{{
`${userDetail.user.displayName}${userDetail.user.cachedBungieGlobalDisplayNameCode ? `#${standardizeBungieGlobalCode(userDetail.user.cachedBungieGlobalDisplayNameCode)}` : ""}`}}
{{getUserClans()}}
12/13/2010 12:28:46 AM
584
Does 1 = .9999999.......
1/3 = .333.....
2/3 = .666.....
So does
3/3 = .999.....
or
3/3 = 1
-
[quote][b]Posted by:[/b] EnderJSV So what? Does that offend you?[/quote] Nope. I'm really surprised that we already have +500 posts into this thread.
-
[quote][b]Posted by:[/b] Trace007 [quote][b]Posted by:[/b] Kkman4evah [quote][b]Posted by:[/b] drummer0702 [quote][b]Posted by:[/b] Kkman4evah It seems that from this math can be easily twisted to reach the required number.[/quote] I don't quite understand what you're saying; can you give an example?[/quote]Like this. 1/3=.3333333.......... (1/3)+(1/3)+(1/3) must equal .999999........ BUT 3/3 also equals 1.[/quote] If 1/3 can equal .3 repeating, why can't 3/3 equal .9 repeating?[/quote] It can. It does.
-
[quote][b]Posted by:[/b] drummer0702 How to get a hot topic- -Ask something that is abstract that many people will have opinions on. -Find multiple people that are dedicated to proving people wrong -Find a dedicated person that likes to play devil's advocat Congratulations you have a hot topic![/quote] So what? Does that offend you?
-
[quote][b]Posted by:[/b] Kkman4evah Like this. 1/3=.3333333.......... (1/3)+(1/3)+(1/3) must equal .999999........ BUT 3/3 also equals 1.[/quote]Those are all true.
-
[quote][b]Posted by:[/b] Kkman4evah [quote][b]Posted by:[/b] drummer0702 [quote][b]Posted by:[/b] Kkman4evah It seems that from this math can be easily twisted to reach the required number.[/quote] I don't quite understand what you're saying; can you give an example?[/quote]Like this. 1/3=.3333333.......... (1/3)+(1/3)+(1/3) must equal .999999........ BUT 3/3 also equals 1.[/quote] If 1/3 can equal .3 repeating, why can't 3/3 equal .9 repeating?
-
[quote][b]Posted by:[/b] drummer0702 How to get a hot topic- -Ask something that is abstract that many people will have opinions on. -Find multiple people that are dedicated to proving people wrong -Find a dedicated person that likes to play devil's advocat Congratulations you have a hot topic![/quote]Thanks for that. Now get to the point at hand or leave >.<
-
[quote][b]Posted by:[/b] Kkman4evah Like this. 1/3=.3333333.......... (1/3)+(1/3)+(1/3) must equal .999999........ BUT 3/3 also equals 1.[/quote] Because they do equal each other. Your calculator wouldn't lie to you, right? [Edited on 12.12.2010 8:30 PM PST]
-
How to get a hot topic- -Ask something that is abstract that many people will have opinions on. -Find multiple people that are dedicated to proving people wrong -Find a dedicated person that likes to play devil's advocat Congratulations you have a hot topic!
-
[quote][b]Posted by:[/b] drummer0702 [quote][b]Posted by:[/b] Kkman4evah It seems that from this math can be easily twisted to reach the required number.[/quote] I don't quite understand what you're saying; can you give an example?[/quote]Like this. 1/3=.3333333.......... (1/3)+(1/3)+(1/3) must equal .999999........ BUT 3/3 also equals 1.
-
[quote][b]Posted by:[/b] Kkman4evah [quote][b]Posted by:[/b] drummer0702 [quote][b]Posted by:[/b] Kkman4evah Hm. One of my solutions proven wrong. But if .9 continued forever it would never add that last tiny insignificant 1 that it needs to actually REACH 1. You could easily round it to 1. But it would never completely touch 1.[/quote] What's 1-.999...=? Also, just add up (1/3) three times. Aka (.333...)+(.333...)+(.333...)=.999...=3/3=1[/quote]It seems that from this math can be easily twisted to reach the required number.[/quote]How was math twisted?
-
[quote][b]Posted by:[/b] m3andak [quote][b]Posted by:[/b] Disambiguation [quote][b]Posted by:[/b] m3andak they may be considered the same, but .99999 repeating does not equal 1. .99999 repeating equals .99999 repeating, but rounds to 1[/quote] There's no rounding involved here. People are confusing the difference between a finite number of 9's and an infinite number of 9's.[/quote] you could have and infinite number of 9's and it still wont equal one.[/quote] Yes, it will.
-
[quote][b]Posted by:[/b] Kkman4evah [quote][b]Posted by:[/b] drummer0702 [quote][b]Posted by:[/b] Kkman4evah Hm. One of my solutions proven wrong. But if .9 continued forever it would never add that last tiny insignificant 1 that it needs to actually REACH 1. You could easily round it to 1. But it would never completely touch 1.[/quote] What's 1-.999...=? Also, just add up (1/3) three times. Aka (.333...)+(.333...)+(.333...)=.999...=3/3=1[/quote]It seems that from this math can be easily twisted to reach the required number.[/quote]Lol, that's called mathematical proof. He did no ''twisting,'' he used accepted rules.
