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Articles: My Thoughts | | Ramanujam's Calculation | |
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| i^3 = -1 elaaga?
Posted by: Mr. Aandhrudu At: 26, Apr 2006 5:55:56 PM IST
Andhurudu garu,
The equation would be void after the following step.
subtract 'b^2' from bothsides
a^2-b^2 = ab - b^2
as per formula
(a-b)(a+b) = b(a-b)
When you divide both sides with (a-b), the above euation will be 0=0 because, we have started this proof with an assumption that a=b. So, this is not correct.
Ila ayithe manam any number equal to any number ani prove cheyachu kada ;-).
thanks.
Posted by: Mr. Sarma Anantha At: 19, Apr 2006 1:12:52 AM IST
a=b
multiply with 'a' bothsides
a^2=ab
subtract 'b^2' from bothsides
a^2-b^2 = ab - b^2
as per formula
(a-b)(a+b) = b(a-b)
dividing bothsides with '(a-b)'
(a+b) = b
substitting 'a ' with ' b ' as per first equation
2b=b
dividing bothsides with 'b'
2=1...............?????
Posted by: Mr. Aandhrudu At: 14, Apr 2006 5:14:49 PM IST
can you prove 1=2 ?
Posted by: Mr. Aandhrudu At: 14, Apr 2006 5:07:35 PM IST
(3-5/2)(3-5/2) = (2-5/2)(2-5/2)
The positive square root for (2-5/2)(2-5/2) would be 0.5 only. You can never take the square root for a negative number :-). So, this is the trick here. Hence the above equation turns out like this.
3-5/2=5/2-2
1/2=1/2
That's it.
thanks.
Posted by: Mr. Sarma Anantha At: 11, Apr 2006 9:29:15 PM IST
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