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#1 |
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Take the function:
f(x) = (1/x)(x^3 - x^2) So: f(x) = (x^3/x) - (x^2/x) = x² - x. We proved that: (1/x)(x^3 - x^2) = x² -x. But if x= 0: -> x² - x = 0² - 0 = 0 -> (1/x)(x^3 - x^2) = (1/0)(0^3 - 0^2) : this is impossible (division per zero). So these two functions are the same, but they're not the same (O_O) OMG OMG WTF?!? |
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#2 | |
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If f(x) = x² - x = f(x) = 0² - 0 then the division is the same. But 0² = 0. You're dividing 0 by 0. Not possible unless 0 = x OMG OMG WTF?!? |
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#3 | |
expwy. to yr skull
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a pole at x=0 what's so strange about that? edit: Just realized that it isn't a pole. Anyways, it's late |
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#4 |
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tit, you dividing by 0, what's new?
-- the moral: do not solve math equations while stoned. wow, man, did you see that? it's like... like... wow... |
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#5 | |
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ps-
youre doing [x^3 - x^2]/x, if x= 0 then it's 0/0 Quote:
you stated it was IMPOSSIBLE, which was wrong, hence the logical loophole. |
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#6 |
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Please, can you stop speaking chinese, I don't understand...
![]() Logical ? What does it mean ? |
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#7 |
expwy. to yr skull
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what he is struggling with is 1/0, not 0/0. meaning that they are not equal because y is undefined when x=0. can the first one even be considered a parabola since it has the break in it?
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#8 | |
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yeah but in the case of x=0 that formula amounts to 0/0. he's ignoring the value of the numerator, sez eye. |
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#9 |
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#10 |
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What !@#$%! said.
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Sab Kuch Tick Tock Hai |
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#11 |
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Should be able to make sense out of that? Cause I can't
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#12 | |
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sorry for my bad english, what i meant was that his logic was short-circuiting. yes? he was reaching a self-contradictory conclusion. but it is because his premises were wrong. |
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#13 |
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Basically, he created a self-contradictory loop, which feeds back onto itself, thereby creating an infinite paradox, which if questioned begins to melt down the logic circuits of the brain...
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#14 | |
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^^ what he said. 2 thumbs up. |
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#15 |
expwy. to yr skull
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The function is undefined at x=0. I forget what those are called, but they always have to be looked out for when you do this kind of thing.
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#16 |
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I think you might have to factor in limits in this equation. I think it's not so much undefined, but the proper term is indeterminate, I believe. If you factor in limits in the above equation and let the limits of x approach zero, then you effectively illustrate that the equation is unsolvable algebraically, thereby removing the conundrum that has been put in place by attempting to solve the equation further.
"...To avoid this little inconsistency the epsilon-delta limit device is used to establish that the LIMIT of the ratio dy/dx is 2ax + b as dx APPROACHES zero."
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#17 | ||
expwy. to yr skull
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this function never crosses the y axis because when x=0 y is undefined, the reduced version is a full parabola so to speak that actually crosses the y axis at (0,0)
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#18 |
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I think you forgot to carry the 1.
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#19 |
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