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# 155

Facebook Friends - January 25, 2010
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Yup, that's what it looks like.

Nor are the communative...

Argh--it must be Mnday. I can't even spelll. Much less remember my math facts.

Make that: Nor are they symmetric.

Isn't the a null function by definition symmetric?

Whats a mathmatician clowns favorite trick?
A Pi in the face!
(gotta make this a comic somehow lol)

Will it sting if I get pi in my i?

I imagine it would

Pi^i is messy

If you think, THAT is messy, what do you think i^pi is?

But e^πi is not messy at all--it negates everything...

e^(e^(e^(...)*pi/2)*pi/2)*pi/2 = sqrt(e^(e^(e^(...)*pi/2)*pi/2)*pi)

I find that rather amazing.

Aww... someone needs a friendship request.

Nope.. this doesn't work..

If the result of making FB-relations transitive is as shown, the person must have at least one friend (other than himself).

If they were, we'd finally find out whether social networks are non-trivially partitioned. I always suspected there could be good probability, that such networks do not have only one weak component. Of course you are not supposed to count the isolated nodes ... like ... you know ... you.

Psychology Today: Why Your Friends Have More Friends Than You Do.

http://www.psychologytoday.com/blog/the-scientific-fundamentalist/200911/why-your-friends-have-more-friends-you-do

You need some imaginary friends -- http://xkcd.com/410/

But... but... the Mike class is implicitly friends with itself! It can't be 0.

Relationship Status is also not reflexive.

Pfft, love isn't symmetric either.

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