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**HallsofIvy** Yes, your argument is correct. If a is a "scalar" and v is a "vector" then av must be a vector. If you multiply a "differentiable real-valued function defined on the real line" by a complex number, the result would no longer be "differentiable real-valued function defined on the real line". Now, it would be possible to have the space of "differentiable real-valued functions defined on the real line" over the field of **rational** numbers since the product of a rational and a real number is a real number- but that would be very unusual.