# Thread: Scalar by vector multiplication

1. ## Scalar by vector multiplication

Show why scalar multiplicative inverse property (∀a ≠ 0) (∃ b) ab=1 has no analogue using vector-by-scalar multiplication.

I'm not sure how to go about this. Help would be appreciated I know what the two symbols ∀ ∃ represent, but not sure what to do for this question.

2. I don't know what the question is even asking. There is no notion of the "multiplicative inverse" of a vector (well... not in the sense of the vector space structure anyway).

3. Originally Posted by topspin1617
I don't know what the question is even asking. There is no notion of the "multiplicative inverse" of a vector (well... not in the sense of the vector space structure anyway).
Yup, I don't understand either. Anyone else know? Really don't know how to do this.

4. Originally Posted by brumby_3
Show why scalar multiplicative inverse property (∀a ≠ 0) (∃ b) ab=1 has no analogue using vector-by-scalar multiplication.

I'm not sure how to go about this. Help would be appreciated I know what the two symbols ∀ ∃ represent, but not sure what to do for this question.
1. If you have real numbers you know that for every number $a \in \mathbb{R} \setminus \{0\}$ exists a number $\frac1a$, called the reciprocal of a, such that $a \cdot \frac1a = 1$.

2. If you have the scalar product of two vectors the result is not a vector but a single real number. This relation $2\ vectors \longrightarrow real\ number$ is not invertible.

As a practical consequence you are not allowed to calculate

$\vec a \cdot \vec b = 1~\implies~\vec a = \dfrac1{\vec b}$

5. But this question was about "vector by scalar multiplication" (which I take to mean scalar multiplication), not the dot product.

In order to have a multiplicative inverse, you must first have a multiplicative identity- but there is no vector, v, such that, for every scalar, a, av= a because the product is a vector, not a scalar. There is a scalar multiplicative identity, 1, so that 1v= 1 for every vector, v, but we cannot get that as the result of a multiplication because, again, the product is a vector, not a scalar.