By using this site, you agree to our Privacy Policy and our Terms of Use. Close
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

 

yeah called imaginary numbers in England too!