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I'm new to this very interesting world of mathematics, and I'm trying to learn some linear algebra from Khan academy.

In the world of vector spaces and fields, I keep coming across the definition of $\mathbb R^2$ as a vector space ontop of the field $\mathbb R$.

This makes me think, Why can't $\mathbb R^2$ be a field of its own? Would that make $\mathbb R^2$ a field and a vector space?

Thanks

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    Vondip - Perhaps this is at a slight tangent, but a significant difference between R and C is that R is an ordered field and C is not. e.g. 5 is larger than 3, but which is "larger", 4 + 7i or 6 + 5i ? (Answer: well, defining how "large" or "the length" a complex number is is not as obvious as for the reals. In fact, there are many different ways of defining the length of a complex number). Just something to think about.2012-12-04

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If you define:

$(a,b)+(x,y):=(a+x,b+y)$

$(a,b)\cdot (x,y):=(ax-by,ay+bx)$

then the set $\,\Bbb R^2=\Bbb R\times\Bbb R\,$ turns into a field, and a rather well known and important one. Can you identify it?

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    Indeed, every finite-dimensional vector space $V$ over a field $\mathbb{F}$ is isomorphic to $\mathbb{F}^n$, where $n$ is the dimension of $V$. This fact impressed me a lot when I first learned it.2019-05-26
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It is important to understand that a set on its own has no algebraic structure. By defining operators on $\mathbb{R}^2$ you could turn it into (almost) anything you like.

The natural operators on $\mathbb{R}^2$, namely $(x, y) + (a, b) \mapsto (x+a, y+b)$ and $(x, y) \cdot (a, b) \mapsto (x\cdot a, y\cdot b)$ do not define a field as $(0, 1)$ has no multiplicative inverse.

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Usually in mathematics one defines these structures as tuples

A field is a triple $(K,+,\cdot)$ such that $K$ is a set and [...] and $\cdot:K \times K \rightarrow K$

A Vectorspace is a triple $(V,+,\cdot)$ such that $V$ is a set and [...] and $\cdot: K \times V \rightarrow V$

So your question is meaningless: A set (say $\mathbb R^2$) cannot be a field or a vectorspace or a group or anything - only if you add some additional structure (most of the time operations) you can ask this question.

For example $\mathbb R^2$ can be the set used in the definition of a field, as well as the underlying set used in the definition of a vectorspace. And we are happy, the addition operation $ +:\mathbb R^2 \times \mathbb R^2 \rightarrow \mathbb R^2 $ is the same, and the multiplicaton for "the" vectorspace structure $ \mathbb R \times \mathbb R^2 \rightarrow \mathbb R^2 $ is "compatible" with the multiplication for "the" field structure $ \mathbb R^2 \times \mathbb R^2 \rightarrow \mathbb R^2 $

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Adding to the above answer. With the usual exterior multiplication of the $\mathbb{R}-\{0\}$ as a ring with the natural addition and multiplication you can not make a field out of $\mathbb{R}^{2}-(0,0)$ \ But there may exist other products such as the one in the answers which can make a field out of ${\mathbb{R}\times\mathbb{R}}-\{ 0\}$ \ According to one of the theorems of Field theory every field is an Integral domain. So by considering : ${\mathbb{R}\times\mathbb{R}}-\{ 0\}$ With the following natural product: $(A,B)*(C,D)=(AB,CD) $ We see that $(1,0)*(0,1)=(0,0) $ Which means that $\mathbb{R}^{2} $is not an integral domain and hence not a field.

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    Yes I noticed my latex failure , I edited the comment . In fact this was the first time I placed star and it didn't go well :))2015-02-23