Abstract
Examples of hypotheses about the number of planets are frequently used to introduce the topic of (actual) truthlikeness but never analyzed in detail. In this paper we first deal with the truthlikeness of singular quantity hypotheses, with reference to several ‘the number of planets’ examples, such as ‘The number of planets is 10 versus 10 billion (instead of 8).’ For the relevant ratio scale of quantities we will propose two, strongly related, normalized metrics, the proportional metric and the (simplest and hence favorite) fractional metric, to express e.g. the distance from a hypothetical number to the true number of planets, i.e. the distance between quantities. We argue that they are, in view of the examples and plausible conditions of adequacy, much more appropriate, than the standardly suggested, normalized absolute difference, metric.
Next we deal with disjunctive hypotheses, such as ‘The number of planets is between 7 and 10 inclusive is much more truthlike than between 1 and 10 billion inclusive.’ We compare three (clusters of) general ways of dealing with such hypotheses, one from Ilkka Niiniluoto, one from Pavel Tichý and Graham Oddie, and a trio of ways from Theo Kuipers. Using primarily the fractional metric, we conclude that all five measures can be used for expressing the distance of disjunctive hypotheses from the actual truth, that all of them have their strong and weak points, but that (the combined) one of the trio is, in view of principle and practical considerations, the most plausible measure.
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Acknowledgements
I like to thank the two referees for their stimulating questions and suggestions, David Atkinson for his elaborate mathematical support. Finally, I like to thank, Gustavo Cevolani, Alfonso García-Lapeña, Jan-Willem Romeijn and Tom Sterkenburg for their stimulating comments.
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Appendices
Appendix 1
This appendix presents two proofs and some research questions. In the first section it is proved that fractional similarity to the nth power, and hence the basic one in the text, i.e. s*, (n = 1), satisfies (the similarity version of) the triangle inequality. In the second section it is proved that (basic) proportional (relative) distance, i.e. d# in the text, also satisfies the triangle inequality. The second proof is fully due to David Atkinson. The first proof, though much more general, is much less complicated and was highly inspired by the second proof.
1.1 Proof of Triangle Inequality for Fractional Similarity to the n th Power
Definition of the ‘candidate’ metric
where x, y and z are real positive numbers. Note that s1*(x, y) = s*(x, y) and ds1*(x, y) = d*(x, y), i.e. the basic versions of fractional similarity and distance. Note also that dsn*(x, y) essentially differs from [d*(x, y)]n for n > 1.
It is easy to check that both functions satisfy the corresponding conditions of adequacy for being a metric CA1, CA2, CA3 and the relevant versions of the special conditions CA5 and CA6. To be proved is that they satisfy (the relevant version of) the triangle inequality CA4. We start with a lemma.
1.2 Lemma 1
Given three distinct positive numbers, designate the smallest one by x, the intermediate one by y and the largest one by z, so 0 < x < y < z. Then, using symmetry, and simplifying the notation during the proof to dsn*(x, y) = df d(x, y) and sn*(x, y) = df s(x, y), the following three inequalities hold:
-
1.
\(\begin{array}{lll}{\mathrm{d}}({\mathrm{x}},\mathrm{ y})+{\mathrm{d}}({\mathrm{y}},\mathrm{ z})>{\mathrm{d}}({\mathrm{x}},\mathrm{ z})& (1-{\mathrm{s}}({\mathrm{x}},\mathrm{ y}))+(1-{\mathrm{s}}({\mathrm{y}},\mathrm{ z}))>1-{\mathrm{s}}({\mathrm{x}},\mathrm{ z})& 1+{\mathrm{s}}({\mathrm{x}},\mathrm{ z})>{\mathrm{s}}({\mathrm{x}},\mathrm{ y})+{\mathrm{s}}({\mathrm{y}},\mathrm{ z})\end{array}\)
-
2.
\(\begin{array}{lll}{\mathrm{d}}({\mathrm{x}},\mathrm{ z})+{\mathrm{d}}({\mathrm{y}},\mathrm{ z})>{\mathrm{d}}({\mathrm{x}},\mathrm{ y})& \left(1-{\mathrm{s}}\left({\mathrm{x}},\mathrm{ z}\right)\right)+\left(1-{\mathrm{s}}\left({\mathrm{y}},\mathrm{ z}\right)\right)>1-{\mathrm{s}}({\mathrm{x}},\mathrm{ y})& 1+{\mathrm{s}}({\mathrm{x}},\mathrm{ y})>{\mathrm{s}}({\mathrm{x}},\mathrm{ z})+{\mathrm{s}}({\mathrm{y}},\mathrm{ z})\end{array}\)
-
3.
