Krz Posted February 17, 2006 Posted February 17, 2006 a friend of mine said there was a way to prove 4 is more than 5 could anyone please explain?
silkworm Posted February 17, 2006 Posted February 17, 2006 I'd poke your friend in the eye and give him a titty twister. -4 can be considered as more than -5 in the sense that if you owed $4 instead of $5 you'd have more money, but who cares?
Krz Posted February 17, 2006 Author Posted February 17, 2006 haha i know that, but apparently there's a way to prove it using logs.
Sisyphus Posted February 17, 2006 Posted February 17, 2006 You can "prove" anything if you're willing to break the rules, which you would have to do to prove 4>5.
Krz Posted February 17, 2006 Author Posted February 17, 2006 You can "prove" anything if you're willing to break the rules, which you would have to do to prove 4>5. so.. would this be considered breaking the rules? : x=.999999repeating 10x=9.999999repeating 10x-x = 9.99999 - .99999 9x=9 x=1
silkworm Posted February 17, 2006 Posted February 17, 2006 Is that just saying 0.9999999/0.99999999 = 1?
NeonBlack Posted February 17, 2006 Posted February 17, 2006 haha i know that, but apparently there's a way to prove it using logs. Yes, usually by dividing by the log of 1, you can prove just about anything.
Dave Posted February 17, 2006 Posted February 17, 2006 There's an awful lot of mathematical trickery out there. That doesn't mean that it's true. Clearly there's no way that 4 > 5, unless of course you're considering some crazy field other than the reals which has a different ordering. But since he's talking about logs, you can pretty much discount that. Bottom line: ignore him
chilehed Posted February 19, 2006 Posted February 19, 2006 Yes, usually by dividing by the log of 1, you can prove just about anything.In my first algebra class the instructor showed a proof that 1=2 and challenged us to find the flaw. The smart kid took finally figured out that it involved dividing by zero. log 1 = 0
Klaynos Posted February 19, 2006 Posted February 19, 2006 no, and it's true, .999.... = 1 if you want to be 100% acurate then your statement is wrong. If you're prepared to round stuff then you introduce errors and your acuraccy moves alway from 1 and you can eventually prove pretty much anything...
Connor Posted February 19, 2006 Posted February 19, 2006 no, there is absolutely no number between .999... and 1 they are mathematically equivilant, as can be seen with Krz's proof 1) [math]x=.999...[/math] 2) [math]10x=9.999...[/math] [math] x= .999...[/math] 3) [math]10x-x=9.999... - .999...[/math] [math]9x=9[/math] [math]x=1[/math] no logical fallicies
Dave Posted February 19, 2006 Posted February 19, 2006 Oh God, not another one of these threads. Please, for the love of the most Holy One, read the substantial thread on this that already exists, otherwise I run the risk of having to gauge my eyes out with a rusty spoon.
clarisse Posted February 19, 2006 Posted February 19, 2006 Oh God, not another one of these[/i'] threads. Please, for the love of the most Holy One, read the substantial thread on this that already exists, otherwise I run the risk of having to gauge my eyes out with a rusty spoon. Lol, poor Dave. I can only imagine his suffering.
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