## Let's go.........RANDOM!

xander

- Captain Jean-Luc Picard
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- bert_the_turtle
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You forgot to tell us what the topological definition of a continuous function is, and what the topology induced by a metric is.

And you made me hate math with the pure legwork proof you made me go through in my head. Congratulations, that is quite an accomplishment! I'll need to read some Lie Algebra stuff now.

And you made me hate math with the pure legwork proof you made me go through in my head. Congratulations, that is quite an accomplishment! I'll need to read some Lie Algebra stuff now.

What? To prove De Morgan's Laws you just need a quick n' dirty truth table. How could you POSSIBLY mess that up?!Xocrates wrote:Today in class we proved Morgan's laws... by using Morgan's laws. I did a double take at that.

Granted, the point was to practice a method that relied on Morgan's laws, so what we were proving was irrelevant. Still weird though.

And like I said, the method (or more precisely, the conversion of the equation into canonical normal form) assumes the laws are true. So the goal of the exercise wasn't to prove the laws to be true, but to learn to apply the method.

But since you're curious:

Hipothesis:

~( A || B) -> ~A && ~B

proof by contradiction, so we'll deny it:

~(~( A || B) -> ~A && ~B)

Convert into CNF (skipping several steps here - which includes applying de morgan's laws)

~A && ~B && (A || B)

or, alternative representation:

{{~A},{~B},{A,B}} -- this essentially means that in order for the hipothesis to be true, all internal groups must be true

Proof:

1. {~A} Premise

2. {~B} Premise

3. {A,B} Premise

4. {B} Resolution, (1,3) -- this essentially means that for ~A to be true, the remaining symbols in {A,B} - in this case B - must be true

5. {} Resolution, (2,4) -- contradiction, since it means both ~B and B must be true.

So the hipothesis must be true.

Edit: Uh, does my avatar work? I think I may have messed something up, it doesn't appear on my screen

bert_the_turtle wrote:You forgot to tell us what the topological definition of a continuous function is, and what the topology induced by a metric is.

And you made me hate math with the pure legwork proof you made me go through in my head. Congratulations, that is quite an accomplishment! I'll need to read some Lie Algebra stuff now.

It was assumed that you knew what those were. :P And yes, it is a pure legwork proof, and a clever undergrad should be able to figure it out. It was a nice break from covering r-balls with rho-balls.

xander

- paktsardines
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