jheadley 0 #1 December 16, 2007 Does .999... (repeating) equal 1.0? Quote Share this post Link to post Share on other sites
grannyinthesky 0 #2 December 16, 2007 I can do the math that shows it. Sorta boggles the mind."safety first... and What the hell..... safety second, Too!!! " ~~jmy POPS #10490 Quote Share this post Link to post Share on other sites
kallend 2,138 #3 December 16, 2007 QuoteDoes .999... (repeating) equal 1.0? Let s = 0.9999... then 10s = 9.99999... then (10s - s) = 9 9s = 9 therefore s = 1... The only sure way to survive a canopy collision is not to have one. Quote Share this post Link to post Share on other sites
grannyinthesky 0 #4 December 16, 2007 Yep. That's it!!"safety first... and What the hell..... safety second, Too!!! " ~~jmy POPS #10490 Quote Share this post Link to post Share on other sites
SpeedRacer 1 #5 December 17, 2007 Quotethen (10s - s) = 9 10s - s should equal 9s, not 9. Speed Racer -------------------------------------------------- Quote Share this post Link to post Share on other sites
grannyinthesky 0 #6 December 17, 2007 He was subtracting one equation for the other. The work is correct."safety first... and What the hell..... safety second, Too!!! " ~~jmy POPS #10490 Quote Share this post Link to post Share on other sites
SpeedRacer 1 #7 December 17, 2007 oh i get it now. Speed Racer -------------------------------------------------- Quote Share this post Link to post Share on other sites
shropshire 0 #8 December 17, 2007 So, could you please explain where the error is introduced (because clearly 0.99999 != 1) ? (.)Y(.) Chivalry is not dead; it only sleeps for want of work to do. - Jerome K Jerome Quote Share this post Link to post Share on other sites
JackC 0 #9 December 17, 2007 QuoteSo, could you please explain where the error is introduced (because clearly 0.99999 != 1) ? It works because the value of 0.99999... extends to infinity (the ellipsis is used to imply the continuation of a series) . As soon as you tuncate the value of 0.99999... to something finite, the proof fails. Infinities are best avoided, they're usually trouble. For example the series a=1+2+3+4+... adds up to infinity. The series b=1+2+4+8+16+... also adds up to infinity. Clearly for any finite series containing n terms, a so does that mean b a bigger infinity than a? Riddle me that one Batman. Quote Share this post Link to post Share on other sites
shropshire 0 #10 December 17, 2007 Ta!.... I think but my head still hurts. (.)Y(.) Chivalry is not dead; it only sleeps for want of work to do. - Jerome K Jerome Quote Share this post Link to post Share on other sites
grannyinthesky 0 #11 December 17, 2007 Yes, but it's the kind of thing that makes math almost as much fun as skydiving!"safety first... and What the hell..... safety second, Too!!! " ~~jmy POPS #10490 Quote Share this post Link to post Share on other sites
jheadley 0 #12 December 17, 2007 Here's another proof : 1/3 = .333.... .333.... *3 = .999... 1/3 * 3 = 1 so 1 = .999... Quote Share this post Link to post Share on other sites
Andy9o8 2 #13 December 17, 2007 Math nerds should be tarred and feathered. (Now where's some grammar I can correct?) Quote Share this post Link to post Share on other sites
shropshire 0 #14 December 17, 2007 that I can correct?)>> (.)Y(.) Chivalry is not dead; it only sleeps for want of work to do. - Jerome K Jerome Quote Share this post Link to post Share on other sites
Royd 0 #15 December 17, 2007 Absolutely. I've never paid the 9/10ths at the gas pump unless I pumped 10 gal. Quote Share this post Link to post Share on other sites
KelliJ 0 #16 December 18, 2007 Yes...but with reservations. I can understand how if x=.9999.... and 10X=9.9999... then 10X-X=9 so x=1. What gives me trouble, maybe someone can explain, is how that can be true but contradicted by 1-.9999...=.000...1 Quote Share this post Link to post Share on other sites
Everon 0 #17 December 18, 2007 QuoteYes...but with reservations. I can understand how if x=.9999.... and 10X=9.9999... then 10X-X=9 so x=1. What gives me trouble, maybe someone can explain, is how that can be true but contradicted by 1-.9999...=.000...1 .000...1 is a finite non-zero real number. 1-.9999... = 0, so it is not contradicted at all since your equation is false. Some mathematical results don't seem logical, but can easily be proven. It just goes with the territory. Quote Share this post Link to post Share on other sites
rehmwa 2 #18 December 18, 2007 "........" <=== it's pesky little rascal, isn't it? But worth it's weight in gold ... Driving is a one dimensional activity - a monkey can do it - being proud of your driving abilities is like being proud of being able to put on pants Quote Share this post Link to post Share on other sites
KelliJ 0 #19 December 27, 2007 Sorry to bring this thread back to the front but my mind wouldn't let me stop pondering the question. I have to change my vote. .999... comes infinitely close but never equals 1. If x=.999... then 10x=9.999... and 10x-x=9x. If x=.99 then 9x=8.91 If x=.999 then 9x=8.991 If x=.9999 then 9x=8.9991 Since, regardless of how many decimal places you give x, 9x would always be less than 9. It therefore stands to reason that x would never be equal to 1. Infinitely close, close enough that 1 could be used in place of .999... , but not equal. Quote Share this post Link to post Share on other sites