-
[quote][b]Posted by:[/b] m3andak [quote][b]Posted by:[/b] Disambiguation [quote][b]Posted by:[/b] m3andak they may be considered the same, but .99999 repeating does not equal 1. .99999 repeating equals .99999 repeating, but rounds to 1[/quote] There's no rounding involved here. People are confusing the difference between a finite number of 9's and an infinite number of 9's.[/quote] you could have and infinite number of 9's and it still wont equal one.[/quote] Actually, if you have an infinite number of 9's it would equal 1. Please read the previous 500+ posts in this thread :) Thanks.
-
[quote][b]Posted by:[/b] Kkman4evah It seems that from this math can be easily twisted to reach the required number.[/quote] I don't quite understand what you're saying; can you give an example?
-
[quote][b]Posted by:[/b] Mister Math [quote][b]Posted by:[/b] Dream053 [quote][b]Posted by:[/b] drummer0702 [quote][b]Posted by:[/b] Dream053 Doesn't the same [i][b]general[/b][/i] concept lie with asymptotes: They're constantly approaching but never reaching 1, and are simply considered the same out of convenience?[/quote] They reach the line at infinity. [/quote]Is that a fact? I was taught that asymptotes were never reached.[/quote]0.999... is a limit. It would be the limit of the asymptote of the function of x as x goes to infinity. [/quote] Yep. Exactly this. Well put.
-
[quote][b]Posted by:[/b] drummer0702 [quote][b]Posted by:[/b] Kkman4evah Hm. One of my solutions proven wrong. But if .9 continued forever it would never add that last tiny insignificant 1 that it needs to actually REACH 1. You could easily round it to 1. But it would never completely touch 1.[/quote] What's 1-.999...=? Also, just add up (1/3) three times. Aka (.333...)+(.333...)+(.333...)=.999...=3/3=1[/quote]It seems that from this math can be easily twisted to reach the required number.
-
[quote][b]Posted by:[/b] Disambiguation [quote][b]Posted by:[/b] m3andak they may be considered the same, but .99999 repeating does not equal 1. .99999 repeating equals .99999 repeating, but rounds to 1[/quote] There's no rounding involved here. People are confusing the difference between a finite number of 9's and an infinite number of 9's.[/quote] you could have and infinite number of 9's and it still wont equal one.
-
So I take it this thread is finally winding down?
-
[quote][b]Posted by:[/b] Dream053 [quote][b]Posted by:[/b] drummer0702 [quote][b]Posted by:[/b] Dream053 Doesn't the same [i][b]general[/b][/i] concept lie with asymptotes: They're constantly approaching but never reaching 1, and are simply considered the same out of convenience?[/quote] They reach the line at infinity. [/quote]Is that a fact? I was taught that asymptotes were never reached.[/quote] Nope. You're right. asymptotes never reach 1. But an asymptote is a series. Its a process of something. Like, when graphed, it's the process of forever increasing x and watching where where it's graphed on a 2 dimensional plane. But .999 is NOT a process. It's not a number that gets increasingly closer to one as you continue to add 9s. The infinite series of nines ALREADY exist. They are all there. I know, it's a hard concept. Lot of people don't like it. [Edited on 12.12.2010 8:27 PM PST]
-
[quote][b]Posted by:[/b] m3andak they may be considered the same, but .99999 repeating does not equal 1. .99999 repeating equals .99999 repeating, but rounds to 1[/quote] There's no rounding involved here. People are confusing the difference between a finite number of 9's and an infinite number of 9's.
-
[quote][b]Posted by:[/b] Kkman4evah Hm. One of my solutions proven wrong. But if .9 continued forever it would never add that last tiny insignificant 1 that it needs to actually REACH 1. You could easily round it to 1. But it would never completely touch 1.[/quote] What's 1-.999...=? Also, just add up (1/3) three times. Aka (.333...)+(.333...)+(.333...)=.999...=3/3=1
-
No. --------------- .99999... and 1 are different numbers. .99999... is a decimal, and 1 is not.
-
they may be considered the same, but .99999 repeating does not equal 1. 0.99999 repeating equals .99999 repeating, but rounds to 1 [Edited on 12.12.2010 8:25 PM PST]
-
[quote][b]Posted by:[/b] Kkman4evah Hm. One of my solutions proven wrong. But if .9 continued forever it would never add that last tiny insignificant 1 that it needs to actually REACH 1. You could easily round it to 1. But it would never completely touch 1.[/quote]It IS one, look at the proofs.
-
[quote][b]Posted by:[/b] Dream053 Is that a fact? I was taught that asymptotes were never reached.[/quote] How do you usually find an asymptote of a function? You usually use a limit that goes to infinity and the result you get is where the asymptote is located.
-
[quote][b]Posted by:[/b] Dream053 [quote][b]Posted by:[/b] drummer0702 [quote][b]Posted by:[/b] Dream053 Doesn't the same [i][b]general[/b][/i] concept lie with asymptotes: They're constantly approaching but never reaching 1, and are simply considered the same out of convenience?[/quote] They reach the line at infinity. [/quote]Is that a fact? I was taught that asymptotes were never reached.[/quote]0.999... is a limit. It would be the limit of the asymptote of the function of x as x goes to infinity.
