# “Treatise on the Arithmetical Triangle” by Blaise Pascal (1654)

“Treatise on Arithmetical Triangle” by Blaise Pascal (1654)

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•  1st consequence – In every arithmetical triangle, all the cells of the first row & first column are the same as the generating cell. Each cell in the triangle is equal to the sum of the immediately preceding row & column

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• 2nd consequence – In every arithmetical triangle, each cell is equal to the sum of all the cells of the preceding row from its own column to the first, inclusive

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• 3rd consequence – In every arithmetical triangle, each cell is equal to the sum of all the cells of the preceding column from its own row to the first, inclusive

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• 4th consequence – In every arithmetical triangle, each cell exceeds by unity the sum of all the cells within its rows & columns exclusive

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• 5th consequence – In every arithmetical triangle, each cell is equal to its reciprocal

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• 6th consequence – In every arithmetical triangle, the corresponding cells of rows & columns having the same exponents are equal

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• 7th consequence – In every arithmetical triangle, the sum of the cells of each base is double that of the preceding base
• 8th consequence – In every arithmetical triangle, the sum of the cells of each base is the number of the double progression beginning with unity whose exponent is the same as that of the base

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• 9th consequence – In every arithmetical triangle, each base exceeds by unity the sum of all the preceding bases

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• 10th consequence – In every arithmetical triangle, the sum of any number of contiguous cells of the same number of cells of the preceding base together with the same number except one

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• Definition: Cells of a bisector = those that cut diagonally of the right angle

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• 11th consequence – each cell of the bisector is double the preceding cell in its row or column

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• 12th consequence – In every arithmetical triangle, of 2 contiguous cells in the same base the upper is to the lower as the number of cells from the upper to the top of the base is to the number of cells from the lower to the bottom of the base, inclusive

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• 13th consequence – In every arithmetical triangle, of 2 contiguous cells in the same column the lower is to the upper at the exponent of the base of the upper is to the exponent of its row

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• 14th consequence – In every arithmetical triangle, of the 2 contiguous cells in the same row, the greater is to the lesser as the exponent of the base of the lesser is to the exponent of its column

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• 15th consequence – In every arithmetical triangle, the sum of the cells of any row is to the last cell of the row of the exponent of the triangle is to the exponent of the row

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• 16th consequence – In every arithmetical triangle, any row is to the next lower row as the exponent of the next lower row is to the number of its cells

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• 17th consequence – In every arithmetical triangle, any cell together with all those of its row as the number of perpendicular cells taken is to the number of parallel cells taken

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• 18th consequence – In every arithmetical triangle, 2 rows equidistant from the extremities are to each other as the number of their cells

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• 19th consequence – In every arithmetical triangle, of 2 contiguous cells in the bisector the lower is to 4x the higher as the exponent of the base of the higher is to a number exceeding it by unity

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Use of Arithmetical Triangle of Orders of Numbers

 1 2 3 4 5 etc. Units Order 1 1 1 1 1 1 etc. Natural Numbers Order 2 1 2 3 4 5 etc. Triangular Numbers Order 3 1 3 6 10 15 etc. Pyramidal Numbers Order 4 1 4 10 20 35 etc. etc.

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Use of Arithmetical Triangle for Combinations

Combination of 1 of 4 things can be made 4 ways: A, B, C, D

Combination of 2 of 4 things can be made 6 ways: AB, AC, AD, BC, BD, CD

Combination of 3 of 4 things can be made 4 ways: ABC, ABD, ACD, BCD

Combination of 4 of 4 things can be made 1 way: ABCD

Lemma 1 – No combinations of a number in a smaller number, e.g. 4 can’t be combined into 2.

Lemma 2 – 1 combination of 1 in 1, 1 combination of 2 in 2, 1 combination of 3 in 3. In general, there is only one combination of a number in an equal number.

