Show that the set of all positive numbers with operations
$$\vec{x}+\vec{y}=\vec{x}\vec{y}\quad (1)$$ $$\lambda \vec{x}=\vec{x}^{\lambda},\lambda\in R\quad (2)$$ is a vector space. What is the zero element?

This new definition vector addition and scalar multiplication works for the axioms of a vector space except for $$\vec{x}+\vec{0}=\vec{x}\quad (3)$$ $$\vec{x}+(-\vec{x})=\vec{0}\quad (4)$$ unless we define $1$ as the zero vector.

  1. Is $\vec{0}=1 $ what the question calls the zero element?
  2. Can you define the zero vector any way you like as long as (3) and (4) are satisfied?


  1. Is the definition of the zero vector the conditions (3) and (4) for all vector spaces?
  • $\begingroup$ Yes (to both questions). $\endgroup$ Jan 10, 2022 at 9:37

1 Answer 1


In a Vector space or a Ring or a Field. $0$ is not really a "number". It is a symbol to denote the additive identity. Here the term additive might also mislead you. It does not necessarily mean "addition" in the usual sense. It means the binary operation wrt which the set forms a commutative group.

So You might as well denote it by $e$ which is the symbol for identity in a group or $\mathbf{0}$ to make it distinct from usual $0$. and $"+"$ by something like $\oplus$ to differentiate it from the "usual" addition.

So in reality you are looking at the set $\mathbb{R}_{>0}$ over the field $\mathbb{R}$. with the operation $\oplus$ as addition.

$x\oplus \mathbf{0}= x \implies \mathbf{0}=1$.

And the multiplicative and distributive axioms follow from the laws of indices.

And yes. As long as you have the axioms of vector space is satisfied for any set, it forms a vector space.


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