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You are not logged in. #1 20071110 23:22:38
ContradictionsSometimes, It is the subtle mistakes that make the biggest mistakes (this was in our exam, I thought it was pretty easy for us, until I asked my classmates what they answered...) Last edited by JohnnyReinB (20071110 23:51:28) "There is not a difference between an inlaw and an outlaw, except maybe that an outlaw is wanted" Nisi Quam Primum, Nequequam #2 20071110 23:36:03
Re: Contradictionsfor anyone that doesn't spot it. the first line states x < 0, so x is negative, at step 4, the equation is being divided by x, a negative number, so that sign has to change direction to 2x/x > x/x, which then leads to the right result of 2 > 1 The Beginning Of All Things To End. The End Of All Things To Come. #3 20071110 23:40:57
Re: ContradictionsHeres an interesting one from wikipedia which cannot be done in the same way as yours to show its false. Last edited by lucadeltodesco (20071110 23:42:05) The Beginning Of All Things To End. The End Of All Things To Come. #4 20071110 23:47:46
Re: ContradictionsI remember one tricky question based on contradictions. Character is who you are when no one is looking. #5 20071111 00:36:47
Re: Contradictionsnote quite johnny, its perfectly valid to only substitue into parts of the equation, the problem is something more ambiguous. The Beginning Of All Things To End. The End Of All Things To Come. #6 20071111 02:31:41
Re: Contradictions
Last edited by JaneFairfax (20071111 02:33:23) #7 20071119 20:57:01
Re: ContradictionsOk, so it's like this one: by dividing both sides by x And, by substituting to the first equation, you get... simplified, is multiplying both sides by x... finding the cube root... So now, to check, substitute again... simplified, is Which is ,obviously, wrong... Ok...right? Last edited by JohnnyReinB (20071119 20:57:30) "There is not a difference between an inlaw and an outlaw, except maybe that an outlaw is wanted" Nisi Quam Primum, Nequequam #8 20071119 23:43:39
Re: ContradictionsYep, it’s the same. That’s because in the the original equation, x is complex. Therefore, x should be a complex solution of x^{3} = −1, not x = −1. One must always be wary of such psychological pitfalls in mathematics. Last edited by JaneFairfax (20071119 23:49:47) #9 20071123 16:18:56
Re: ContradictionsSo, how do you solve these equations? "There is not a difference between an inlaw and an outlaw, except maybe that an outlaw is wanted" Nisi Quam Primum, Nequequam 