Lemma 3 – 1 combination of 1 in 1. 2 combinations of 1 in 2. 3 combinations of 1 in 3. In general, there are as many combinations of 1 in any number as there are units in the number

Lemma 4 – If there are 4 numbers, the 1st of which is arbitrary, the 2nd exceeding the 1st by 1, the 3rd any number not smaller than the 2nd, the 4 exceeding the 3rd by 1, the number of combinations of the 1st & 3rd together with the number of combinations of 2nd in the 3rd is equal to the number of combinations of the of the 2nd in the 4th.

e.g. A=1, B=2, C=3, D=4

• The number of combinations of 1 in 3 together with the number of combinations of 2 in 3 is equal to the number of combinations of 2 in 4
• 6 combinations of A,B,C,D in pairs: AB, AC, AD, BC, BD, CD
• We must demonstrate the number of combinations of 1 in 3 & of 2 in 3 = number of combinations of 2 in 4. Easy for combinations of 2 in 4 are made up of the combinations of 1 in 3 & 2 in 3
• A, B, C (3) + AB, BC, AC (3) = AB, AC, AD, BC, BD, CD (6)
• Number of combinations of 29 in 40 & number of combinations of 30 in 40 = Number of combinations of 30 in 41
• Number of combinations of 15 in 55 & number of combinations of 16 in 55 = Number of combinations of 16 in 56, etc.

Proposition 1 – In every arithmetical triangle the sum of any parallel row of cells is equal to the number of combinations of the exponent of the row in the exponent of the triangle

• Take any triangle, e.g. GDλ
• Sum of cells of any row, e.g. 2nd ϕ + ψ + θ = number of combinations of 2 in the exponent of the 2nd row in 4th exponent of the triangle
• Lemma – in 1st triangle –> sum of cells in only row G – number of combinations of 1, the exponent (column) in I, exponent of the triangle
• Lemma – if an Arithmetical Triangle is found with the equality, whatever row we take, the sum of the cells = number of combinations of the exponent of the row in the exponent of the triangle
• Lemma – Let any triangle be taken & we suppose that the equality (sum of 1st row G + σ + π) = number of combination of 1 in 3 & sum of cells in 2nd row (ϕ + ψ) = combinations of 2 in 3 & sum of cells in the 3rd row, A = combinations of 3 in 3. 4th triangle will have the same equality, & the sum of cells of the 2nd row, ϕ + ψ + θ (by hypothesis) = ϕ + ψ (number of combinations of 2 in 3)    + ϕ (= G + σ + π –> number of combinations of 1 in 3)

Proposition 2 – The number of any cell is equal to the number of combinations of a number less by 1 than its parallel exponent in a number less by 1 than the exponent of its base.

• Take any cell, e.g. F, in 4th parallel row & in the 6th base. Equal to number of combinations of 3 in 5, less by 1 than 4 & 6, because it equals cells A + B + C

Problem 1, Proposition 3

• 2 numbers having been set out, to find by the arithmetical triangle how many combinations there are of one in the order
• Let numbers set out be 4 & 6. Let’s find out the number of combinations of 4 in 6
• 1st way – take the sum of the cells of the 4th row in 6th triangle & it will satisfy the problem
• 2nd way – take the 5th cell of 7th base because numbers 5, 7 exceed by 1 the numbers 4 & 6, the number of this cell is the required number

Conclusion – Easy to see the relation between cells & rows of the Arithmetical Triangle & combinations of all that is proved of the former applies to the latter analogously.

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Use of the Arithmetical Triangle

• To determine how the stakes should be divided between 2 players playing for a set of games
• Once players ante up, the money’s no longer their but they can expect a chance to get some or all of the pot according to agree upon rules
• All voluntary & can back out if they want but giving up the chance of winning
• Principle to divide winnings
• If a player’s position is where a portion of the pot will be his if he wins or loses, he should take it from the whole as if it’s assured (don’t use expected value)
• If 2 players’ position are that if 1 wins, he’ll get a certain amount & if he loses it’ll go to the other player
• If it’s pure chance & it’s even to both players, you can avoid the game & negative outcomes by splitting the pot

1st Corollary – If 2 players plays a game of pure chance on condition that player 1 wins, a certain sum will be his & if he loses a smaller sum will be his & if they wish to separate without playing & each take what belongs to him, player 1 will take what would be his if he lost & half the difference between & what he would have won.

