
| -flood- #446830 10 months ago |
A Derpy one~ Nice!
But I still ain't changin my Pinkie Pie :D |
| Anonymous #446833 10 months ago |
New desktop background forever |
| Anonymous #446835 10 months ago |
Hello new wallpaper |
| Homfrog #446878 10 months ago |
Fancy mathematics |
| Anonymous #447089 10 months ago |
.9 repeating does equal one...
assume x=.9999... then 10x = 9.99999... -now subtract the two- you get 9x = 9... then divide by nine - you get x=1... using the law of substitution x=1 and x=.9999... plug in .999... into 'x' for x=1 and you get .9999...=1 |
| -flood- #447126 10 months ago |
^
You were already wrong at "-now subtract the two- " Just cuz you've got seperate givens doesn't mean you could equate them together. |
| AndroidAR #447733 10 months ago |
A simpler way is:
1/3 = 0.333... 1/3 + 1/3 + 1/3 = 0.999... 1/3 + 1/3 + 1/3 = 3/3 = 1 therefore 0.999... = 3/3 = 1 Though something deep inside me feels that 0.999... and 1 are practically the same, but technically different. |
| Glamador #448633 10 months ago |
What we have here is a meaningless proof. By extension of the same logic you can equate any infinitely repeating decimal to the next highest non-repeating real number. I.E. in the proof above instead of subtracting to get 9x = 9 just multiple directly to get 9x = 8.999... implying 8.999.... = 9.
Should you continue to use these decimal substitutions in equations involving non-repeating real numbers you'll notice that all results will be in terms of infinitely repeating decimals which can then be substituted for their non-repeating counterparts. All this proof does is rename the integers we know with inexpressible repeating digits, simplifying nothing and providing no new information. I have seen proofs that use the equality 1 = 0.999... but I can think of very few in which it was a necessary step. And I have my suspicions that they could be proven in other ways. |
| Glamador #448658 10 months ago |
However, I would posit that 1/3 =/= 0.333... since 1/3 is inexpressible as a series of digits. An infinite sequence cannot be expressed on paper so we use the closest available estimate (I.E. 0.xxxxxx until we're blue in the face).
But just as one can never actually count to infinity, no matter how many 9's or 3's we tack onto the end of number expressed as a decimal it will never EQUAL any non-repeating digit. That equality can only be expressed as a limit as the # of digits approaches infinity. Therefore I suggest it is more correct to say the limit of the expression 0.999... as the number of digits approaches infinity = 1. |
| Anonymous #1454214 3 weeks ago |
The value of 0.999... and other infinitely repeating decimals is not subject to debate. They are all representations of finite rational numbers--you are not "counting to infinity." In some bases, decimal expansions go on indefinitely, and in others, they don't. If you don't believe me, try writing 1/5 in base 2 sometime.
If you haven't seen the Wikipedia article on "0.999..." yet, I'd encourage you to read it. Especially see http://en.wikipedia.org/wiki/Talk:0.999.../Arguments, which is full of incredulous people making similar non-arguments to the one you made above. |