Recent content by amits

yes, you are right. a number always has 2 square roots, 1 positive & 1 negative

it isn't an error. i added a^2 + b^2 to both sides -ab = -ab a^2 + b^2 - ab = a^2 + b^2 - ab a^2 - a^2 - ab = b^2 - b^2 - ab also a^2 - a^2 = 0 & b^2 - b^2 = 0, so adding that doesn't make a difference

101 = (4!*4) + [sqrt(4)/.4] 102 = (4!*4) + 4 + sqrt(4) 103 = (4! + sqrt 4)*4 - sqrt(sqrt(sqrt(sqrt...(sqrt(4))...))) 104 = (4!*4) + 4 + 4 105 = (4! + sqrt 4)*4 + sqrt(sqrt(sqrt(sqrt...(sqrt(4))...))) 106 = (4! + sqrt 4)*4 + sqrt(4) 107 = (4 + (4!/4!))! + sqrt(4) double factorials...

here are the answers - 1 = 44/44 2 = (4/4) + (4/4) 3 = [(4*4) -4]/4 4 = (4/.4) - (4!/4) 5 = (4!/4) - (4/4) 6 = (4!/4)*(4/4) 7 = 4 + 4 - (4/4) 8 = 4 + 4 + 4 - 4 9 = 4 + 4 + (4/4) 10 = (44 - 4)/4 11 = (4/.4) + (4/4) 12 = (44 + 4)/4 13 = 4! - (44/4) 14...

in 1 post in the puzzles community on (nowadays inactive)orkut, i remember doing it. it involved the usage of double factorials too. ex 7! = 1*3*5*7 = 105, 8! = 2*4*6*8 = 384, etc i was able to find the solutions of all numbers from 1 to about 150 using the double factorial thing alongwith...

All numbers are equal :D heres the latest proof and it doesn't involve 0 anywhere. no term like (a-b) is involved unlike other similar proofs -ab = -ab => a^2 - a^2 - ab = b^2 - b^2 - ab => a^2 - a(a+b) = b^2 - b(a+b) => a^2 - a(a+b) + (a+b)^2/4 = b^2 - b(a+b) + (a+b)^2/4 as (x-y)^2 = x^2 -...

himanshu, thanks for ur reply. i c the rules of this forum. what i want to know was some tips on how to study physics. also what books u used to study physics ? HC Verma ? SL Gupta ? Resnick-Halliday ? Interactive Physics ? or any other

i am 18 from New Delhi, India

no matter how hard i try, but i am not able to study physics. i have the IIT JEE exams in 2004, and my physics isn't even to the basic level. can anybody help me on how to study physics [?]