I win !!
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Maybe math on your side of globe if different, but in Poland you must not divide by zero!A1 wrote:Lets see a proof:Code: Select all
Let, a = b =>a^2 = ba // multiplying 'a' on both side =>a^2 - b^2 = ba - b^2 // adding -b^2 on b.s =>(a + b) (a - b) = b (a - b) =>(a + b) = b //dividing by (a - b) from both side => 2b = b // as a = b => 2 = 1 // ?? :o
A equals b therefore (a-b) equals zero. So you can't divide by (a-b) !
Simple? I'll bet it is
Maybe you have got something more difficult Nope. We define i to be a "number" such that i*i = -1. By adding i to the set of all real numbers and closing it under addition and multiplication we get the complex numbers -- all numbers of the form a+bi. For example, (7i)^2 = -49, (3+i) * (2-i) = (6 - 3i + 2i - i^2) = 7-i, etc.Andrew_PL wrote:As far as I remember the domain of function of square root is {x in R, x >= 0 }. So you can not write "sqrt(-1)". Error is in transformation "sqrt( (-1) * (-1) ) = sqrt(-1) * sqrt(-1)".misof wrote:1 = sqrt(1) = sqrt( (-1) * (-1) ) = sqrt(-1) * sqrt(-1) = i * i = -1
You did say you wanted somethin more difficult
Man, I have been using complex numbers for 2 years in math and it's applications in electronics! So don't tell me what it is, all right :>misof wrote:Nope. We define i to be a "number" such that i*i = -1. By adding i to the set of all real numbers and closing it under addition and multiplication we get the complex numbers -- all numbers of the form a+bi. For example, (7i)^2 = -49, (3+i) * (2-i) = (6 - 3i + 2i - i^2) = 7-i, etc.
BTW: Don't say i (or j) is a "number", it is a number. Complex numbers are generalization of real numbers.
You didn't say what I said wrong! I still saying that an error is in transformation "sqrt( (-1) * (-1) ) = sqrt(-1) * sqrt(-1)". You can't just flip between domains just like thatmisof wrote:You did say you wanted somethin more difficult
Well, yes, the transformation is wrong, but still it isn't necessarily because it's about flipping domains. One can easily generalize sqrt() to complex numbers: for example by defining sqrt( r * (cos phi + i sin phi) ) = sqrt(r) * (cos (phi/2) + i sin (phi/2) ) [ for phi in <0,2*pi) ].Andrew_PL wrote: You didn't say what I said wrong! I still saying that an error is in transformation "sqrt( (-1) * (-1) ) = sqrt(-1) * sqrt(-1)". You can't just flip between domains just like that
(In human words, for other than positive real numbers: out of the two possible complex "square roots" we take that one with a non-negative imaginary part.)
The problem with the "equation" above is that no possible generalization is reasonable enough to be distributive w.r.t. multiplication anymore.
misof wrote:Well, yes, the transformation is wrong
I agree.misof wrote:One can easily generalize sqrt() to complex numbers
In other words it is the same what I saidmisof wrote:The problem with the "equation" above is that no possible generalization is reasonable enough to be distributive w.r.t. multiplication anymore.
BTW: What w.r.t means ?
nice & interesting post!!!!!!!!!!!!1
Hi,
Thank's A1 for this Interesting post.Nice rule for the infinite game.
Ok now i am the winner of the game...
Thank's in advance
Rocky
Thank's A1 for this Interesting post.Nice rule for the infinite game.
Ok now i am the winner of the game...
Thank's in advance
Rocky
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Ali Arman Tamal
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