.Atlantis Share Posted October 1, 2003 Uhh, explain why I can't subtract 0.9 recurring from 9.9 recurring? Dazzla sleep already, do math when your brain is fresh! Link to post Share on other sites

OPaul Share Posted October 1, 2003 All these proof are all well and dandy but I still don't see how something that approaches 1 can be 1. It never reaches 1, it isn't a real number. Link to post Share on other sites

Dazzla Veteran Share Posted October 1, 2003 you can do that. that part is not invalid. your 9x part is. Hang on hang on, if you get that then what's the problem? x = 0.9 recurring right? so 10x = 9.9 recurring right? With me? You're fine with 9.9 recurring - 0.9 recurring = 9 right? So what's the problem? Link to post Share on other sites

Samoa Share Posted October 1, 2003 you can do that. that part is not invalid. your 9x part is.10x-x=9x 10(.9999) - (.9999) = 9 * (.9999) 9.9999 -.9999 = 8.991 9 = 8.991 that is the flaw this is the problem DAZZLA 9 times 9 is 81 Link to post Share on other sites

bangbang023 Veteran Share Posted October 1, 2003 ok I'm going to try to break this down but i have to hurry: x = 0.9999... 10x = 9.9999... 10x - x = 9x 9.9999... - 0.9999... = 9 9 = 9x 1 = x = 0.9999... 1) now he sets x = 0.9999... 2) he now claims, corrctly, that 10 * 0.999.... = 9.9999.... so 10x = 9.9999... 3) 10x - x equals 9x. Plugin 0.999... for x and you are left with 9. so 9x does equal 9 4) the rest is simple eidt: I think thats what he meant. I started and put on music and lost my train of thought lol. Link to post Share on other sites

Samoa Share Posted October 1, 2003 ok I'm going to try to break this down but i have to hurry:x = 0.9999... 10x = 9.9999... 10x - x = 9x 9.9999... - 0.9999... = 9 9 = 9x 1 = x = 0.9999... 1) now he sets x = 0.9999... 2) he now claims, corrctly, that 10 * 0.999.... = 9.9999.... so 10x = 9.9999... 3) 10x - x equals 9x. Plugin 0.999... for x and you are left with 9. so 9x does equal 9 4) the rest is simple your math leaves of 9 times .9999 what does the 9x part just disappear? Link to post Share on other sites

Dazzla Veteran Share Posted October 1, 2003 ok I'm going to try to break this down but i have to hurry:x = 0.9999... 10x = 9.9999... 10x - x = 9x 9.9999... - 0.9999... = 9 9 = 9x 1 = x = 0.9999... 1) now he sets x = 0.9999... 2) he now claims, corrctly, that 10 * 0.999.... = 9.9999.... so 10x = 9.9999... 3) 10x - x equals 9x. Plugin 0.999... for x and you are left with 9. so 9x does equal 9 4) the rest is simple ^^ Bingo. Do you also disagree with this Samoa? 0.3 recurring = 1/3 0.3 recurring * 3 = 0.9 recurring 1/3 *3 = 1 Link to post Share on other sites

OPaul Share Posted October 1, 2003 All these proofs are all well and dandy but I still don't see how something that approaches 1 can be 1. It never reaches 1, it isn't a real number. Can someone respond to this without giving a proof. Dazzel or who ever is says 1 = .99... tell me how that can be if the number never gets there. If it got there it would be a real number and would no longer approach 1. Link to post Share on other sites

bangbang023 Veteran Share Posted October 1, 2003 ok I'm going to try to break this down but i have to hurry:x = 0.9999... 10x = 9.9999... 10x - x = 9x 9.9999... - 0.9999... = 9 9 = 9x 1 = x = 0.9999... 1) now he sets x = 0.9999... 2) he now claims, corrctly, that 10 * 0.999.... = 9.9999.... so 10x = 9.9999... 3) 10x - x equals 9x. Plugin 0.999... for x and you are left with 9. so 9x does equal 9 4) the rest is simple your math leaves of 9 times .9999 what does the 9x part just disappear? ok so 10x - x if you plugin the value for x yo uhave 10(0.999...) - 0.9999 qhich equals 9 but 10(x) - x = 10(0.9999....) - 0.999.... 9x = 9 x = 1 Link to post Share on other sites

