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I know I should (oops, I meant shouldn't) say this, all things given, but I have to say that the above emoticon is the finest creation in the whole of the universe. Never has something so simple caused so much hassle and yet so much joy to a bunch of nerds on forums...
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Given how many previously banned members have been let back in, perhaps its time to welcome Gamergod back into the fold. or not. |
Did you know? 2=1..
http://boobers.nion.be/motivators/mi...r-calculus.jpg
How long did it take for you to crack it? http://www.discussworldissues.com/fo...es/tongue1.gif |
Too bad its messed up!
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The thing is, how long does it take to..crack it? [yes] |
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where is the mistake in the calculation? I've gone over it twice and it seems to be right [help]
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The mistake is the firs line...
a=b The very nature of algebraic equation is to use a single letter to represent a single integer thus allowing an unknown number to be represented as said letter. a cannot equal b because algebra states that a must only equal a and b must only equal b. that aside... in the fourth line: (a+b) (a-b) = b (a-b) this suggests that (a+b) is equal to b which again at the top of the equation is stated that a=b so how can a+b = b There's a whole sh-ugar storm of assmption disguised by the infinitely daring leap-of-faith attitude that if u put 1+2=3 as letters you can just re-write the laws of physics as you see fit. Unfortunately 1=1 which is why your gross montly paycheck didn't just double after reading neinstein's theorem above. Good show though a+b/b for effort |
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And this isn't calculus, it's algebra. |
I could have concluded it was zero if the last line was omitted.
I totally agree on the calculus/algebra thing. You cannt have two distinctly different given values (x and y for example) in algebra for the same number though. You can on one occasion have two different numericals represented with one universal value (say x) however*. *This exception being a square root... eg. SqRoot of 9 = x x = either +3 or -3 or indeed as the answer is in this case 0 |
There is still this one big mistake noone has pointed out.
When going from 4th line to 5th line (a+b)(a-b) = b(a-b) to (a+b) = b you are essentially dividing by (a-b), which equals zero (as first line states a=b). Division by zero is not allowed, so all calcuations after that are irrelevant. ...moving back to the shadows... |
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When dividing by (a-b) you're stating: A+b = b OR a-b = 0 The last one is often forgotten, but is true in this case, since a = b |
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Point is the OR statements are left out. Same for the 2nd to last line, you're dividing by a. Either that's possible, or a = 0. That's what caused the mess. |
Heh noticed you removed your "or divide by 0", but that isnt possible because that will equal nothing, so as soon as you divide a or b by 0 it equals nothing (my maths is a bit sketchy, but I think thats correct). Even Windows Calculator wont allow you to divide by 0 as the answer would be nothing, which it cannot generate - look for yourself!
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I see the point about the step where essentially there is a div/0.
Isn't there however an actual algebraic error done prior which is what allowed that situation? I think it lies along these lines.....b does not equal the square root of b^2. |
It's a fun prank to give to someone that just started algebra that might miss that "divide by zero" situation at 4th line..[thumbup]
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