Question: If we accept that scalar multiplication is effectively multiplying each component of the vector by the scalar value, it becomes much more straightforward to multiply

 If we accept that scalar multiplication is effectively multiplying each component

of the vector by the scalar value, it becomes much more straightforward

If we accept that scalar multiplication is effectively multiplying each component of the vector by the scalar value, it becomes much more straightforward to multiply by more than just integer-valued scalars.

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