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Anonymous

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Posted: Sun December 03, 2006 Post subject: 1 equals 2 |
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a = b
a^2 = a*b
a^2-b^2 = a*b-b^2
(a+b)(a-b) = b(a-b)
(a+b) = b
a+a = a
2a = a
2 = 1
it's old i know but still awesome....
do u like it
post ur replies |
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Josh Site Admin

Joined: 15 Nov 2006 Posts: 1648 Location: Sydney, Australia
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Posted: Sun December 03, 2006 Post subject: |
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a^a is not the same as a*a
For example
2^3 = 8
2*3 = 6
However its good, you nearly got me. |
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Anonymous

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Posted: Sun December 03, 2006 Post subject: |
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You have to define your variables uditbhargava.
It is ok josh, thats not the error in that equation. Sine a = b you can also say that it says a^2 = a*a. The problem is that you can't divide by zero. (a-b)
This one is better:
-1/1 = 1/-1
sq(-1/1) = sq(1/-1)
sq(-1)/sq(1) = sq(1)/sq(-1)
i/1 = 1/i
i/2 = 1/(2i)
i/2 + 3/(2i) = 1/(2i) + 3/(2i)
i(i/2 + 3/(2i)) = i(1/(2i) + 3/(2i))
i^2/2 + 3i/(2i) = i/(2i) + 3i/(2i)
-1/2 + 3/2 = 1/2 + 3/2
2/2 = 4/2
2 = 1
This should be way below your level of math josh. Good luck solving.  |
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Josh Site Admin

Joined: 15 Nov 2006 Posts: 1648 Location: Sydney, Australia
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Posted: Sun December 03, 2006 Post subject: |
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Yes its divide by zero error.
Last edited by Josh on Sun December 03, 2006; edited 2 times in total |
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Josh Site Admin

Joined: 15 Nov 2006 Posts: 1648 Location: Sydney, Australia
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Posted: Sun December 03, 2006 Post subject: |
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I think the problem is in the square root.
sq(-1) = +/- i
EDIT: Actually I don't think that would make a difference.
OK I've got it the square root function SQRT(A*B) = SQRT(A)*SQRT(B) if A and B are both real.
Last edited by Josh on Sun December 03, 2006; edited 1 time in total |
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Anonymous

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Posted: Sun December 03, 2006 Post subject: |
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| That be correct, and sq(1) = +/- 1 too. |
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Josh Site Admin

Joined: 15 Nov 2006 Posts: 1648 Location: Sydney, Australia
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Posted: Sun December 03, 2006 Post subject: |
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| OK I've got it the square root function SQRT(A*B) = SQRT(A)*SQRT(B) if A and B are both real and greater than zero. Same with division. |
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Anonymous

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Posted: Sun December 03, 2006 Post subject: |
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| I never even thought of that part, good job. |
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Josh Site Admin

Joined: 15 Nov 2006 Posts: 1648 Location: Sydney, Australia
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Posted: Sun December 03, 2006 Post subject: |
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The you can't split up the square root unless both terms are positive. |
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Anonymous

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Posted: Sun December 03, 2006 Post subject: |
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| sq(1) = +/- 1 So that should be correct. Just a false root? |
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Josh Site Admin

Joined: 15 Nov 2006 Posts: 1648 Location: Sydney, Australia
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Posted: Sun December 03, 2006 Post subject: |
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No. The answer is that
This principle is only correct when y is a positive number.
http://en.wikipedia.org/wiki/Invalid_proof
sq(-1/1) = sq(1/-1)
sq(-1)/sq(1) = sq(1)/sq(-1)
This part is wrong. You can't split the squareroot up.
Last edited by Josh on Sun December 03, 2006; edited 1 time in total |
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Anonymous

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Posted: Sun December 03, 2006 Post subject: |
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| This is confusing.... |
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Anonymous

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Posted: Sun December 03, 2006 Post subject: |
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| I concur, thats true. However, if it wasn't you'd have to balance the formula out using multiple roots. |
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Hobo Dirty Mothertrucker

Joined: 21 Nov 2006 Posts: 323 Location: Boonies
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Posted: Mon December 04, 2006 Post subject: |
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| This must be a joke for smart people. I don't get it or want to get it. |
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Anonymous

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Posted: Mon December 04, 2006 Post subject: |
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Your not the only one, this makes no sense at all.  |
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