How is a set $\mathbb{R}$ a field if for a=0$\in \mathbb{R}$, $\nexists a^{-1}$ such that $a\cdot a^{-1}$=1

Question

I was reading a chapter concerning the definition and examples of fields.

It states that examples of some known fields include $F=\mathbb{R}, \mathbb{Q}, \mathbb{C}$

One of the 9 properties of a field is that for $\forall a\in F, \exists a^{-1}$ such that $a\cdot a^{-1}=1$ (existence of inverse)

How is a set $\mathbb{R}$ a field if for a=0$\in \mathbb{R}$, $\nexists a^{-1}$, such that $a\cdot a^{-1}$=1?

Also, if a set $\mathbb{R}$ is a field, can we say that the Cartesian product of sets $\mathbb{R}\times\mathbb{R}$ is also a field?

Answer 1

$\forall a\in F, \exists a^{-1}$ such that $a\cdot a^{-1}=1.$ This statement is true for all elements of the field except $0.$

Answer to the second question: No for a general situation. Don't be confused with the fact that $\mathbb{R^2}$ is treated a different way to create $\mathbb{C}.$ Note that in $\mathbb{R^2},$ we don't have multiplication. However, in $\mathbb{C}$, we define a multiplication, which is the multiplication of complex numbers. This leads $\mathbb{C}$ to be a field. Please don't forget that $\mathbb{C}$ and $\mathbb{R^2}$ are not technically the same thing.

Some extra comments: In $\mathbb{R^2},$ we can define the so-called inner product/dot product of two vectors. Note that this definition in fact does not give a binary operation on $\mathbb{R^2}.$ I mean this inner product gives a number in $\mathbb{R}$ (not in $\mathbb{R^2}$) when you compose two vectors in $\mathbb{R^2}.$

On the other hand, we have a standard cross/vector multiplication in $\mathbb{R^3}.$ This definition is not that bad. At least this gives a binary operation on $\mathbb{R^3}.$ That means you have the closure axiom for your multiplication. The sad thing is that the vector/cross multiplication is not associative, which is also true for dot multiplication. So, it does not help us.

However, in $\mathbb{R^2}$ the cross multiplication is not even a binary operation. we have the so-called right-hand screw rule for vector/cross multiplication. This reveals the fact this multiplication does not have the closure property. That means $\mathbb{R^2}$ is not a field if we consider the usual dot and cross multiplications of vectors.