[QUOTE=Mr. P-1;248189]Forget partitions. You don't know what they are, as we haven't done them yet.

[COLOR="Red"]Sorry partitions came first in the book.[/COLOR]

Yes, that's a binary relation. What set is this relation [i]on[/i]?



Transitivity is a claim about [i]all[/i] possible combinations of x, y, and z. You can use a single counterexample to disprove such a claim, but you cannot use a single example to prove it.

Let x, y, and z be any squares on a chess board and let c() be the colour function on x

Then x is the same colour as y implies that c(x) = c(y). (This is not really an implication, more a restatement).

y is the same colour as z implies that c(y) = c(z).

But c(x) = c(y) AND c(y) = c(z) IMPLIES c(x) = c(z)

This is true because equality is a transitive relation. (This is an elementary property of equality; we don't need to prove it.)

And so x is the same colour as z.

Can you prove reflexivity and symmetry, using the above as a model? Hint: equality is more than just transitive. Equality is an equivalence relation.[/QUOTE]
I kinda get that last part now.

[QUOTE=davar55;248147]RDS seems to assume a more advanced student.
Some students struggling with the definitions of sets, relations,
equivalences, and partitions are only at the junior high level.[/QUOTE]

Where I live I don't remember using most of anything explained here in high school.

[QUOTE=Mr. P-1;248181]That was my assessment of sm88's level earlier in the thread.



This would have been much easier for all of us, including sm88, if we'd started with the basics of set theory, and worked our way through unions, intersections, differences, complements, De Morgan's laws, cartesian products, and so on, and only then did relations. As it is, I'm not sure if sm88 even knows what it means for two sets to be equal.

But the relations thing has achieved a certain momentum, so I'm running with it.[/QUOTE]


union: the collection of sets into a bigger set such that all subsets are represented.

intersection: having a element in common between sets.

and I've now heard of everything except De Morgans law, but can't really define them all.

looking up difference got me to [url]http://en.wikipedia.org/wiki/Inequality_(mathematics[/url])

but looking up complement I got to : [url]http://en.wikipedia.org/wiki/Complement_(set_theory[/url]) and [url]http://en.wikipedia.org/wiki/Difference_(set_theory)#Relative_complement[/url]

[QUOTE=R.D. Silverman;248203]It might be useful if he also gained some understanding of quantifiers
and got some practice in the use of variables.[/QUOTE]

Meaning maybe looked up Symbolic Logic and elementary algebra.

Texts, google, our forum, plus pen and paper thinking -- great resources.

the De Morgan law I'm having trouble understanding is NOT (P AND Q) = (NOT P) OR (NOT Q) I get the other one.

[QUOTE=science_man_88;248302]the De Morgan law I'm having trouble understanding is NOT (P AND Q) = (NOT P) OR (NOT Q) I get the other one.[/QUOTE]

The OR here is inclusive or, so just use truth tables.

Binary logic.

[QUOTE=davar55;248306]The OR here is inclusive or, so just use truth tables.

Binary logic.[/QUOTE]

In other words its like saying to not have both p and q you need to either have a thing not include p or not include q .

[QUOTE=science_man_88;248314]In other words its like saying to not have both p and q you need to either have a thing not include p or not include q .[/QUOTE]

Exactly. Like the Law of Excluded Middle.

[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]

look back to this davar55 I solved it but I couldn't explain it I think now I could, lets see:

2^25 binary relations I believe it would stem from the fact that each square has 2 possible relations ~ "is black" or "is white" so could it be the number of (relations per pair(square))^ (the number of squares) ? because each relation can be cast out for each square independent of ordering so you have 2^25 paths you could take through the different relations. As to the combinatorics of it I have formula from number freak but I don't know all the numbers to plug in.

[QUOTE=science_man_88;248337]look back to this davar55 I solved it but I couldn't explain it I think now I could, lets see:

2^25 binary relations I believe it would stem from the fact that each square has 2 possible relations ~ "is black" or "is white" so could it be the number of (relations per pair(square))^ (the number of squares) ? because each relation can be cast out for each square independent of ordering so you have 2^25 paths you could take through the different relations. As to the combinatorics of it I have formula from number freak but I don't know all the numbers to plug in.[/QUOTE]

The wording is poor but comprehensible.

You're definitely on the right track.

I don't yet know what to suggest except try re-wording.

[QUOTE=science_man_88;248302]the De Morgan law I'm having trouble understanding is NOT (P AND Q) = (NOT P) OR (NOT Q) I get the other one.[/QUOTE]

I actually meant De Morgan's laws as they apply to sets.