When determining if a subset is in a vector space, can the scalar for the axiom

Question

When determining if a subset is in a vector space, can the scalar for the axiom regarding scalar multiples being in the vector space be irrational?

For example, consider the question "is the set of all polynomials p(t)= at2 where a ? R a subspace of Pn ?". When testing whether or not this is a subspace, can the scalar multiple in the axiom stating "the subspace is closed under multiplication by scalars" be an irrational number such as 3i, making the given set of polynomials not a subspace because ca ? R ? a ? R?

Explanation / Answer

No, we cannot take scalar as irrational number because in that case it will not be a vector space

As per definition of scalar multiplication if we take irrational number as a scalar then by scalar multiplication resultunr elements cannot contained in given set.so, it cannot be a subspace

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