The Definition of Points
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From: RLG on 24 Mar 2007 20:40 "Tony Orlow" <tony(a)lightlink.com> wrote in message news:460520e5(a)news2.lightlink.com... > > 1=0.999...? 1 = 0.999... Multiply both sides of this equation by 10 to get: 10 = 9.999... Substract the original equation from this one to get: 9 = 9. So 0.999... is simply another way of writing the number 1. If you don't like that then write 0.999... as follows: 9*( 1/10 + 1/100 + 1/1000 + ...) Notice that we have an infinite geometric series with x=1/10: x + x^2 + x^3 + x^4 + ... = x/(1-x) This gives us (1/10)/(1-(1/10)) = 1/9 which gives us: 9*( 1/10 + 1/100 + 1/1000 + ...) = 9*(1/9) = 1 R
From: The Ghost In The Machine on 24 Mar 2007 20:44 In sci.physics, RLG <Junk(a)Goldolfo.com> wrote on Sat, 24 Mar 2007 16:40:04 -0800 <nM6dnRdlcfJKK5jbnZ2dnUVZ_rWnnZ2d(a)comcast.com>: > > "Tony Orlow" <tony(a)lightlink.com> wrote in message > news:460520e5(a)news2.lightlink.com... >> >> 1=0.999...? > > > 1 = 0.999... > > Multiply both sides of this equation by 10 to get: > > 10 = 9.999... > > Substract the original equation from this one to get: > > 9 = 9. A little more complicated than that, as one could either use that logic, resulting in 9.0000..., or assume an infinite borrow, resulting in 8.9999... . The proper method of handling this would be using limits, which you implicitly do below: > > So 0.999... is simply another way of writing the number 1. If you don't > like that then write 0.999... as follows: > > 9*( 1/10 + 1/100 + 1/1000 + ...) > > Notice that we have an infinite geometric series with x=1/10: > > x + x^2 + x^3 + x^4 + ... = x/(1-x) > > This gives us (1/10)/(1-(1/10)) = 1/9 which gives us: > > 9*( 1/10 + 1/100 + 1/1000 + ...) = 9*(1/9) = 1 A geometric series of n terms results in a * (1 - r^(n+1))/(1-r); if -1 < r < 1, lim(n->+oo) r^(n+1) will be 0. It's a bit like Zeno's paradox, which one can write 1/2 + 1/4 + 1/8 + ... Zeno argued this was infinite and that motion was impossible (obviously his logic was a little off, since in order to speak or write one has to move something); modern math would simply write this as 1 = 1/2 + 1/4 + 1/8 + ... for reasons similar to 0.999... = 1. In fact, 0.111...(2) = 1, in binary. > > > R > > -- #191, ewill3(a)earthlink.net -- insert random pedancy here Useless C++ Programming Idea #104392: for(int i = 0; i < 1000000; i++) sleep(0); -- Posted via a free Usenet account from http://www.teranews.com
From: RLG on 24 Mar 2007 22:59 "The Ghost In The Machine" <ewill(a)sirius.tg00suus7038.net> wrote in message news:v8tgd4-1qv.ln1(a)sirius.tg00suus7038.net... > In sci.physics, RLG > <Junk(a)Goldolfo.com> > wrote > on Sat, 24 Mar 2007 16:40:04 -0800 > <nM6dnRdlcfJKK5jbnZ2dnUVZ_rWnnZ2d(a)comcast.com>: >> >> "Tony Orlow" <tony(a)lightlink.com> wrote in message >> news:460520e5(a)news2.lightlink.com... >>> >>> 1=0.999...? >> >> >> 1 = 0.999... >> >> Multiply both sides of this equation by 10 to get: >> >> 10 = 9.999... >> >> Substract the original equation from this one to get: >> >> 9 = 9. > > A little more complicated than that, as one could either use that logic, > resulting in 9.0000..., or assume an infinite borrow, resulting in > 8.9999... . The proper method of handling this would be using limits, > which you implicitly do below: Just curious, how would an infinite borrow be done? The method I used had 10-1=9 and: 9.99999... 0.99999... ----------- 9.00000... R
From: The Ghost In The Machine on 25 Mar 2007 01:07 In sci.logic, RLG <Junk(a)Goldolfo.com> wrote on Sat, 24 Mar 2007 18:59:26 -0800 <Gfednb222eDiSpjbnZ2dnUVZ_ragnZ2d(a)comcast.com>: > > "The Ghost In The Machine" <ewill(a)sirius.tg00suus7038.net> wrote in message > news:v8tgd4-1qv.ln1(a)sirius.tg00suus7038.net... >> In sci.physics, RLG >> <Junk(a)Goldolfo.com> >> wrote >> on Sat, 24 Mar 2007 16:40:04 -0800 >> <nM6dnRdlcfJKK5jbnZ2dnUVZ_rWnnZ2d(a)comcast.com>: >>> >>> "Tony Orlow" <tony(a)lightlink.com> wrote in message >>> news:460520e5(a)news2.lightlink.com... >>>> >>>> 1=0.999...? >>> >>> >>> 1 = 0.999... >>> >>> Multiply both sides of this equation by 10 to get: >>> >>> 10 = 9.999... >>> >>> Substract the original equation from this one to get: >>> >>> 9 = 9. >> >> A little more complicated than that, as one could either use that logic, >> resulting in 9.0000..., or assume an infinite borrow, resulting in >> 8.9999... . The proper method of handling this would be using limits, >> which you implicitly do below: > > Just curious, how would an infinite borrow be done? The method I used had > 10-1=9 and: > > 9.99999... > 0.99999... > ----------- > 9.00000... > > > R > Well, that's just it...there's no last digit. However, were there a last digit one might run into either .....999999999 .....999999999 which has no borrow, or .....999999990 .....999999999 which will need one. After all, we multiplied by 10... Of course that's why limits need to be used anyway; my strawman logic verges on the ridiculous. :-) But a few paradoxes get messy without them: -1 = 1 + 2 + 4 + 8 + ... since if x = 1 + 2 + 4 + 8 + ... then 2x = x-1 ? = 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + ... could be 0, 1, 1/2. log 2 = 1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + .... = (1 + 1/3 + 1/5 + ...) - (1/2 + 1/4 + 1/6 + ...) = infinity - infinity 1 = 1/2 + 1/4 + 1/8 + 1/16 + ... (Zeno's Paradox) and of course 1.00000... = 0.99999... Makes life interesting. :-) -- #191, ewill3(a)earthlink.net Useless C++ Programming Idea #12398234: void f(char *p) {char *q = strdup(p); strcpy(p,q);} -- Posted via a free Usenet account from http://www.teranews.com
From: Bob Kolker on 25 Mar 2007 02:17
The Ghost In The Machine wrote: > and of course > > 1.00000... = 0.99999... > > Makes life interesting. :-) You can see what a service that Bishop Berkeley did for mathematics by savaging the illogic of Newton and Leibniz and ultimately forcing mathematicians to define limits and convergence properly. That is the only way of beating Zeno. Bob Kolker > |