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#101 |
FFR Veteran
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![]() either 2 or 3.
Duh. Haven't really learned about series and such yet.
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#102 |
FFR Player
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Yeah. Parentheses are like commas in mathematics.
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#103 | |
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![]() Quote:
---------- Firstly, remember that we can't really interpret 0.000...1 as a finite decimal value, but rather, the point that by definition bounds the limit of 1/n as n gets infinitely large. This point is of course 0 (which will be subsequently proven). For any two values x1 and x2, x1/n can be made arbitrarily close to x2/n by taking n to be a sufficiently large number. Thus, lim (n→∞) x1/n = lim (n→∞) x2/n, for all x Since x is a constant, we can extend the notion and say that lim (n→∞) x1/n = lim (n→∞) x2/n for all x => x1[lim (n→∞) 1/n] = x2[lim (n→∞) 1/n] for all x Thus, taking x2 to be 0, x1[lim (n→∞) 1/n] = x2[lim (n→∞) 1/n] => x1[lim (n→∞) 1/n] = 0[lim (n→∞) 1/n] => x1[lim (n→∞) 1/n] = 0 => lim (n→∞) x1/n = 0 And that's why lim (n→∞) x/n = 0 for all x. Proceeding with the proof... Taking x1 to be 1, lim (n→∞) x1/n = 0 => lim (n→∞) 1/n = 0 => From the aforementioned interpretation, 0.000…1 is equivalent to lim (n→∞) 1/n. Thus, lim (n→∞) 1/n = 0 => 0.000...1 = 0 And finally getting back to the problem at hand... Since lim (n→∞) 1/n = 0.000...1 = 0 = lim (n→0) n, 15.68 - lim (n→∞) 1/n = 15.68 - lim (n→0) n => 15.68 - 0.000...1 = 15.68 - 0 => 15.67999... = 15.68 ---------- IT'S SUPER EFFECTIVE! Last edited by ieatyourlvllol; 10-19-2009 at 04:05 PM.. Reason: fixed a slight omission |
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#104 |
Cutie&Handshaking Sounds
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But its only 1 answer, unless stated otherwise.
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#105 | |
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ieat wouldn't 10^-n have been easier? :P
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![]() Last edited by bluguerrilla; 10-19-2009 at 12:29 PM.. |
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#106 |
Cutie&Handshaking Sounds
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![]() If x is any number,
x ^ 0 = 1 Explain please. |
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#107 |
I am leonid
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![]() The reason why I suggested we please not get into this is that we were going to discuss a thing that's of the same level as "IS ONE PLUS ONE TWO OR THREE"
Last edited by leonid; 10-19-2009 at 12:41 PM.. |
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#108 |
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fixed, because you just can't divide by 0.
x² x x² = x^(2+2) = x^4 x² / x² = x^(2-2) = x^0 so, x^0 = x / x, it's like dividing x by itself, or just 1. Last edited by PhaeL v2; 10-19-2009 at 12:43 PM.. |
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#109 |
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#110 |
Cutie&Handshaking Sounds
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![]() But can I say that 1 = 2?
0 x 1 = 0 0 x 2 = 0 Thus 0 x 1 = 0 x 2 and then 0/0 x 1 = 0/0 x 2 so finally 1 = 2 |
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#111 | |
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But yeah whatever. The only reason you can't rearrange them is because they are already infinite and it doesn't make sense. |
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#112 |
I am leonid
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![]() @hakim: You just divided by zero
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#113 | |
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Priorizing multiplication, instead of division, using parentheses, you have 0 / (0 x 1) = 0 / (0 x 2), then, 0 / 0 = 0 / 0 but you can just ignore the parentheses using and priorize the multiplication, it will be right anyway. That's why I think if you eliminate the 0 division, you eliminate the whole line. |
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#114 |
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#115 |
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#116 | |
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And I meant to say absolutely in my above post, sry.
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#117 |
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actually, this is typically the proof used to define 0/0 as an indeterminate form
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#118 |
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![]() yup pretty much
0*whatever = 0 => 0*whatever1 = 0*whatever2 => 0 / 0 = (whatever1 / whatever2)^sqrt(1) = whatever wala you did math |
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#119 |
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#120 |
Retired
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![]() Are you all in math enriched? God damn, I dont understand anything of those things and I'm ''ok'' in Math( im not in enriched)
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