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08-22-2009, 05:00 PM
| | Registered User | | Join Date: Jun 2008 Location: Fredericksburg, Virginia | | | what just happened here?(wanring: contains math)
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Given:a=b
a2=ab
a2-b2=ab-b2
(a+b)(a-b)=b(a-b)
(a+b)=b
(a+a)=a
2a=a
2=1
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08-22-2009, 05:10 PM
|  | Master of Reality | | Join Date: Jul 2006 Location: San Diego, CA | | | Aren't there rules against variables having the same value? If not, there certainly should be.
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08-22-2009, 05:24 PM
| | Registered User | | Join Date: Jun 2008 Location: Fredericksburg, Virginia | | Quote:
Originally Posted by Unrepresented  Aren't there rules against variables having the same value? If not, there certainly should be. | Well, techinically, it's if a=b, it doesn't matter what you do. if 1=1, then all the rest is just conjecture and is incorrect.
math is weird.
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08-22-2009, 05:28 PM
| | Registered User | | Join Date: Apr 2008 Location: Leuven, Belgium | | | (a+b)(a-b)=b(a-b)
(a+b)=b
Didn't you divide by zero there (since a-b = 0)? |
08-22-2009, 05:29 PM
| | Registered User | | Join Date: Jun 2008 Location: Fredericksburg, Virginia | | Quote:
Originally Posted by drteeth (a+b)(a-b)=b(a-b)
(a+b)=b
Didn't you divide by zero there (since a-b = 1-1)? | 
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08-22-2009, 05:29 PM
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08-22-2009, 05:31 PM
| | Registered User | | Join Date: Aug 2007 Location: Big Island | | Quote:
Originally Posted by drteeth (a+b)(a-b)=b(a-b)
(a+b)=b
Didn't you divide by zero there (since a-b = 0)? |
He just imploded the universe! 
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08-22-2009, 05:39 PM
| | Master of Reality | | Join Date: Jul 2006 Location: San Diego, CA | | Quote:
Originally Posted by Thunderscreech | That was my point. Having two variables with the same value creates a situation where such circumstances are potential.
Both variables shouldn't be given different the presumption of different values if they're not different.
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08-22-2009, 06:23 PM
| | Registered User | | Join Date: Sep 2007 Location: Richmond, Virginia | | Quote:
Originally Posted by drteeth (a+b)(a-b)=b(a-b)
(a+b)=b
Didn't you divide by zero there (since a-b = 0)? | ^ Correct |
08-22-2009, 09:32 PM
|  | Registered User | | Join Date: Jul 2008 Location: Eh? | | | Yup, that's the classic hidden zero.
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08-22-2009, 09:36 PM
| | Registered User | | Join Date: Aug 2008 Location: Montreal, Canada | | Quote:
Originally Posted by Unrepresented That was my point. Having two variables with the same value creates a situation where such circumstances are potential.
Both variables shouldn't be given different the presumption of different values if they're not different. | I don't think 2 variables being equal is a problem. Just do it again by replacing every "b" by "a" and you will come to the same exact conclusion.
Whenever you see those weird mathematical solvings (usually it ends up with 1+1=3 or something like that), there's always an obvious mistake to anyone used to maths.
To spot them quickly, it is almost always a division by 0 or, in some cases, a bad solving of a polynome (ex : a^2+2b+c) when there is 2 theoretical answers but only one is part of real numbers (and the wrong one is displayed on the next line ...). These are the 2 tricks to spot those quickly.
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08-23-2009, 01:47 AM
| | Registered User | | Join Date: Jun 2009 Location: QLD, Australia | | | Or a rounding error:
1 = 3/3
1 = 0.3333.... * 3
1 = 0.99999....
the .333 and .999 are recurring
i know a good one.. ill have to try and remember it.
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Last edited by Simo98 : 08-23-2009 at 02:00 AM.
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08-23-2009, 06:50 AM
|  | Registered User | | Join Date: Nov 2005 Location: Palo Alto, CA | | | Technically, isn't .999 recurring the same thing as 1 though?
-Dash
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08-23-2009, 07:26 AM
|  | Registered User
Maker of HPF-Pre upright bass preamp | | Join Date: Mar 2004 Location: Madison WI | | Quote:
Originally Posted by Dash Rantic Technically, isn't .999 recurring the same thing as 1 though?
-Dash | Yes. |
08-23-2009, 07:54 AM
| | Registered User | | Join Date: Jun 2008 Location: Belgium | | Quote:
Originally Posted by Unrepresented That was my point. Having two variables with the same value creates a situation where such circumstances are potential.
Both variables shouldn't be given different the presumption of different values if they're not different. | Well, that's the whole point of variables... they're variable.
Meaning they can be any value. In the off chance that 2 different variables are the same value the universe shouldn't implode.
a2-b2=ab-b2 , just proves 0 = 0  when you have nothing you shouldn't go on.
But you do and you divide by zero.
(a+b)=b
(a+a)=a
a=a-a
a=0
therefore b=0
???
I'm searching if a and b can be another value
Last edited by René_Julien : 08-23-2009 at 07:59 AM.
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08-23-2009, 08:01 AM
| | Registered User | | Join Date: Jul 2008 Location: Eh? | | | No, because as along as a=b, you can't divide by (a-b), since it means it is 0.
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Originally Posted by tom once dead Also to prove my Australianism, I've been stung by an irukandji jellyfish before, while snorkelling at an island looking at stingrays. | |
08-23-2009, 08:09 AM
| | Registered User | | Join Date: Jun 2008 Location: Belgium | | Quote:
Originally Posted by L-A No, because as along as a=b, you can't divide by (a-b), since it means it is 0. | Yup, I've been thinking about it again.
Since these comparisons started with the fact that a=b you have to keep respecting that.
Going from line 3 to 4 is dividing by zero.
Doesn't mean that in a formula some variables can't be the same value.
But this is not a formula... it's just a comparison worked out wrong.
BTW, is it allowed to divide zero by zero?  I know it's not allowed. |
08-23-2009, 10:00 AM
| | Registered User | | Join Date: Apr 2002 Location: West Side SA | | | i'm starting school tomorrow, i'm taking this to my math Prof..
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08-23-2009, 11:04 AM
| | Registered User | | Join Date: Apr 2002 Location: West Side SA | | Quote:
Originally Posted by René_Julien it's just a comparison worked out wrong.
I know it's not allowed. | sounds like this is it..
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