Jtdylktuy

Preliminaries

The focus of Choie et al 2007 (references below) is Theorem 1 in Robin 1984 which states that the Riemann Hypothesis is true if and only if there are no numbers $gt$ 5040 that violate Robin’s Inequality, $sigma(n)lt e^gamma nloglog n$.

There are some minor errors in Choie et al 2007, for example, on p.357 the inclusion of n = 1 in the domain of Robin’s Inequality when Robin’s Inequality is undefined at n = 1, or in Theorem 1.4 where 72 is omitted from the list of squareful numbers that violate Robin’s Inequality.

Proposition 5.1

However my question is about Proposition 5.1 on page 367 which for brevity I will refer to as just "5.1".

When 5.1 is reformulated in terms of the sets involved then it appears to be equivalent to a statement of the form, if $Acap B = emptyset $ then $B = emptyset $.

The statement of 5.1 in Choie et al 2007 is, "If Robin’s inequality holds for all Hardy–Ramanujan integers n with 5041 $leq $ n $leq $ x it holds for all integers in this same range."

Hardy–Ramanujan integers are defined in Choie et al 2007, p.367 but it is not necessary to know their definition to understand my question.

Define the following sets,

H = set of all Hardy-Ramanujan numbers in ℕ

X = set of integers n such that 5041 ≤ n ≤ x

V = set of numbers in ℕ that violate Robin's Inequality

$emptyset $ = empty set

then 5.1 may be re-stated as,

5.1a : If $(Hcap X)cap V = emptyset $ then $Xcap V = emptyset $

and so 5.1 iff 5.1a (equivalence 1).

Set intersection is associative so 5.1a may be re-stated,

5.1b : If $Hcap (Xcap V) = emptyset $ then $Xcap V = emptyset $

and so 5.1b iff 5.1a (equivalence 2).

Equivalences(1,2) imply 5.1 iff 5.1b (equivalence 3).

5.1b has the form, if $Acap B = emptyset $ then $B = emptyset $, and hence 5.1b is false and so equivalence 3 implies 5.1 is false.

Corollary : If 5.1 is false, and since any argument that claims to prove a false statement must itself be false, then the particular proof of 5.1 in Choie et al 2007 is false.

Note that the conclusions of 5.1a and 5.1b, $Xcap V = emptyset $, is equivalent to the Riemann Hypothesis via Theorem 1 in Robin 1984 because 5.1 does not impose any upper bound on x in the definition of X and therefore x can be arbitrarily large $geq $ 5041.

My Question

Is the above correct? Does it show 5.1 is equivalent to 5.1b and hence 5.1 is false?

The reformulation of 5.1 as a statement about the sets involved and the steps that lead to 5.1 being equivalent to 5.1b, seem straight forward.

On the other hand, there is some reason for skepticism as 5.1 appears in a paper co-authored by four professional mathematicians and published in a peer-reviewed journal. This makes it seem probable that if there was a problem with 5.1 then it would have been noticed.

References

Robin,G., “Grandes Valeurs de la Fonction Somme des Diviseurs et Hypothèse de Riemann”, J. de Math. pures et appl., 63 (1984), 187-213.

YoungJu Choie, Nicolas Lichiardopol, Pieter Moree and Patrick Solé, "On Robin's Criterion for the Riemann Hypothesis" "Journal de Théorie des Nombres de Bordeaux" 19 (2007), 357-372.

End

Reply on 181123 to Hagen von Eitzen

Thanks for commenting. I agree some statements that are false in general can have specializations that are true but it seems in the case, if $Acap B = emptyset $ then $B = emptyset $, there is no specialization of A and B that will change it from a false statement to a true statement. For brevity I will refer to this "if-then" statement as $S(A,B)$.

There seems to be a problem with the specialization you describe to show how $S(A,B)$
could be true.

Define the set of zeroes of $f(z)$ as $Z(f) = {z| f(z) = 0}$ and $Z(f)^C $ as the complement of $Z(f)$ where $f : ℂ mapsto ℂ $ .

You define $A = { frac 1n | nin ℕ }$ and, given the above definition, your $B$ = $Z(f)^C $ so that $Acap B = emptyset $ is equivalent to $A subseteq Z(f)$.

The specialization of $S(A,B)$ is then, if $A subseteq Z(f)$ then $Z(f)^C = emptyset $, and so the specialization has not changed $S(A,B)$ from false to true.

End

Reply on 181124 to lalala

The statement, if $Acap B = emptyset $ then $B = emptyset $, is false for any choice of $A$ and $B$ because there is not enough information in the premise, $Acap B = emptyset $, to draw the conclusion, $B = emptyset $.

For example, if there is a box of apples and a separate box of oranges, the information that the boxes are separate is not enough to determine how many apples are in one box or how many oranges are in the other.

If you add the premise, $B = emptyset $, then the statement becomes, if $Acap B = emptyset $ and $B = emptyset $ then $B = emptyset $, which is true but it is not the original statement.

End

Reply on 181125 to user120527

You are wrong when you assert that, “Hagen von Eitzen's comment is already a full answer”. I explain in my reply to his comment why his example does not make S(A,B) true.

