Thread: Why .9r = 1
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Old 10-16-2004, 11:25 AM   #18 (permalink)
Slavakion
Mjollnir Incarnate
 
Location: Lost in thought
Quote:
Originally Posted by Paradise Lost
The problem looks like this.

x = .9r
10(.9r)=9.9r
10(.9r)-.9r=9.9r-.9r
8.1r=9.0r

r=1.11r
1=1.11 (Which is the correct answer, although of course, the not true one... something like that.)
Well, um, yeah. That's where you went wrong. How did you get 8.1?

Quote:
Originally Posted by AngelicVampire
(because if we are dealing with say any number of 9s the multiplied number will have one less dp than the original number, so will produce a 1 at the end of that number...)
What do you mean? When I multiply .9r by 10, I get .9almostr or .9r with one less decimal place? I think that inifinity minus any non-infinite number is still infinity. Especially one, since one is so... oneish. This is when you get into those freaky (*cough* obsessive *cough*) metaphysical debates over the nature of infinity.

Last edited by Slavakion; 10-16-2004 at 11:29 AM..
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