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Okay if this was rounding them then just .9 or even .999999999999999999999999999 would be = to 1,but if it was no rounding, then .9 would be closest to one then the more 9's you get the farther you get from a one
he has eyebrows of authority + 5. Are you kidding?
Quote from: Wolf_Man on November 05, 2006, 11:36:06 amOkay if this was rounding them then just .9 or even .999999999999999999999999999 would be = to 1,but if it was no rounding, then .9 would be closest to one then the more 9's you get the farther you get from a oneWrong, you get closer with more numbers following in the decimal..99 is not closer to 1 then .9
Dascoo's post reminds me one of the ali g's show where he was interviewing nasa lol
I like this algebraic proof my university math professor showed us:a = 0,999...Multiply by 10: 10a = 9,999...Substract a: 10a - a = 9Divide by 9: a = 1http://en.wikipedia.org/wiki/0.99999End of discussion!
0.333… = 1⁄33 × 0.333… = 3 × 1⁄30.999… = 1
c = 0.999…10c = 9.999…10c − c = 9.999… − 0.999…9c = 9c = 1
.99999999--> will continually get closer to 1, but never reach it. draw a graph of y=1/x to see a visual example.
Quote from: Plonkoon on November 05, 2006, 11:26:39 am.99999999--> will continually get closer to 1, but never reach it. draw a graph of y=1/x to see a visual example.Plonkoon wins for giving a tangible explination. Applaud.
Students of mathematics often reject the equality of 0.999… and 1, for reasons ranging from their disparate appearance to deep misgivings over the limit concept and disagreements over the nature of infinitesimals. There are many common contributing factors to the confusion: * Students are often "mentally committed to the notion that a number can be represented in one and only one way by a decimal." Seeing two manifestly different decimals representing the same number appears to be a paradox, which is amplified by the appearance of the seemingly well-understood number 1.[12] * Some students interpret "0.999…" (or similar notation) as a large but finite string of 9s, possibly with a variable, unspecified length. If they accept an infinite string of nines, they may still expect a last 9 "at infinity."[13] * Intuition and ambiguous teaching lead students to think of the limit of a sequence as a kind of infinite process rather than a fixed value, since a sequence need not reach its limit. Where students accept the difference between a sequence of numbers and its limit, they might read "0.999…" as meaning the sequence rather than its limit.[14] * Some students regard 0.999... as having a fixed value which is less than 1 but by an infinitely small amount. * Some students believe that the value of a convergent series is an approximation, not the actual value.
There's no such thing as .000~1. That's called zero. And no, you can't prove 1=2. They divided by zero. MATH IS MATH, AND IT IS THE LAW. MATH DOESN'T NEED ANYONES OPINION. YOU CAN'T PROVE SOMETHING THAT ISN'T TRUE, PANZER. MATH IS NOT INACCURATE. THE NUMBER SYSTEM IS PERFECT BY DEFINITION.
0.9999~ is impossible to graph, at least, completely. It's like trying to see where the y = x line ends by making a graph out of it.
And swarmer:0.333… = 1⁄3is still rounding. Just go with fractions people.
it would be .000000 infinity zeros and then 1 as the last number since infinity zeros go on forever
Quote from: Nfsjunkie91 on November 05, 2006, 11:52:48 amQuote from: Wolf_Man on November 05, 2006, 11:36:06 amOkay if this was rounding them then just .9 or even .999999999999999999999999999 would be = to 1,but if it was no rounding, then .9 would be closest to one then the more 9's you get the farther you get from a oneWrong, you get closer with more numbers following in the decimal..99 is not closer to 1 then .9I'm confused. Are you saying that .9 is closer to 1 than .99 or what?.99 is closer to 1 because it is .09 more than .9 (or .90) and so on and so forth. It's like saying that 99 is closer to 100 than 90.