Samoa Share Posted October 1, 2003 bangbang, you're out of your depth, don't bother.Proof 1: x = 0.9999... 10x = 9.9999... 10x - x = 9x 9.9999... - 0.9999... = 9 9 = 9x 1 = x = 0.9999... Proof 2: 0.111... = 1/9 0.111...*9 = 0.999... since 0.111.... = 1/9 -> 1/9 * 9 = 0.999... 1/9 * 9 = 9/9 = 1 Proof 3: 0.3 recurring = 1/3 0.3 recurring * 3 = 0.9 recurring 1/3 *3 = 1 Disprove... :sleep: http://sakuramb.com/showthread.php?s=&threadid=266 http://forums.philosophyforums.com/showthr...38&page=1&pp=25 I'm off to bed now, time to unsubscribe and leave you talking to a brick wall... Proof 2: 0.111... = 1/9 0.111...*9 = 0.999... since 0.111.... = 1/9 -> 1/9 * 9 = 0.999... 1/9 * 9 = 9/9 = 1 0.1111.... does equal 1/9 [ 0.111... = 1/9] .1111..... times 9 equals 1 not [0.111...*9 = 0.999...] you cannot divide something then multiply by the same number and arrive at a different original number. once again a mistake. however this is correct, but is not proving .9999 is equal to 1 Proof 3: 0.3 recurring = 1/3 0.3 recurring * 3 = 0.9 recurring 1/3 *3 = 1 Link to post Share on other sites

un4given1 Share Posted October 1, 2003 10x = 9.9999...10x - x = 9x Where in the hell did you get this? that's not proper math. x=0.9999... 10x=9.9999... 10x/x=9.9999.../x (I don't know where you came up with the rule that you can subtract x from 10*x) 10=9/x 10=9/.9999... 10=10.0000...9 This is a FALSE PROOF! let's try this with a known variable... x=5 10*5=10*5 10/5=10/5 10=10 and Dazzla's method with the same variable x=5 10*5=50 10*5-5=45 45=10/5 50=50 your math is WRONG! Link to post Share on other sites

Dazzla Veteran Share Posted October 1, 2003 How is it not proving .999 = 1? 0.3 recurring = 1/3 0.3 recurring * 3 = 0.9 recurring 1/3 * 3 = 1 ^^^^That you 100% agree with, can also read (seeing as 0.3 recurring = 1/3) as: 1/3 * 3 = 0.9 recurring But on the final line we've shown that 1/3 * 3 = 1, and above we've shown 1/3 * 3 = 0.9 recurring: 1/3 * 3 = 0.9 recurring 1/3 * 3 = 1 Wow, what do you think that means? Link to post Share on other sites

applelover Share Posted October 1, 2003 Here is the way I would: 0.99999999... = 0.9 + 0.09 + 0.009 + 0.0009 + ... This is a geometric series where the nth term can be represented by a(n) = a(0)*0.1^(-n) for n=1,2,3,4..., where a(0) is the first term of the series. As such, the sum of the terms is S = a(0)/(1-r) where r=0.1 and a(0)=0.9 So, 0.9/(1-0.1) = 0.9/0.9 = 1 Link to post Share on other sites

bangbang023 Veteran Share Posted October 1, 2003 you guys are missing the point! x = 0.999... right? ok we got that. now slow down a bit 10x - x = 9.999... - 0.9999 now what is 9.999... - 0.999....? that's right, 9. 10x - x = 9 9x = 9 x = 1 Link to post Share on other sites

Dazzla Veteran Share Posted October 1, 2003 Here is the way I would:0.99999999... = 0.9 + 0.09 + 0.009 + 0.0009 + ... This is a geometric series where the nth term can be represented by a(n) = a(0)*0.1^(-n) for n=1,2,3,4..., where a(0) is the first term of the series. As such, the sum of the terms is S = a(0)/(1-r) where r=0.1 and a(0)=0.9 So, 0.9/(1-0.1) = 0.9/0.9 = 1 That's the way I was taught it at college by my Math professor. Link to post Share on other sites

