Jay-qu Posted September 2, 2005 Report Share Posted September 2, 2005 my friend showed me this little math riddle the other day and i have been trying to find an error to somehow prove him wrong, but i cant.. maybe someone here can shed some light on it for me... x = 0.999...(recurring)10x = 9.999...10x - x = 99x = 9x = 1 ???? :eek2: its got me... Quote Link to comment Share on other sites More sharing options...
Buffy Posted September 2, 2005 Report Share Posted September 2, 2005 x = 0.999...(recurring) 10x = 9.999... (2)10x - x = 9 (3)9x = 9 (4)x = 1 (5)The fallacious line is actually (3). Restating this to eliminate the 'recurring' specification, which leads to the fallacy: x = 1 - e (e is epsilon)10x = 10 - 10e10x - x = (10 - 10e) - (1 - e)9x = 10 - 10e - 1 + e9x = 9 - 9ex = 1 - e Or rather you really need to think of 10x - x as being 9.000(recurring)0000001with an infinite number of zeros in the middle... Cheers,Buffy Quote Link to comment Share on other sites More sharing options...
Qfwfq Posted September 2, 2005 Report Share Posted September 2, 2005 The fallacious line is actually (3).:D Hey, Buf! Don't be confusing poor li'l JQ! :) He's a great guy, an inquisitive mind that can learn a lot of good stuff! :D There's nothing fallacious at all about line 3, or about the whole thing, x is equal to 1. If you want a simpler "fallacious riddle" here it is: x = 1/ 3 = 0.3333333333333333333333....... y = 3x = ? How much is y? Is y = 0.999999999999999999 or is y = 1? ;) :eek2: :D Quote Link to comment Share on other sites More sharing options...
chatlack Posted September 2, 2005 Report Share Posted September 2, 2005 I think its not thrue but we all know 0.999(recurring) is equal 1. But there is a mystery with 0 and infinity. We must solve infinity in philoshopy maybe than control everything we thought "we solved" before. First is relativity. Because in einstein's and Lorent's formuleas everthing is determined depending on 0 and infinity. Damn... I get really out of here again... :eek2: Quote Link to comment Share on other sites More sharing options...
rockytriton Posted September 2, 2005 Report Share Posted September 2, 2005 I'm still trying to figure out how line 2 could be right, if line 1 is:x = 9.999...(recurring) since when is10x = 9.999... (10 * 9.999... = 9.999... ???????) Quote Link to comment Share on other sites More sharing options...
Qfwfq Posted September 2, 2005 Report Share Posted September 2, 2005 I'm quite sure Buffy is right that he meant x = 0.9999999.... and not 9.999999.... Quote Link to comment Share on other sites More sharing options...
CraigD Posted September 2, 2005 Report Share Posted September 2, 2005 I'm still trying to figure out how line 2 could be right, if line 1 is:x = 9.999...(recurring) since when is10x = 9.999... (10 * 9.999... = 9.999... ???????)Good catch! In our haste to explain this riddle (which practically everybody with a math or science background encounters in the teens of early 20s), we’ve overlooked a typographical error (actually, Buffy caught it too, but still, good catch). The riddle should read: x = 0.999... 10x = 9.999...10x - x = 99x = 9x = 1 As various folk have already explained, the answer to the riddle is that 0.999… actually is equal to 1. The moral of the riddle, IMHO, is to be selective in choosing representations of numbers that are un-confusing to your intended reader, eg:1/3 + 2/3 = 3/3 = 1rather than0.333… + 0.666… = 0.999… = 1 Quote Link to comment Share on other sites More sharing options...
Qfwfq Posted September 2, 2005 Report Share Posted September 2, 2005 Rocky had overlooked Buffy. Hers was the very first reply.The moral of the riddle, IMHO, is to be selective in choosing representations of numbers that are un-confusing to your intended readerThere is a systematic way of finding numerator and denominator given any periodic number in a given base. It's explainable by the fact that a periodic number is equal to a factor times the sum of a geometric series. Consider: a = 0.(713) which is a way of writing 0.713713713713713... Sometimes the period is overstruck but I can't do that here. The above number a can be written as 0.713 times the geometric series of ratio 0.001, which is: 1.001001001001... Divide 713 by 999 and you will get a: 0.713713713713... For 0.(9) it is 9/9 = 1. For a base b other than ten, just replace each of the nines in the denominator with b - 1. Quote Link to comment Share on other sites More sharing options...
Buffy Posted September 2, 2005 Report Share Posted September 2, 2005 You're all right. Of course representation is the key, and I'd even argue that to justify saying 1=0.9999(recurring) You really need to use my notation and specify: lim (e->0) 1 - e = 1 Or you'll get the question wrong on your math quiz! Sorry that I got caught up in my love of epsilon! :eek2: Computationally yours,Buffy Quote Link to comment Share on other sites More sharing options...
Jay-qu Posted September 3, 2005 Author Report Share Posted September 3, 2005 opps, sorry bout the typo I was in a hurry... thanks for the explanation guys Quote Link to comment Share on other sites More sharing options...
Dark Mind Posted September 3, 2005 Report Share Posted September 3, 2005 Not a problem :eek2:. Quote Link to comment Share on other sites More sharing options...
Dark Mind Posted September 5, 2005 Report Share Posted September 5, 2005 ...I edited out the error from the first two posts, since there was such a big deal being made about it :lol:. Quote Link to comment Share on other sites More sharing options...
Qfwfq Posted September 5, 2005 Report Share Posted September 5, 2005 I'd even argue that to justify saying 1=0.9999(recurring) You really need to use my notation and specify: lim (e->0) 1 - e = 1Yup! Likewise lim(e->0) 9 + e = 9 and 10 - 10e is much the same as 10 - e. But which question will I get wrong on my math quiz??? :lol: Quote Link to comment Share on other sites More sharing options...
Buffy Posted September 6, 2005 Report Share Posted September 6, 2005 But which question will I get wrong on my math quiz??? :lol:The one where you leave out the intermediate steps in your proof! That's what always bit me in math classes: several "required" steps in the middle just seemed too obvious to mention... Cheers,Buffy Quote Link to comment Share on other sites More sharing options...
Qfwfq Posted September 6, 2005 Report Share Posted September 6, 2005 That happened to me, the odd time, but I got through my courses all the same! :lol: I was usually pedantic enough when it was a math test and not a physics one. This was neither, though I could have given the proof that the sum of a geometric series is 1/(1 - x). :) I'm no longer a student anyway now, I could easily be a high-school teacher, like many of my friends, if I had gone through the rigmarole to become one. Quote Link to comment Share on other sites More sharing options...
Dark Mind Posted September 13, 2005 Report Share Posted September 13, 2005 Here's another riddle :lol:. Well, sort of... A weightless and perfectly flexible rope is hung over a weightless, frictionless pulley attached to the roof of a building. At one end is a weight which exactly counterbalances a monkey at the other end. If the monkey begins to climb, what will happen to the weight? It will remain stationary.It will rise.It will fall. My answer: It will rise. Anyone disagree? :hihi: Quote Link to comment Share on other sites More sharing options...
Dark Mind Posted September 13, 2005 Report Share Posted September 13, 2005 Is this a trick question ? You have a ten centimeter by ten centimeter square with a flap on the top edge. Unfolding this flap reveals a second flap, and unfolding this second flap reveals a third, as shown. If there are an infinite number of these flaps, what is the area of the entire figure (in square centimeters) when they are all unfolded? My answer was 150 :hihi:. Quote Link to comment Share on other sites More sharing options...
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