>>13256133>>13256115From Wikipedia on imaginary numbers
Care must be used when working with imaginary numbers that are expressed as the principal values of the square roots of negative numbers:[11]
{\displaystyle 6={\sqrt {36}}={\sqrt {(-4)(-9)}}\neq {\sqrt {-4}}{\sqrt {-9}}=(2i)(3i)=6i^{2}=-6.}{\displaystyle 6={\sqrt {36}}={\sqrt {(-4)(-9)}}\neq {\sqrt {-4}}{\sqrt {-9}}=(2i)(3i)=6i^{2}=-6.}
That is sometimes written as:
{\displaystyle -1=i^{2}={\sqrt {-1}}{\sqrt {-1}}{\stackrel {\text{ (fallacy) }}{=}}{\sqrt {(-1)(-1)}}={\sqrt {1}}=1.}{\displaystyle -1=i^{2}={\sqrt {-1}}{\sqrt {-1}}{\stackrel {\text{ (fallacy) }}{=}}{\sqrt {(-1)(-1)}}={\sqrt {1}}=1.}
The fallacy occurs as the equality {\displaystyle {\sqrt {xy}}={\sqrt {x}}{\sqrt {y}}}{\displaystyle {\sqrt {xy}}={\sqrt {x}}{\sqrt {y}}} fails when the variables are not suitably constrained. In that case, the equality fails to hold as the numbers are both negative, which can be demonstrated by:
{\displaystyle {\sqrt {-x}}{\sqrt {-y}}=i{\sqrt {x}}\ i{\sqrt {y}}=i^{2}{\sqrt {x}}{\sqrt {y}}=-{\sqrt {xy}}\neq {\sqrt {xy}},}{\displaystyle {\sqrt {-x}}{\sqrt {-y}}=i{\sqrt {x}}\ i{\sqrt {y}}=i^{2}{\sqrt {x}}{\sqrt {y}}=-{\sqrt {xy}}\neq {\sqrt {xy}},}
where both x and y are non-negative real numbers.