\(\begin{array}{lll}{\mathrm{d}}({\mathrm{x}},\mathrm{ y})+{\mathrm{d}}({\mathrm{x}},\mathrm{ z})>{\mathrm{d}}({\mathrm{y}},\mathrm{ z})& (1-{\mathrm{s}}({\mathrm{x}},\mathrm{ y}))+(1-{\mathrm{s}}({\mathrm{x}},\mathrm{ z}))>1-{\mathrm{s}}({\mathrm{y}},\mathrm{ z})& 1+{\mathrm{s}}({\mathrm{y}},\mathrm{ z})>{\mathrm{s}}({\mathrm{x}},\mathrm{ y})+{\mathrm{s}}({\mathrm{x}},\mathrm{ z})\end{array}\)
Proof of Lemma 1
Using 0 < x < y < z we prove the similarity version of these claims in only slightly different ways.
-
1.
1 + s(x, z) > s(x, y) + s(y, z), that is 1 + xn/zn > xn/yn + yn/zn
multiplying by ynzn we get ynzn + ynxn > xnzn + y2n, and hence
ynzn − xnzn > y2n − ynxn, which is equivalent to (yn – xn) zn > (yn – xn) yn, hence OK.
-
2.
1 + s(x, y) > s(x, z) + s(y, z), that is 1 + xn/yn > xn/zn + yn/zn
multiplying by ynzn we get ynzn + xnzn > xnyn + y2n, and hence
ynzn − y2n > xnyn − xnzn, which is equivalent to yn (zn − yn) > 0 > xn (yn − zn), hence OK.
-
3.
1 + s(y, z) > s(x, y) + s(x, z), that is 1 + yn/zn > xn/yn + xn/zn
multiplying by ynzn we get ynzn + y2n > xnzn + xnyn, and hence
ynzn − xn zn > xnyn − y2n, which is equivalent to (yn − xn)zn > 0 > (xn − yn)yn, hence OK.
1.2.1 Theorem 1: Triangle Inequality of Fractional Similarity to the n th Power
Recall that sn*(x, y) = [s*(x, y)]n = df s(x, y).
Given three positive numbers (not necessarily distinct), labelled x, y and z.
-
1.
1 + s(x, z) ≥ s(x, y) + s(y, z)
-
2.
1 + s(x, y) ≥ s(x, z) + s(y, z)
-
3.
1 + s(y, z) ≥ s(x, y) + s(x, z)
that is, the similarity version of CA4.
1.3 Proof
If x = y ≠ z, then.
-
1.
1 + s(x, z) = s(x, y) + s(y, z)
-
2.
1 + s(x, y) > s(x, z) + s(y, z)
-
3.
1 + s(y, z) = s(x, y) + s(x, z)
and analogously if x = z ≠ y or y = z ≠ x.
If x = y = z then trivially
-
1.
1 + s(x, z) = s(x, y) + s(y, z)
-
2.
1 + s(x, y) = s(x, z) + s(y, z)
-
3.
1 + s(y, z) = s(x, y) + s(x, z)
If x, y and z are distinct, we use the lemma. This concludes the proof of Theorem 1. QED.
We noted already that dsn*(x, y) (= 1 − sn*(x, y) = 1 − [s*(x, y)]n, now proved to satisfy the triangle inequality, essentially differs from [d*(x, y)]n for n > 1. Again it is easy to check that the latter measure satisfies CA1—CA6, except perhaps the triangle inequality CA4 for n > 1. An interesting question for further research.
1.4 Proof of Triangle Inequality for (Basic) Proportional Distance
Definition of the ‘candidate’ metric
where x, y and z are positive real numbers.
Again it is easy to check that d(x, y) satisfies CA1—CA6, except perhaps the triangle inequality CA4.
1.5 Lemma 2
Given three distinct positive numbers, designate the smallest one by x, the intermediate one by y and the largest one by z, so 0 < x < y < z. Then
-
1.
d(x, y) + d(y, z) > d(x, z)
-
2.
d(x, z) + d(z, y) > d(x, y)
-
3.
d(y,x) + d(x, z) > d(y, z).
Proof of lemma 2
Define
We need to show that m1, m2 and m3 are all positive.
Set
with i = 1, 2, 3, and
After some straightforward algebra, one can show that
which are all manifestly positive. The lemma is thereby proved.
1.5.1 Theorem 2: Triangle Inequality of Proportional Distance
Recall that d(x, y) =|x − y| / (x + y) (= d#(x, y) in the text).
Given three positive numbers (not necessarily distinct), labelled x, y and z,
-
1.
d(x, y) + d(y, z) ≥ d(x, z)
-
2.
d(x, z) + d(z, y) ≥ d(x, y)
-
3.
d(y, x) + d(x, z) ≥ d(y, z).