• E(V)= αL + 1/2(W-L)(1-α)
• α = chances of losing, (1-α) = chances of winning, L=payout for a loss, W=payout for a win, α = (1-α)
• E.G. – Winner gets 8 pistoles (coins), loser gets 2
• 1 game W=8, L=2 ==> (8-2)/2 = 3
• W = 8 = 6 + 2
• L = 2
• 1/2(W+L) = 1/2((W-L)+L)=5

2nd Corollary – If playing the same game, division can be: winner’s share & loser’s shared combined – player 1 take 1/2 the sum (2+8=10, /2 = 5)

• Case 1 – If 1 player has the required number of games & the other doesn’t, he gets all the pot
• Case 2 – If 1 player lacks 1 game & the other does too, they should split the pot. Same goes if both are missing 2, 3, etc.
• Case 3 – If 1 player lacks 1 game & the other 2, division should be (without playing the next game):
• Scenario A: Lacking 1 game, P1 – win next game, win all 8 –> only P1 wins
• Scenario B: Lacking 2 games, P1 – win next game, lack only 1 game
• P1 & P2 should split
• Scenario A –> Win Game 1 –> 8, Lose Game 1 –> 8/2=4
• Without playing Game 1 via corollary 2 –> 3/4 (8) – 6
• (8 + 4)/2 = 6 out of a possible 8
• Case 4 – If one player lacks one win & the other 3, division is similar…
• Win game 1 –> get all 8
• Lose game 1 –> P2 lacks 2 & P1 lacks 1 –> all 6 pistoles to P1
• Win Game 2 –> get all 6
• Lose Game 2 –> P2 lacks 1 & P1 lacks 1 –> split last 2 pistoles –> 6 + 2/2 = 7
• Case 5 – If 1 player lacks game & the other 4
• Win G1 –> 8 pistoles
• Lose G1 –> P2 lacks 3 games & P1 lacks 1
• 7 of 8 belong to P1
• 8/2 + 7/2 –> 7.5
• Case 6 – Similar if P1 lacks 1 & P2 lacks 5, ad infinitum
• Case 7 – If P1 lacks 2 & P2 lacks 3 –> use same method
• P1 wins G1 = P1 needs 1 & P2 needs 3
• P1 gets 7 of 8
• P2 wins G1 = P1 needs 2 & P2 needs 2
• P1 & P2 each get 4
• P1 wins, gets 7, loses gets 4
• E(V) – (7 + 4) / = 5.5

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Method for Dividing the Stakes Between 2 Players Playing for a Set of Games by Means of the Arithmetical Triangle

Lemma – 2 players playing a game of pure chance & P1 wins, he gets a part of a sum (in a fraction) & P1 loses, he’ll get a smaller fraction of the sum, to separate without playing

• Let fractions be reduced to a common denominator
• Let a fraction be taken with a numerator as a sum of the 2 numerators & denominator as double the common denominator. This is P1’s part of the sum
• E.G. P1 gets 3 for a win –> 4/10 for expected value
• Same if 1/2 & 2×2 = 1/4
• 1 + 0 & 2×2 = 1/4
• W = 3/3, L = 0/3 –> (3 + 0) / (3 x 2)

Problem 1, Proposition 1 – Given 2 players needing a certain number of games to complete set, to find by the Arithmetical Triangle. What the division should be with respect to games needed to win it all

