(𝐾 ∈ (0...𝑁) → (𝑁C𝐾) ∈ ℝ+)
 
Theorembccmpl 13091 "Complementing" its second argument doesn't change a binary coefficient. (Contributed by NM, 21-Jun-2005.) (Revised by Mario Carneiro, 5-Mar-2014.)
((𝑁 ∈ ℕ0𝐾 ∈ ℤ) → (𝑁C𝐾) = (𝑁C(𝑁𝐾)))
 
Theorembcn0 13092 𝑁 choose 0 is 1. Remark in [Gleason] p. 296. (Contributed by NM, 17-Jun-2005.) (Revised by Mario Carneiro, 8-Nov-2013.)
(𝑁 ∈ ℕ0 → (𝑁C0) = 1)
 
Theorembc0k 13093 The binomial coefficient " 0 choose 𝐾 " is 0 for a positive integer K. Note that (0C0) = 1 (see bcn0 13092). (Contributed by Alexander van der Vekens, 1-Jan-2018.)
(𝐾 ∈ ℕ → (0C𝐾) = 0)
 
Theorembcnn 13094 𝑁 choose 𝑁 is 1. Remark in [Gleason] p. 296. (Contributed by NM, 17-Jun-2005.) (Revised by Mario Carneiro, 8-Nov-2013.)
(𝑁 ∈ ℕ0 → (𝑁C𝑁) = 1)
 
Theorembcn1 13095 Binomial coefficient: 𝑁 choose 1. (Contributed by NM, 21-Jun-2005.) (Revised by Mario Carneiro, 8-Nov-2013.)
(𝑁 ∈ ℕ0 → (𝑁C1) = 𝑁)
 
Theorembcnp1n 13096 Binomial coefficient: 𝑁 + 1 choose 𝑁. (Contributed by NM, 20-Jun-2005.) (Revised by Mario Carneiro, 8-Nov-2013.)
(𝑁 ∈ ℕ0 → ((𝑁 + 1)C𝑁) = (𝑁 + 1))
 
Theorembcm1k 13097 The proportion of one binomial coefficient to another with 𝐾 decreased by 1. (Contributed by Mario Carneiro, 10-Mar-2014.)
(𝐾 ∈ (1...𝑁) → (𝑁C𝐾) = ((𝑁C(𝐾 − 1)) · ((𝑁 − (𝐾 − 1)) / 𝐾)))
 
Theorembcp1n 13098 The proportion of one binomial coefficient to another with 𝑁 increased by 1. (Contributed by Mario Carneiro, 10-Mar-2014.)
(𝐾 ∈ (0...𝑁) → ((𝑁 + 1)C𝐾) = ((𝑁C𝐾) · ((𝑁 + 1) / ((𝑁 + 1) − 𝐾))))
 
Theorembcp1nk 13099 The proportion of one binomial coefficient to another with 𝑁 and 𝐾 increased by 1. (Contributed by Mario Carneiro, 16-Jan-2015.)
(𝐾 ∈ (0...𝑁) → ((𝑁 + 1)C(𝐾 + 1)) = ((𝑁C𝐾) · ((𝑁 + 1) / (𝐾 + 1))))
 
Theorembcval5 13100 Write out the top and bottom parts of the binomial coefficient (𝑁C𝐾) = (𝑁 · (𝑁 − 1) · ... · ((𝑁𝐾) + 1)) / 𝐾! explicitly. In this form, it is valid even for 𝑁 < 𝐾, although it is no longer valid for nonpositive 𝐾. (Contributed by Mario Carneiro, 22-May-2014.)
((𝑁 ∈ ℕ0𝐾 ∈ ℕ) → (𝑁C𝐾) = ((seq((𝑁𝐾) + 1)( · , I )‘𝑁) / (!‘𝐾)))
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