Samoa Share Posted October 1, 2003 ok I'm going to try to break this down but i have to hurry:x = 0.9999... 10x = 9.9999... 10x - x = 9x 9.9999... - 0.9999... = 9 9 = 9x 1 = x = 0.9999... 1) now he sets x = 0.9999... 2) he now claims, corrctly, that 10 * 0.999.... = 9.9999.... so 10x = 9.9999... 3) 10x - x equals 9x. Plugin 0.999... for x and you are left with 9. so 9x does equal 9 4) the rest is simple your math leaves of 9 times .9999 what does the 9x part just disappear? ok so 10x - x if you plugin the value for x yo uhave 10(0.999...) - 0.9999 qhich equals 9 but 10(x) - x = 10(0.9999....) - 0.999.... 9x = 9 x = 1 9x = 9 x = 1 yes that would be correct if X=1, and only if x=1 we already said it did not with x=.9999 but let's use another interger [y] to show the point y=.9999 x=1 9xy = 9y? 9 (1) (.9999) = 9 (.9999) 8.991=8.991 this holds true but in no way says 1 is equal to .9999 Link to post Share on other sites

un4given1 Share Posted October 1, 2003 x=1 y=.9999... 1000000x=1000000y 1000000 = 999999.9999... if x=1 and y=.9999... but x should = y, then the statement above should be correct, right? Looks a little off to me Link to post Share on other sites

Dazzla Veteran Share Posted October 1, 2003 Sammy, you choose to ignore this: 0.3 recurring = 1/3 0.3 recurring * 3 = 0.9 recurring 1/3 * 3 = 1 ^^^^That you 100% agree with, can also read (seeing as 0.3 recurring = 1/3) as: 1/3 * 3 = 0.9 recurring But on the final line we've shown that 1/3 * 3 = 1, and above we've shown 1/3 * 3 = 0.9 recurring: 1/3 * 3 = 0.9 recurring 1/3 * 3 = 1 Link to post Share on other sites

John Veteran Share Posted October 1, 2003 in regards to the title: no. 0.999... is 1 - an infinitely small number (ie. less than 0.000...1, which doesn't exist anyway, cause you can't have a digit AFTER an infinite number of digits) someone can explain this with limits, but i don't really want to right now... i'm 100% sure the answer is no. edit: AFAIK, you can't solve this with just algebra, so dazzla's method, although it looks correct, isn't correct... Link to post Share on other sites

OPaul Share Posted October 1, 2003 Can someone respond to this without giving a proof. Dazzel or who ever is says 1 = .99... tell me how that can be if the number never gets there. If it got there it would be a real number and would no longer approach 1. Alright, enough proofs. Proofs are all well and dandy and I see what there doing and such. My problem is this; 0.999999.... is approaching 1 indefinitely, it is not a real number, it never reaches 1. So how does it equal 1 if it never gets there? Can someone explain that without giving another proof. Link to post Share on other sites

Samoa Share Posted October 1, 2003 (edited) anyone who says .9999 is equal to one cannot be correct in dazzla's equation 9x is repeated left out. When you can show me 9x is not 8.991, when x=.9999 then the world will fall apart. you can show me over an over that 1 is equal to 1 all you want. Cause I know it is true. 10x-x when x=1 is going to be 9. There is no argument. But when X=.9999, then 10x-x also equal 9 10x-x 10(.9999)-.9999 9.9999-.9999 9 10(1)-(1) = 9 = 10(.9999)-(.9999) that is no way proves .9999 equals 1, started to fall for that flawed reasoning as well for a moment but... Edited October 1, 2003 by Samoa Link to post Share on other sites

Dazzla Veteran Share Posted October 1, 2003 Chosen to ignore applelovers post and mine 3 posts up? Link to post Share on other sites

OPaul Share Posted October 1, 2003 I think the problem is you guys are trying to solve a calculus problem with plain algebra, your ignoring the fact the limit of something doesn't equal what it approaches. That's just simple Calculus I. Link to post Share on other sites

.Atlantis Share Posted October 1, 2003 I'm gonna have to agree on both sides of the argument :| I really need to ask our Math proffessor tommorow... -Grant Link to post Share on other sites

Samoa Share Posted October 1, 2003 (edited) Chosen to ignore applelovers post and mine 3 posts up? no I am showing where an how your calculation is flawed, Please consider my post again. 10x-x 10(.9999)-.9999 9.9999-.9999 9 9 = 10(.9999)-(.9999) 10(1)-(1) = 9 then, 10(1)-(1) = 9 = 10(.9999)-(.9999) 10(1)-(1)=9=10(.9999)-(.9999) if x=1, z=.9999 then 10(x)-(x)=9=10(z)-(z) 10(x)-(x)/10=9/10=10(z)-(z)/10 (x)-(x)=.9=(z)-(z) plug in the values for x and z it don't work. (1)-(1)=.9=(.9999)-(.9999) 0=.9=0 Edited October 1, 2003 by Samoa Link to post Share on other sites

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