• Take a triangle base with number of cells 2 players need to win
• e.g. P1 needs 2 & P2 need 4 –> base with 6 cells
• Pδ = cells = P, M, F, ω, S, δ
• P1 lacks P, M & P2 lacks F, ω, S, δ
• P1’s odds to P2 odds = F + ω + S + δ : P + M
• Sum of 4 cells of P1: δ + S + ω + F (lacks 2 games)
• Sum of 2 cells of P2: P + M (lacks 4 games)
• P1’s share = (F + ω + S + δ) / (P + M + F + ω + S + δ)
• P2’s share = (P + M) / (P + M + F + ω + S + δ)
• e.g. from 4th base Dλ
• P1 needs 1, P2 needs 3
• P1’s share: (D + B + θ) / (D + B + θ + λ)
• P2’s share: λ / (D + B + θ + λ)
• e.g 5th base Hμ
• P1 needs 2, P2 needs 3
• P1’s share (H + E + C) / (H + E + C + R + μ)
• to know what belongs to 2 players both need games, we must take the faction belonging to P1 if winning & if losing
• Make in a common denominator with sum in numerator & double denominators
• If P1 needs 2 with 1 game to pay & wins, he’ll need 1 & P2 will need 3, together, they’ll need 4
• WIN, then P1’s share (D + B + θ) / (D + B + θ + λ)
• LOSE, then P1’s share (D + B) / (D + B + θ + λ)
• TOGETHER: (D + B + θ + D + B) / (2D + 2B + 2θ + 2λ)
• EQUAL TO: (H + E + C) / (H + E + C + R + μ)

Problem 2, Proposition 2 – Given 2 players with same sum on a given number of games, to find in the Arithmetic Triangle the part of the loser’s money the last game is worth to the winner

• e.g. P1 & P2 each stake 3 pistoles on 4 games. What parts of the loser’s 3 pistoles the last game is wroth to the winner?
• Let there be taken the fraction whose numerator is 1 & the denominator is the sum of cells from the 4th base (because there are 4 games)
• P1 has won 3 & P2 has won 0
• P1’s share from first 3 games: (H + E + C + R) / (H + E + C + R + μ)
• Numerator = 1st 4 cells of the 5th base
• Denominator = all of the 5th base
• μ / (H + E + C + R + μ) –> remains to evaluated from the last game for P1
• 1 / (2D + 2B + 2θ + 2λ) –> each stake P1 & P2 of (D + B + θ + λ)*1/2

Problem 3 Proposition 3 – Given 2 players each needing the same number of games, find in the Arithmetical Triangle the part of the loser’s money represented by the value of the 1st game

• P1 & P2 each stake 3 pistoles on 4 games. What’s the loser’s stake of the 1st game worth to the winner?
• Add 3 (=4-1) to 4 (=no. of games) = 7
• Use Arithmetical Triangle with base of 7
• Fraction with denominator of cells of 7th base & number of bisector of base
• (ρ) / (V + Q + K + ρ + ξ + N + ζ)
• If 2 players play for a set of 4 games, P1 has 1 game to his credit & must win 4 games (3 more) & P2 needs 4 more
• P1’s share = (V + Q + K + ρ) / (V + Q + K + ρ + ξ + N + ζ)
• Denominator – all 7 cells of the base
• V + Q + K + ρ belongs to him of the whole amount of 2 stakes of V + Q + K + ρ + ξ + N + ζ
• P1 has staked his half of (V + Q + K + ρ + ξ + N + ζ), which is V + Q + K + 1/2ρ
• He has 1/ρ = ω more than he had started with & has won so far: ω / (V + Q + K + ρ + ξ + N + ζ) & his portion of P2’s stakes = ρ / (V + Q + K + ρ + ξ + N + ζ)

Corollary – Part of loser’s stake constituting the value of Game 1 of 2 is expressed by 1/2

• Take fraction using cells of 3rd base because 2 games + (2-1) = 3
• Number of cell ψ = 2 & number of cells A, ψ, π are 1, 2, 1
• 2 / (1 + 2 + 1) = 1/2

Problem 4 Proposition 4 – Given 2 players who stake the same sum on a given number of games, find in the Arithmetical Triangle the value of 2 games with respect to the loser’s stake

• 4 games 0 what’s the value of the 2nd game with respect to loser’s stake
• With 2 game to P1’s credit:
• (P + M + F + ω) / (P + M + F + ω + S + δ)
• Sum of 1st 4 of base 6 / base 6 of arithmetical triangle
• Staked: (P + F + M) / (P + M + F + ω + S + δ) = Half of total staked
• Winnings: P / (V + Q + K + ρ + ξ + N + ζ)
• Loser’s stake: 2ρ / (V + Q + K + ρ + ξ + N + ζ)