kallend 2,138 #20 December 27, 2007 QuoteSorry to bring this thread back to the front but my mind wouldn't let me stop pondering the question. I have to change my vote. .999... comes infinitely close but never equals 1. If x=.999... then 10x=9.999... and 10x-x=9x. If x=.99 then 9x=8.91 If x=.999 then 9x=8.991 If x=.9999 then 9x=8.9991 Since, regardless of how many decimal places you give x, 9x would always be less than 9. It therefore stands to reason that x would never be equal to 1. Infinitely close, close enough that 1 could be used in place of .999... , but not equal. Wrong. There is one-to-one correspondence between the decimal places of 9.999... and 0.999... hence when one is subtracted from the other you are left with 9.0 exactly... The only sure way to survive a canopy collision is not to have one. Quote Share this post Link to post Share on other sites
KelliJ 0 #21 December 27, 2007 I'm not a math wiz but I still don't buy it. .999... always has been and always will be an infinitely small amount away from 1. 1 can be substituted since any error would also be infinitely small. Look at it this way. 1 is finite, it is defined with no error. .999... is not finite, it continues on and as such any error resulting from its rounding up to one, though infinitely small, is also very real. Quote Share this post Link to post Share on other sites
JackC 0 #22 December 27, 2007 QuoteThere is one-to-one correspondence between the decimal places of 9.999... and 0.999... hence when one is subtracted from the other you are left with 9.0 exactly Isn't that a bit misleading? For any finite number of decimal places there isn't a one to one correlation between x and 10x, there's always one extra decimal place in x compared to 10x. The proof only works because there is no final decimal place in an infinite series. For every decimal place of 0.9999... there is always a corresponding one for 9.9999... As soon as you truncate x the proof fails. Thinking of it the other way around by taking 0.9999... and multiplying it by 10 gives you 9.999... This is asymptotically equal to 10 because you cannot take any finite number away from 10 to give you 9.999... At least that's how I'd think of it. Quote Share this post Link to post Share on other sites
KelliJ 0 #23 December 27, 2007 Not to steal the Professors chance to respond, but I just need to get clarification. Is what you are saying something along these lines... x=.999... 10x=9.999... 10x-9x=x=? To solve we must multiply .999... by 9 Hmmmm....this stuff can make your head hurt. But the debate is fun. Quote Share this post Link to post Share on other sites
JackC 0 #24 December 27, 2007 Eh? Nope, the original proof was right: if x=0.999... then 10x=9.999... and 10x-x=9, 9x=9 and x=1 I was trying to think of it another way: if x=0.999... then 10x=9.999... which is asyptotically equal to 10. This is because there is no finite number that you can subtract from 10 to give you 9.999... Thinking in terms of finite numbers is confusing and leads to wrong answers. For most people thinking of infinities is equally confusing. What I do is think of what a number tends to. For example 0.9999... tends towards 1 as more decimal places are added. 9.999... tends towards 10, 1/x tends to zero as x tends to infinity, etc. Quote Share this post Link to post Share on other sites
kallend 2,138 #25 December 27, 2007 QuoteQuoteThere is one-to-one correspondence between the decimal places of 9.999... and 0.999... hence when one is subtracted from the other you are left with 9.0 exactly Isn't that a bit misleading? For any finite number of decimal places there isn't a one to one correlation between x and 10x, there's always one extra decimal place in x compared to 10x. The proof only works because there is no final decimal place in an infinite series. For every decimal place of 0.9999... there is always a corresponding one for 9.9999... As soon as you truncate x the proof fails. Thinking of it the other way around by taking 0.9999... and multiplying it by 10 gives you 9.999... This is asymptotically equal to 10 because you cannot take any finite number away from 10 to give you 9.999... At least that's how I'd think of it. infinity + 1 = infinity. infinity - 1 = infinity. The proof IS correct. If you don't like that one, the one posted earlier in this thread using 1/3 = 0.33..., 3 x 0.33... = 0.99..., and 3 x 1/3 = 1 therefore 0.99... = 1 exactly, is just as good. There are lots of non intuitive things about infinity (like there's more than one (aleph-naught and C), there are as many numbers between 0 and 1 as there are between 0 and 1,000,000 there are MORE real numbers between 0 and 1 than there are rational numbers. there are an infinite number of irrational real numbers between every adjacent pair of rational numbers. there are as many integers as there are rational numbers between 0 and 1. etc. Ask any mathematician.... The only sure way to survive a canopy collision is not to have one. Quote Share this post Link to post Share on other sites
