R.D Silverman's number theory homework

[science_man_88]

Originally Posted by R.D. Silverman

Then your question was not completely stated.

Beef it out in PM people! why else would PM be in this forum?

[Mr. P-1]

Originally Posted by science_man_88

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.

Lets use a different symbol for xilman's variant. Lets call his relation #. So Son ~ Son, but Son !# Son.

To show that a property (symmetry, reflexivity, transitivity) does not hold, you only need to find a single counter example. So Son # Daughter AND Daughter # Son BUT Son !# Son proves that # is not transitive. And you've already shown that it's not reflexive. Is it symmetric?

To show that a property does hold, you need to show this for every combination of variables. To show that ~ is symmetric, you need to show that for every combination of values of x and y, IF x ~ y THEN y ~ x. So far, you've only looked at one particular instance: x = Son, y = Daughter. To show that ~ is transitive, you need to show that for every combination of values of x, y, and z, IF x ~ y AND y ~ z THEN x ~ z. Again, so far, you've only looked at one particular instance: x = z = Son, y = Daughter.

[science_man_88]

Originally Posted by Mr. P-1

Lets use a different symbol for xilman's variant. Lets call his relation #. So Son ~ Son, but Son !# Son.

To show that a property (symmetry, reflexivity, transitivity) does not hold, you only need to find a single counter example. So Son # Daughter AND Daughter # Son BUT Son !# Son proves that # is not transitive. And you've already shown that it's not reflexive. Is it symmetric?

To show that a property does hold, you need to show this for every combination of variables. To show that ~ is symmetric, you need to show that for every combination of values of x and y, IF x ~ y THEN y ~ x. So far, you've only looked at one particular instance: x = Son, y = Daughter. To show that ~ is transitive, you need to show that for every combination of values of x, y, and z, IF x ~ y AND y ~ z THEN x ~ z. Again, so far, you've only looked at one particular instance: x = z = Son, y = Daughter.

Well since son~daughter is reversible and then still true I would think yes. The relation of sibling is a reversible relation so in this case = reverses to = not to and hence both sides hold so it is confirmed.

[axn]

Originally Posted by xilman

Your analysis is correct, given your definition of "sibling".

Originally Posted by science_man_88

2) in my definition of sibling it can roughly be defined as having at least one parent in common(at home or biologically)

His analysis is correct, insofar as the set is a very small one lacking any counterexamples. However, if we start with a more realistic set, by his definition, the relation is not transitive (think of step siblings sharing a half sibling).

[Mr. P-1]

Originally Posted by science_man_88

Well since son~daughter is reversible and then still true I would think yes. The relation of sibling is a reversible relation so in this case = reverses to = not to and hence both sides hold so it is confirmed.

You're still not expressing yourself clearly.

One way to do show what you need to show would be by brute force: list every single combination of x and y, or x, y, and z, and verify that the conditions hold.

That's only practical in this case because the underlying set S is so small. Here's a smarter way:

Proposition: ~ is symmetric.

Proof: we need to show that x ~ y IMPLIES y ~ x. Consider the case where x = y. Then we can substitute x for y in that statement:

x ~ x IMPLIES x ~ x

which is trivially true. Now consider the case where x != y. There are just two cases where this is true: Son ~ Daughter and Daughter ~ Son, and each implies the other is a true statement.

Proposition: ~ is transitive.

Proof: we need to show that x ~ y AND y ~ z IMPLIES x ~ z. Consider the case where x = y. Substituting, we get

x ~ x AND x ~ z IMPLIES x ~ z

which is trivially true. Can you complete this proof? (Hint: what other cases do you need to consider?)

[science_man_88]

Originally Posted by Mr. P-1

You're still not expressing yourself clearly.

One way to do show what you need to show would be by brute force: list every single combination of x and y, or x, y, and z, and verify that the conditions hold.

That's only practical in this case because the underlying set S is so small. Here's a smarter way:

Proposition: ~ is symmetric.

Proof: we need to show that x ~ y IMPLIES y ~ x. Consider the case where x = y. Then we can substitute x for y in that statement:

x ~ x IMPLIES x ~ x

which is trivially true. Now consider the case where x != y. There are just two cases where this is true: Son ~ Daughter and Daughter ~ Son, and each implies the other is a true statement.

Proposition: ~ is transitive.

Proof: we need to show that x ~ y AND y ~ z IMPLIES x ~ z. Consider the case where x = y. Substituting, we get

x ~ x AND x ~ z IMPLIES x ~ z

which is trivially true. Can you complete this proof? (Hint: what other cases do you need to consider?)

Proposition: ~ is reflexive

Proof: we need to show that x~x. I based my statement on :

if x "is a child of" y and x "is a child of" y implies x "is a sibling of"x but I know this isn't acceptable.

[CRGreathouse]

sm, I think P-1 was asking for you to finish the proof of transitivity, not to prove reflexivity.

[science_man_88]

Originally Posted by CRGreathouse

sm, I think P-1 was asking for you to finish the proof of transitivity, not to prove reflexivity.

Well then, now considering the case x!=y

we get:

x~y and y~z implies x~z ? If not then no I can't complete the proof.

[Mr. P-1]

Originally Posted by CRGreathouse

sm, I think P-1 was asking for you to finish the proof of transitivity, not to prove reflexivity.

Yes, I only gave a partial proof. Sorry I wasn't clear.

[Mr. P-1]

Originally Posted by science_man_88

Well then, now considering the case x!=y

Instead of x!=y, try setting some other combination of two variables equal.

BTW, do you understand why

x ~ x AND x ~ z IMPLIES x ~ z

is trivially true?

[science_man_88]

Originally Posted by Mr. P-1

Instead of x!=y, try setting some other combination of two variables equal.

BTW, do you understand why

x ~ x AND x ~ z IMPLIES x ~ z

is trivially true?

because x~x is trivially true by my definition of ~ and x~z implies x~z is also trivially trues hence the whole thing is trivially true ?