dude thats messed up. x = 0.999 10x = 9.999 10x - x = 9.999 -x (your only subtracting 1x from 10x... so where are the other 9x?????) 9X = 9 and x is a constant, not a variable in the original equation.
You're can't solve for x because it is a constant. Subtracting the constant is masterbation but that's what was done originally. x = 0.999 10x = 9.999 10x - x = 9.999 -x 10(0.999) - (0.999) = 9.999 - (0.999) 9.999 - 0.999 = 9.999 - 0.999 9 = 9
well if you substitute 1 for 0.999... x = 1 10x = 10 10x - x = 10-x 10(1) - (1) = 10 - (1) 10 - 1 = 10 - 1 9 = 9 its the same.
Your signature creeps me the fuck out. Sorry, that's all I had to say, there was a couple of these last time I was active at the boards, m onths ago...
no, there are one 2 numbers that could make the answer to that equasion 9=9. .999..., and 1. which make them equal.
x = 0.999 10x = 10 10x - x = 10-x 10(0.999...) - (0.999... ) = 10 - (0.999) 9.9999... - 0.999... = 9.00...001 9 = 9.00...001 Nope. Only one # works in the equation: 1.
it was all good til u did 9x = 9. 10x is equal to 9.999... and 10x - x is equal to one. but if x is .999... then 9x isnt = 9 nice try tho
Well, hopefully I can settle this matter. While soma's equation is correct the logic being used to explain it is a bit flawed. Not a big deal, this thing floats all over the internet and this is just a reposting. Anywho: x = 0.999... 10x = 9.999... 10x - x = 9 9x = 9 x = 1 1 = 0.999... There is nothing wrong with the use of X or anything like that as people have suggested. The issue comes from "your" understanding of a number ACTUALLY continuing forever. An informal proof to get you thinking: How far would you have to travel from 1 on a number line to get to .999....? You might think .000...1, but notice you terminated that number with a 1? Hence it doesn't continue forever, but eventually will terminate not so with .999.... One more informal: .999... is a periodic number, .234234234... is periodic too, as is .19191919 How do we construct these numbers? We take the number we want to period, divided by as many 9s as there are digits in the period (in the case of 234, there are 3 digits so divide by 999). Check it out: .234234234... = 234/999 .151515... = 15/99 then... .999.... = 9/9 = 1 Now for a more mathematical proof: Think about a repeating decimal, what IS that? Well a we use base 10 numbers so lets look at some numbers broken down to their base representation: 458 = 4*100 + 5*10 + 8*1 458.9 = 4*100 + 5*10 + 8*1 + 9/10 458.999... = 4*100 + 5*10 + 8*1 + (9/10 + 9/100 + 9/1000 + ...) Get it? So a repeating decimal, as with all decimals, is a sum; though in their case it repeats forever. So essentially in this case we are taking the limit of the summation starting at n=1^m of 9/10^n to n= infinity with the limit of m going to infinity. This sum is the same as 1. So, .999... = 1
Don't believe me fine how bout these people? http://www.newton.dep.anl.gov/askasci/math99/math99167.htm Sweetheart, the only way you'll ever get into my school is with a bucket and a mop. Shut your tooth.
sourdiesel...lmao! and floydian slip...are you saying what i said, only more eleqoently? the problem is thinking that you can multiply 0.9999 recurring by a number as THOUGH it didnt continue on infinately, when in fact it does...is that right? to the person who replied to my post...yes NORMALLY you can move the decimal place one point to the right to multiply anything by 10. But that (if i've got this right) doesn't work with recurring numbers...