[QUOTE=Mr. P-1;247914]I'm not asking the question you think I am. In particular I did not intend symmetries to be regarded as duplicates.

In fact, my question is just a disguised version of the question asked by CRGreathouse earlier in the thread: if S = {1, 2, 3, 4, 5,}, how many different binary relations are there on S?[/QUOTE]

Well if what I read is any help binary logic should imply 2^25 then.

[QUOTE=science_man_88;247917]Well if what I read is any help binary logic should imply 2^25 then.[/QUOTE]

Correct, which, incidentally is 33,554,432. Now for the hard bit: How is this the same question essentially as "how many binary relations are their on S"?

[QUOTE=Mr. P-1;247919]Correct, which, incidentally is 33,554,432. Now for the hard bit: How is this the same question essentially as "how many binary relations are their on S"?[/QUOTE]

I'm guessing it has to do with the fact that since each one can be express by one relation X and another !(!X) where ! is not since each has square has 2 relations it's binary and and 2^25 = (relations per pair)^pairs.

Set theoretical notation: See your answer to question 1.

[QUOTE]What have you missed?

The words symmetric ?[/QUOTE]

You showed that the symetry property holds for (Son, Daughter). You need to show it holds for every pair in the relation.

[QUOTE]What is the correct implication?

if you want one that doesn't repeat I see no other as the relations in the list I gave number just 2.[/QUOTE]

Assume the transitive property applies. You said Son ~ Daughter AND Daughter ~ Son IMPLIES Son ~ Daughter. Incorrect. What should follow "IMPLIES" here?

[QUOTE=science_man_88;247921]I'm guessing it has to do with the fact that since each one can be express by one relation X and another !(!X) where ! is not since each has square has 2 relations it's binary and and 2^25 = (relations per pair)^pairs.[/QUOTE]

This is gobbledegook. And I think we should just leave this question for now, as we're making progress with the other one.

[QUOTE=Mr. P-1;247922]Set theoretical notation: See your answer to question 1.



You showed that the symetry property holds for (Son, Daughter). You need to show it holds for every pair in the relation.



Assume the transitive property applies. You said Son ~ Daughter AND Daughter ~ Son IMPLIES Son ~ Daughter. Incorrect. What should follow "IMPLIES" here?[/QUOTE]

The only other relation I can see for after implies is Daughter~Son because son~son and daughter~daughter break your definition, and father~son or mother~son or father~daughter or mother~daughter don't hold any sibling relations.

[QUOTE=science_man_88;247925]The only other relation I can see for after implies is Daughter~Son because son~son and daughter~daughter break your definition, and father~son or mother~son or father~daughter or mother~daughter don't hold any sibling relations.[/QUOTE]

The definition of transitive is:

x ~ y AND y ~ z IMPLIES x ~ z.

You wrote

Son ~ Daughter AND Daughter ~ Son IMPLIES something.

To make this fit the definition, what is x? What is y? What is z?

[QUOTE=Mr. P-1;247927]The definition of transitive is:

x ~ y AND y ~ z IMPLIES x ~ z.

You wrote

Son ~ Daughter AND Daughter ~ Son IMPLIES something.

To make this fit the definition, what is x? What is y? What is z?[/QUOTE]

son~son then which disproves the definition hence it's false and either the definition needs changing to be true or it's not an equivalence relation, because proving the transitive is impossible under the definition.

[QUOTE=science_man_88;247929]son~son then which disproves the definition hence it's false[/QUOTE]

We have two definitions: the one you gave in your initial reply to me, and Xilman's variant.

Is son~son true under a your definition, b xilman's variant?

[QUOTE=Mr. P-1;247931]We have two definitions: the one you gave in your initial reply to me, and Xilman's variant.

Is son~son true under a your definition, b xilman's variant?[/QUOTE]

according to sharing at least one parent yes son~son is true, with x!~x then it can't be possible hence any transitive or reflexive forms that need it are disallowed to be equivalence relations.

[QUOTE=Mr. P-1;247914]I'm not asking the question you think I am. In particular I did not intend symmetries to be regarded as duplicates.

[/QUOTE]

Then your question was not completely stated.