OK wtf girls dont poop

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Deny. Your proof shows (if we take all assumptions to be correct which I don't) that there are a countable number of hot non-asian chicks. However, there can be a countable number of hot non-asian chicks plus a set of non-hot non-asian chicks do u c y?

Also, I find it funny that people found that pic "hot." It was a bad angle that hardly shows anything and is far from a flattering picture.

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Deny. Your proof shows (if we take all assumptions to be correct which I don't) that there are a countable number of hot non-asian chicks. However, there can be a countable number of hot non-asian chicks plus a set of non-hot non-asian chicks do u c y?


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Fine, you had to go and nit up my proof.

As long as it's already being nitted up, no I don't CY, unless you're claiming the set of hot chicks is uncountable, or that for some reason we can't index you as hot chick #1.

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Also, I find it funny that people found that pic "hot." It was a bad angle that hardly shows anything and is far from a flattering picture.

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You're a real life girl posting in BBV4L, what'd you expect?

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*sigh* You can have a set of numbers that contains all the integers but also contains a few fractions, yes? Just because you can show all the integers are in the set via your proof doesn't disprove the existence of the fractions.

[ianisakson]

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"When we love a woman, we prefer not to think of the repulsive natural functions to which every woman is subject." -Nietzsche

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quoting nietschze on poop is like quoting richard nixon on foreign policy

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[X] awesome FYP

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*sigh* You can have a set of numbers that contains all the integers but also contains a few fractions, yes? Just because you can show all the integers are in the set via your proof doesn't disprove the existence of the fractions.

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But that set with all the integers plus a few fractions can still be put into a 1-1 correspondence with the integers, yes? Isn't that what it means for it to be a countable set?

Beat: Lamest argument in history of BBV4L.

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But now you're changing your proof. Just because the number of hot chicks and non hot chicks can be put into bijective correspondence with a countable number of hot chicks doesn't magically change the ugly chicks into hot chicks.

The rationals can be put into bijective correspondence with the integers. However, even though you can put them into 1-1 correspondence 1/2 is still NOT AN INTEGER.

You need an isomorphism and all you have is a bijective map.

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But now you're changing your proof. Just because the number of hot chicks and non hot chicks can be put into bijective correspondence with a countable number of hot chicks doesn't magically change the ugly chicks into hot chicks.

The rationals can be put into bijective correspondence with the integers. However, even though you can put them into 1-1 correspondence 1/2 is still NOT AN INTEGER.

You need an isomorphism and all you have is a bijective map.

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Okay, if each non-Asian chick is uniquely identified by some real number that may or may not be an integer, then I agree that you're a million percent correct.

But if the set of non-Asian chicks is simply some countable set with elements that we can index however we please, then we can certainly index them with the integers, right?

Major beat: Lamest argument in history of BBV4L continues.
Variance: Apparently losing argument to a girl.

[divides_by_zero]

Ok, now you've got a different structure.

The problem now is your inductive argument. Sure, hot chicks go to school with other hot chicks (although I don't know why since hello, competition, everyone knows hot chicks hang out with ugly chicks to appear hotter), but even if you accept that doesn't imply that the NEXT chick is hot. You might be able to impy that there is ANOTHER hot chick, but it doesn't have to be the next one in the index. Also, the other hot chicks they go to school with could simply be the first n hot chicks - an equivalence relation sort of.

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Ok, now you've got a different structure.

The problem now is your inductive argument. Sure, hot chicks go to school with other hot chicks (although I don't know why since hello, competition, everyone knows hot chicks hang out with ugly chicks to appear hotter), but even if you accept that doesn't imply that the NEXT chick is hot. You might be able to impy that there is ANOTHER hot chick, but it doesn't have to be the next one in the index. Also, the other hot chicks they go to school with could simply be the first n hot chicks - an equivalence relation sort of.

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I agree. It was hard as hell to make up an inductive step that actually made the slightest bit of sense. That was the best I could do.

Let's just make the whole thing an axiom and call it a day, because you won't convince me that the statement I was trying to prove is wrong.

[divides_by_zero]

All I need is a counterexample. Facebook should provide you with plenty.