Your example does not make S(A,B) true either.

You say, “A proof by (strong) induction consists of establishing,

$Vcap [0,n]=emptyset implies Vcap [0,n+1]=emptyset$.

Let $A=[0,n], B=Vcap [0,n+1]$, it's a statement of the form $Acap B = emptyset $ then $B = emptyset $ “.

Your statement is actually of the form, $Vcap A =emptyset implies B=emptyset$,

and your assertion that it is of the form S(A,B) is equivalent to asserting
$V =emptyset$ or $V={n+1}$, which is shown as follows.

Your assertion is at first equivalent to asserting $V = B$ which, using your definition of $B$, means asserting $V = Vcap [0,n+1]$.

Given the premise, $Vcap [0,n]=emptyset$, and given $[0,n+1]= [0,n]cup {n+1}$ and given the distributivity of $cap$ over $cup$ then asserting $V = Vcap [0,n+1]$, is equivalent to asserting, $V = V cap {n+1}$ which is equivalent to asserting $V =emptyset$ or $V={n+1}$.

There is no reason why a premise that two sets are disjoint should imply one of the sets is empty. S(A,B) is false by virtue of its form which independent of what sets A and B may be.

End

Reply on 181129 to user120527

user120527 says...

@gjh The mistake you make is that you think "A⟹B" says that we can deduce B from A. It's really not. It says that if A is true, then B is true. For example, (1<0⟹1>0) is true.

Reply

It seems the kind of implication user120527 means is what is usually called “material implication” and its use, so far as I know, is limited to the Propositional/Predicate Calculus. The kind of implication I have always meant is “ordinary implication”. The difference is that material implication evaluates the truth-value of an implication solely from the separate truth-values of the premise and the conclusion while ordinary implication does this but in addition it also evaluates the implication on whether the conclusion can be validly deduced from the premise.

One of the problems with material implication is that it evaluates an implication as true when both the premise and conclusion are true but they may be independent of each other and the conclusion may not be deducible from the premise. For example, material implication would evaluate as true the implication, “If some roses are red then some swans are black”.

Books on logic sometimes draw the reader’s attention to this feature of material implication. For example, Basson and O’Connor 1959 (reference below) says about material implications like the example above, “These implications seem to common sense to be neither true nor false but rather to be nonsensical because the ordinary use of “if-then” in linking sentences into implications presupposes that it is the meanings of the sentences so linked (and not their truth-values) which determine whether the implication is true or false.”

Examples like the one above show material implication does not have much use outside the Propositional/Predicate Calculus so I think it is reasonable for me to assume in a math context, without stating it explicitly, that implications are ordinary implications and not material implications.

Reference

Basson, A. and O’Connor, D., “Introduction to Symbolic Logic”, University Tutorial Press, London 1959, p.26.

The answer to your question is that 5.1 and 5.1b are equivalent. The derivation is too short and simple to hide any errors.

It is puzzling why there appears to be disagreement about whether S(A,B) is false for every pair A,B of sets. It is always false due to its form. It doesn’t matter what the sets A,B may be. There is no way to deduce a set is empty solely from the fact it has no members in common with some other set.

There is a statement similar but different to S(A,B) that is true. This is, if $Acap B = emptyset $ for all $A$ then $B = emptyset $. However in 5.1b there is no quantification over $A$ because $A$ is a particular set, namely $H$, which is the set of all Hardy-Ramanujan numbers in ℕ and an infinite subset of ℕ.

Granted 5.1 is false the next question is what results (if any) in the literature have relied on 5.1? I have seen some some papers on t-free numbers and Robin’s Inequality which have Choie et al 2007 as a reference but I haven’t read them closely enough to know if they rely on 5.1.

Reply to user120527

Perhaps you should try to justify the claims that "It is always false due to its form" and "there is no way to deduce...". – user120527

Perhaps I do need to justify these claims and perhaps you need to justify why you down voted the question. Could it be because your counter example was shown to be nonsense?

It is not clear at what level you mean by “justify” but I can explain what I mean by the phrases you quote.

By “It is always false due to its form”, I mean the form of S(A,B) is such that the conclusion does not follow from the premise for any choice of A and B. Some statements are always false “due to their form”. For example, “if P then not P”, where P is some premise.

By “there is no way to deduce …” I mean the same. The premise of S(A,B) is insufficient to deduce the conclusion of S(A,B) for any choice of A and B.

However if you dispute this and know how the mere disjointness of two sets implies one of the sets is empty, then please disclose your argument.

Reply 2 to user120527

By justify, I mean give a proof (good luck with that). I downvoted because you are dead wrong. You should follow a basic course on logic, that would help you to understand why "if P then not P" is true if P is false. – user120527

Reply 2

So you down voted because you think I am dead wrong. That may be a reason for down voting my answer but it does not justify down voting the question which is by someone else and is just a question neither right nor wrong.

A statement “If A then B” is the usual short way of writing the longer statement, “If A is true then B is true”. My example was of a statement that is always false because of its form. When written the long way it is, “If P is true then not P is true” or equivalently “If P is true then P is false”.