1.6 Proof
If x = y ≠ z, then
-
1.
d(x, y) + d(y, z) = d(x, z)
-
2.
d(x, z) + d(z, y) > d(x, y)
-
3.
d(y,x) + d(x, z) = d(y, z)
and analogously if x = z ≠ y or y = z ≠ x.
If x = y = z then trivially
-
1.
d(x, y) + d(y, z) = d(x, z)
-
2.
d(x, z) + d(z, y) = d(x, y)
-
3.
d(y,x) + d(x, z) = d(y, z).
If x, y and z are distinct, we use the lemma. This concludes the proof of Theorem 2 QED.
Section 1 of this appendix suggests ‘power versions’ of proportional distance and proportional similarity, with corresponding questions for further research.
Proportional (relative) distance to the nth power
Proportional (relative) similarity to the nth power
Note again that dn#(x, y) ≠ 1—sn#(x, y) = df dsn#(x, y), that is, for n > 1, the proportional distance to the nth power is different from the distance corresponding to the proportional similarity to the nth power.
Apart from CA4, it is easy to check that the two new measures also satisfy all other conditions of adequacy, i.e. CA1 – CA3 and CA5 and CA6. Hence, the question remains whether they satisfy the relevant version of the triangle inequality. In view of the complexity of the proof of Theorem 2 it seems very complicated to check. We did not even try. The reason is that all genuine power versions, with n > 1, including 1.1, do not seem conceptually very interesting.
Appendix 2
2.1 Theorem: Sufficient Condition for Covariance of TL av and TL α
Given T = {t}, X = {t, a1, a2,.. an, b1, b2, …bm} and Y = {t, b1, b2, …bm}.
Let d(x, y) be a normalized semi-metric on U: 0 ≤ d(x, y) ≤ 1 and d(x, y) = 0 iff x = y and d(x, y) = d(y, x). For the size similarity of sets fractional similarity is presupposed: s*(X, t) = 1/c(X).Then TLav(X, t) = (1 − dav(X, t)) / c(X) < TLav(Y, t) = (1 − dav(Y, t)) / c(Y) if m2 < 1 + n. With the corollary that then also holds TLα(X, t) < TLα(Y, t), in view of the general validity of TLmin(X, t) < TLmin(Y, t) when Y ⊂ X.
2.2 Proof
Given T = {t}, X = {t, a1, a2,.. an, b1, b2, …bm}, Y = {t, b1, b2, …bm}.
Then, c(T) = 1, c(X) = 1 + n + m, c(Y) = 1 + m. (Therefore, s*(X, t) = 1/(1 + n + m), s*(Y, t) = 1/(1 + m).) Hence, TLav(X, t) = (1 − dav(X, t)) / c(X) = (1 − dav(X, t)) / (1 + n + m), and TLav(Y, t) = (1 − dav(Y, t)) / c(Y) = (1 − dav(Y, t)) / (1 + m).
dav(X,t) = ∑k∈X d(k,t))/c(X) = {∑id(ai,t) + ∑id(bi,t)}/(1 + n + m) = df (An + Bm)/(1 + n + m).
Hence,1 − dav(X, t) = (1 + n + m – An – Bm)/(1 + n + m), and therefore:
TLav(X, t) = [(1 + n + m – An – Bm)/(1 + n + m)] x [1/(1 + n + m)] = (1 + n + m – An – Bm)/(1 + n + m)2.
dav(Y,t) = ∑k ∈ Y d(k, t)) / c(Y) = {∑i d(bi, t)}/(1 + m) = Bm/(1 + m).
Hence, 1 − dav(Y, t) = (1 + m – Bm)/(1 + m), and therefore:
TLav(Y, t) = (1 + m – Bm)/(1 + m)2.
Now the question is: under which conditions holds: TLav(X, t) < TLav(Y, t) in any case? Hence:
From 0 ≤ d(x, y) ≤ 1 follows An ≥ 0 and Bm ≤ m, hence m2 < 1 + n is a sufficient condition. QED.
Appendix 3
3.1 Theorem: \(\underset{n\rightarrow\infty}{\mathit{lim}}\frac1N\sum\nolimits_{n=1}^N\frac1n=0\)
Proof (due to David Atkinson).
First we prove a lemma, starting with the well-known fact that the integral from 1 to N of 1/x equals log N:
Using this inequality we can prove another inequality:
Now it is well known that (log N)/N goes to 0 when N goes to ∞. Hence, B(N) goes to 0, and therefore (non-negative) A(N) as well, QED.
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Kuipers, T.A.F. Truthlikeness and the Number of Planets. J Philos Logic 53, 493–520 (2024). https://doi.org/10.1007/s10992-023-09739-y
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DOI: https://doi.org/10.1007/s10992-023-09739-y