chapset said: lets use imaginary numbers to resolve this problem z= ai + b where: i is the imaginary part we can make b = 0, because 0 is a neutral number i = √(-1) i 2 = -1 i4 = 1 i4 + i4 = 2i4 = 2(1)* = 2 i4 + i4 + i4 + i4 = (i4 + i4) + (i4 + i4 )= 2i4 + 2i4 = 2(1) + 2(1) = 2 + 2 = ? but you can put it this way: i4 + i4 + i4 + i4 = 4i4 = 4(1) = 4
and since i4 + i4 + i4 + i4 = i4 + i4 + i4 + i4 then 2 + 2 = 4 *1 is the neutral number of multiplication
note: I just made that one up using the imaginary numbers (i don't know if they are called the same in english) I learnd my math in french and some times the name change but not the numbers ;)
note 2: this is not a real theorem i made that one up
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yeah called imaginary numbers in England too!