Jtdylktuy

Is there some way to prove that exactly $n$ numbers are needed to define a point in a space $mathbb{R}^n$?

Using Cartesian systems, this is intuitive for $n=2$ and $n=3$, but how can we prove it for higher dimensions?

This generally follows from the fact that an $n$-dimensional vector space needs a basis set of $n$ vectors to span it. $mathbb{R}^n$ is, of course, $n$ dimensional. The related proofs can be found in any half-decent introductory text on linear algebra.

The end result is that any point in the vector space, such as the $n$-tuples in $mathbb{R}^n$ you're concerned with, can be considered as a vector, and, as that vector is in the span of the basis set, then you can write it as a linear combination of those basis vectors.

As a simple example that touches on what I mean by this, we typically we represent points in $mathbb{R}^n$ by the $n$-tuple $(x_1, x_2, ..., x_n)$. This can be translated as being the sum of the usual basis for $mathbb{R}^n$, which is a unit vector along each axis:

• $e_1 = (1, 0, 0, ..., 0)$


• $e_2 = (0, 1, 0, ..., 0)$


• $e_3 = (0, 0, 1, ..., 0)$


• $e_k = text{vector of n components, all zero except for a 1 at the k-th place}$


• $e_n = (0, 0, 0, ..., n)$

Thus, considering the $n$ tuple $(x_1, ..., x_n)$ as a vector,

$$(x_1, ..., x_n) = x_1e_1 + x_2e_2 + ... x_ne_n$$

Of course, this is just one way of presenting this $n$-tuple, dependent on this basis. Change the basis set - the $e_k$'s above - and you will need a different calculation. But since

$$dim(V) = n iff text{V's basis set requires n vectors}$$

for any vector space $V$ (here, $V = mathbb{R}^n$), then each point will always be the linear combination of $n$ vectors.