It doesn't pay

Apr 13, 2011 16:21

Some people are just so dense that it doesn't pay to explain to them why they look stupid.

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alpha_strike April 14 2011, 00:02:50 UTC
But I don't understand! Why do you call me stoopid stoped stewped... why do you think I'm not smart?

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carly_sullivan April 14 2011, 00:57:39 UTC
Oh, sweetie. I'm talking about someone who said (and I paraphrase, but not much): I don't care if we lose as long as we win. I tried to point out gently that if you lose repeatedly, you will not win. She said that was ok, she didn't care if we lost a lot as long as we won.

I'm not going another round.

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alpha_strike April 14 2011, 01:43:54 UTC
Oh, I understand what she's talking about: it's a binary state that isn't one of the other two. See, there's 0 (losing) and 1 (winning). She's talking about the condition where 0 = 1, and we all know *that* happens all the time.

Um, we can multiply both sides of the equation by zero without changing anything, right?

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carly_sullivan April 14 2011, 01:50:48 UTC
Sure, and then we can take the square root of -1, accurate to four decimal places, and divide it by the log of a gazillion.

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alpha_strike April 14 2011, 02:21:53 UTC
OK, now you're freaking me out: I thought you're allowed to multiply things by the square root of -1. I think I used to do that while doing vectoral analysis, or calculating phasors, something like that. Isn't it the imaginary component of a complex number? Am I misremembering?

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carly_sullivan April 14 2011, 09:45:17 UTC
LOL. It IS the imaginary unit, beloved by math nerds and electronics geeks. That would be us, right?

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