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4 is more than 5?


Krz

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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

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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 :)

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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

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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...

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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

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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.

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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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