Conclusion – Divisions between 2 players made as proportion between cells in the Treatise on the Arithmetic Triangle have more consequences than noted

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Use of the Arithmetical Triangle to find Powers of Binomials

• To find a power of a binomial with the 1st term, A, & the 2nd term, 1, i.e. (A + 1)⁴, take the 5th base triangle with exponent ((4 + 1) = 5) so that the cells are 1, 4, 6, 4, 1
• Take the first number 1 as a coefficient of A of proposed power, i.e. A⁴
• Then take the 2nd number 4 as the coefficient of A with the next lower base, i.e. 4A³
• Then take the 3rd number 6 as the coefficient of A with the next lower base, i.e. 6A²
• Then take the 4th number 4 as the coefficient of A with the next lower base, i.e., 4A
• Then take the 5th number 1 as it is & together they make
• A⁴ + 4A³ + 6A² + 4A + 1 = (A + 1)⁴
• If A = 1, then = 1 + 4 + 6 + 4 + 1 = 16
• 2⁴ = 16
• If A = 4, then = 256 + 256 + 96 + 16 + 1 = 625
• 5⁴ = 625
• For Binomial (A + 2)⁴
• Original equation A⁴ + 4A³ + 6A² + 4A + 1
• add 1 + 2 + 4 + 8 + 16
• Then, A⁴ + 8A³ + 24A² + 32A + 16
• If A = 1, then = 1 + 8 + 24 + 32 + 16 = 81 = 3⁴
• If A = 2, then = 16 + 64 + 96 + 64 + 16 = 256 = 4⁴
• For Binomial (A + 3)⁴
• Original equation A⁴ + 4A³ + 6A² + 4A + 1
• add 1 + 3 + 9 + 27 + 81
• Then, A⁴ + 12A³ + 54A² + 108A + 81
• For Binomial (A – 1)⁴, alternate signs
• A⁴ – 4A³ + 6A² – 4A + 1

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Treatise of the Orders of Numbers

Proposition 1 –  a number of any order = number preceding it in its own order together with the number corresponding to it in it the preceding order. E.G. 4th, numbers of 4th order = 3rd number of 4th order together with the 4th number of the third order

Proposition 2 – A number of any order = sum of numbers whose root is less than its own by 1 of its own & all preceding orders & consequently, e.g. 4th number of the 4th order = 3rd number of the 4th order + 3rd number of the 3rd order + the natural number plus 3rd number of the units (1)

Proposition 3 – Every number, whichever cell is made of up of as many numbers as there are orders from its own to the 1st, inclusive, each of the numbers representing 1 of the orders. E.g. Number of the 5th order composed of another number of the 5th order, a number of the 4 order, the 3rd, 2nd & 1

Proposition 4, Problem – Given a number of any order, to find a number in each order from the 1st to its own, inclusive, whose sum is equal to the given number. Taken in all orders’ numbers whose root is less by 1 than that of the given number. E.g. in Prop 5

Proposition 5 – 2 numbers of different orders are equal if the root one is the same number as the exponent of the other. Consequently, 3rd number of the 4th order is equal to the 4th number of the 3rd order. 5th number of the 8th order is equal to the 8th number of the 5th order

Proposition 6 – All 4th numbers of all orders the same as all numbers of the 4th order – rows & columns are mirrored

Proposition 7 – A number whatever order is to the next great in the same order as the root of the lesser is to itself together with the column of the order minus 1. This follows the 14th consequence. Fore the column of the base together with the column of the perpendicular row minus 1. The Exponent of the order together with the root minus 1.

Proposition 8 – A number of whatever order is to the corresponding number of the following order as the exponent of the order of the lesser number is to itself together with the common root minus 1. From 13th consequence

Proposition 9 – A number of whatever order is to the number of the preceding order whose root is great than its own by 1 as the root of the 1st number is to the exponent of the order of the 2nd

Proposition 10 – A number of whatever order when multiplied by the preceding root is equal to the exponent of its order multiplied by the preceding number of the following order

Proposition 11 – A number of whatever order, when multiplied by the preceding root & divided by the exponent of its order, gives as its quotient the preceding number of the following order

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