Definition D1
The data arranged in order of size are written x₍₁₎ ≤ ⋯ ≤ x₍ₙ₎. If n is odd, the middle value is the median; if n is even, the mean of the two middle values is the median.
Proof.
(1)By trichotomy and transitivity such an arrangement exists, and the value in the j-th position is uniquely determined.A1, A2
(2)Taking the mean of the two middle values when n is even is not a result of computation but a convention.A4, convention
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Lemma L1
In the data 1, 2, 3, 4, 5, 6, 7, 8, the values Q with exactly 2 data points below them and exactly 6 above them are precisely the values with 2 < Q < 3.
Proof.
(1)Suppose 2 < Q < 3. The only data points below Q are 1 and 2, so there are 2.assumption, N
(2)Under the same assumption, the only data points above Q are 3, 4, 5, 6, 7, 8, so there are 6.assumption, N
(3)Suppose Q ≤ 2. Then the data points below Q are at most 1, so at most 1 in number.assumption, A2, N
(4)Suppose 3 ≤ Q. Then the data points above Q are at most 4, 5, 6, 7, 8, so at most 5 in number.assumption, A2, N
(5)By trichotomy the only cases are Q ≤ 2, 2 < Q < 3, and 3 ≤ Q, and only the middle case satisfies the condition.(1), (2), (3), (4), A1
(6)By density there is always another value between 2 and 3. Therefore this condition does not determine a single value.(5), A3
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Definition D2
Definition ① the halving method. Split the data at the median position; if n is odd, place the middle value in neither half. The median of the lower half is Q₁.
Proof.
This definition applies the median twice, and there is a separate variant that places the middle value in the lower half.K, D1
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Definition D3
Definition ② interpolation method. Set p = 1 + (n − 1)·4⁻¹ and write p = k + t (k an integer, 0 ≤ t < 1); then Q₁ = x₍ₖ₎ + t(x₍ₖ₊₁₎ − x₍ₖ₎).
Proof.
This definition treats the first position as 0 and the last as 1, and takes the one-quarter point between them.K, D1
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Definition D4
Definition ③ interpolation method. Set p = (n + 1)·4⁻¹ and proceed as in D3.
Proof.
This definition imagines one extra position beyond each end of the data and takes the one-quarter point.K, D1
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Theorem T1
On the data 1, 2, 3, 4, 5, 6, 7, 8, the three definitions give three distinct values.
Proof.
(1)n = 8 and the data are already in order of size.assumption, D1, N
(2)Under D2 the lower half is 1, 2, 3, 4 with median (2 + 3)·2⁻¹ = 2.5, so Q₁ = 2.5(1), D2, D1, A4, N
(3)Under D3, p = 1 + 7·4⁻¹ = 2.75, so k = 2, t = 0.75 and Q₁ = 2 + 0.75(3 − 2) = 2.75(1), D3, N
(4)Under D4, p = 9·4⁻¹ = 2.25, so k = 2, t = 0.25 and Q₁ = 2 + 0.25(3 − 2) = 2.25(1), D4, N
(5)2.5, 2.75, and 2.25 are three distinct values.(2), (3), (4), N
(6)All three values are greater than 2 and less than 3, so all satisfy the condition of L1. None of them violates it.(5), L1, N
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Theorem T2
When the data change, which two of the definitions coincide also changes.
Proof.
(1)Consider the data 1, 2, 3, 4, 5, 6, 7, 8, 9. Here n = 9 and the median is the fifth value, 5.assumption, D1, N
(2)Under D2, removing the middle value 5, the lower half is 1, 2, 3, 4 with median 2.5, so Q₁ = 2.5(1), D2, A4, N
(3)Under D3, p = 1 + 8·4⁻¹ = 3, so t = 0 and Q₁ = x₍₃₎ = 3(1), D3, N
(4)Under D4, p = 10·4⁻¹ = 2.5, so k = 2, t = 0.5 and Q₁ = 2 + 0.5(3 − 2) = 2.5(1), D4, N
(5)For n = 8 all three values differed; for n = 9, D2 and D4 agree and only D3 differs.T1, (2), (3), (4)
(6)Meanwhile, in these data the only value with exactly 2 data points below it and 6 above it is Q = 3.A1, N
(7)Whether the dividing position falls between data values or on one is decided by n, and the pattern of disagreement among the definitions changes accordingly.(5), (6), L1
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Theorem T3
Q₁ is not a number determined by the data. It becomes a number only after a definition is chosen.
Proof.
(1)By L1, the position dividing the data 1 : 3 is not one value but every value between 2 and 3.L1
(2)Therefore the data alone do not determine Q₁ as a single number.(1)
(3)The three definitions choose 2.5, 2.75, and 2.25 respectively from that interval.T1
(4)All three values satisfy the condition, so the data contain no grounds for judging any one of them more correct than the others.(3), L1
(5)The choosing rule is set by the person counting, not by the data.D2, D3, D4, K
(6)Therefore 'what is Q₁' is a question with no determined answer until the definition in use is stated.(2), (4), (5)
(7)Statistical packages return different Q₁ from the same data not because the computation is wrong, but because they use different conventions.(6), T2
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Corollary C
Therefore the quartile is not a theorem to prove but a convention to choose.
Proof.
(1)A theorem is something whose answer follows uniquely from what is given; a convention is something added to make the answer unique.convention
(2)By L1, what the data give is an interval, not a single number.L1
(3)Therefore, to make Q₁ a single number, something must be added, and that addition is a definition.(1), (2)
(4)The median stands in the same place. When n is even the middle falls between two values, and fixing it as their mean is a convention, not a proof.D1, A4
(5)The convention is conspicuous for quartiles because more than one variant has survived.(3), T1, T2
(6)Therefore, when stating a quartile, one must state which convention was used along with the value.(5), T3
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