Is there any sort of relationship between the terms "base" ( number systems) and "basis" ( linear algebra) as both have similar definitions.

Question

I was going through the topic of number systems today when I went to see the proper definition of a base.

It said

"the number of digits or combination of digits that a system of counting uses to represent numbers."

I remembered seeing the term basis in linear algebra having a similar definition:

"a set B of vectors in a vector space V is called a basis if every element of V may be written in a unique way as a finite linear combination of elements of B."

So, are the two topics related in any way? Or Are they the same thing? Or are they totally different things?

Answer 1

Basis vectors are a set that spans the space, that is by linear combination you can create every vector in the space from the basis vectors. The digits of a number system are not sufficient by themselves to represent every number in the system. They must be paired with the notion that their place represents a power of the base. There is direct correlation between polynomials as vectors and number systems. Someone more versed in linear algebra would have to speak to it, because a vector space requires its scalars to come from a field. The set of digits could be associated with a ring and I believe such a structure is called a module.

Answer 2

Hmmm... this is interesting in that there is indeed a somewhat loose connection.

To say that $B$ is a basis of vector space $V$ over a field $F$ means that every $v \in V$ can be written uniquely as a linear combination of the form $$v = \sum_{b \in B} r_b \cdot b $$ where the "scalars" $r_b$ are chosen from the field $F$, and all but finitely many of the $r_b$'s are zero.

In the integers, fixing any $N \ge 2$ and letting $B = \{N^i \mid i \ge 0\}$, every non-negative integer $n$ may be written uniquely as a linear combination of the form $$n = \sum_{b \in B} r_b \cdot b $$ where the "scalars" $r_b$ are chosen from the set $\{0,1,\ldots,N-1\}$, and all but finitely many of the $r_b$'s are zero.

There are some serious differences, though. In the vector space situation, addition of two such linear combination expressions is very simple because the set of scalars $F$ is a field and hence is closed under addition: $$\sum_{b \in B} r_b \cdot b + \sum_{b \in B} r'_b \cdot b = \sum_{b \in B} (r_b + r'_b) \cdot b $$ But in the integers, the set of scalars $\{0,1,...,N-1\}$ is not closed under addition, and $r_b + r'_b$ can fall outside of that set. This leads to the grade school algorithm for addition which requires one to "carry".

Answer 3

As others have pointed out there seems to be no direct connection however I am not in the position to be able to answer that with certainty.

Note that the following answer does not necessarily answer your question but rather takes another approach at connecting other 'bases' then base10 with vector spaces.

Anyways an interesting thing to think or talk about is identifying numbers in different Number Systems with Vector Spaces as Larry Cornell pointed out in his answer.

For example for Base 5 Numbers one might be able to identify these with the Vector Space of tuples with finite many entries that are unequal to 0 (and infinitely many zero entries) over the field $\mathbb{F}_5$.

In order to understand what this means let's first look at $\mathbb{F}_5$: This field consists of the Elements $0,1,2,3,4$ or rather the congruence classes $[0], .., [4]$. Basically we have the ring of the integer numbers but identify each number with it's residue when dividing by 5.

That's why the Field consists of 5 elements (since there are 5 possible residues after dividing by 5) and for each congruence class (element) different "representatives" can be choosen. For example $3$ and $8$ would be in the same congruence class and as such $[3] = [8]$.

Another way to talk about these congruence classes would be as an equivalence relation where for $a,b \in \mathbb{Z}: a \sim b \iff a = 5 +b \cdot k$ for $k \in \mathbb{Z}$ (and then we'd define $[a]=[b] \iff a \sim b$) (so iff a is the same as b plus some multiple of 5).

The addition and multiplication for this field would be defined as we know them and it can be proven that they are well defined or in other words independent of the election of representative (which means that they suffice $[a] + [b] = [c] + [b]$ if $[a]=[c]$ or equivalently if $a \sim c$).

Here are some example of addition and multiplication over $\mathbb{F}_5$: $$ [1] + [2] = [3] \qquad [2] \cdot [3] = [6] = [1] \qquad [25] \cdot [2] = [50] = [0] $$

(Sidenote: you might have come across this concept of modular arithmetic before since this is exactly what is used in order to argue that 'sometimes' $2+1 = 0$ or something similar. Maybe you can figure out what $p$ in $\mathbb{F}_p$ would have to be in order for $[2]+[1]$ to be equal to $[0]$ (read the next sentence for a hint)).

It can futhermore be profen that these elements form the structure of a field with said operations which holds true for all $\mathbb{F}_p$ with $p$ being prime. With this knowledge we can look at vector spaces over $\mathbb{F}_5$.

Enough of the modulo arithmetic and now back to the idea of the start: The idea was to construct a vector space of tuples which have finite many non-zero entries. So for example the tuples $$ \begin{pmatrix} 0 & 0 & 1 & 0 & \ldots \end{pmatrix} $$ or $$ \begin{pmatrix} 0 & 1 & 1 & 4 & 0 & 0 & 1 & 0 & \ldots \end{pmatrix} $$ (In this case there would follow infinitely many zeros after the $\ldots$.)

If we now for the entries of the tuples only use Elements of $\mathbb{F}_5$ there is an interesting way to identify numbers of Base5 with these tuples.

For example the number $321$ in Base5, which would be $5^2 \cdot 3 + 5^1 \cdot 2 + 5^1 \cdot 1 = 86$ in Base10 could be identified with the tuple: $$ \begin{pmatrix} 1 & 2 & 3 & 0 & 0 & \ldots \end{pmatrix} $$

In fact one can also identify the tuples of our vector space with numbers in Base5, so for example the tuple $$ \begin{pmatrix} 0 & 1 & 1 & 4 & 0 & 0 & 1 & 0 & \ldots \end{pmatrix} $$ from above could be identified with the number $1004110$ in Base5.

As far as I can tell this identification is unique in the sense that every Base5 number can be identified with exactly one tuple and the other way around. (Of course one would have to take the base5 numbers without 'leading zeros' (except for the number 0).

One may now look at other properties of this sort of 'map' like looking at compatibility with addition, multiplication and so on.

I should add that for the above vector space the addition and multiplication would be defined in the 'canonical' way, meaning that the addition is entrywise and the scalar multiplication would entrywise linearly aswell. For example $( 0 \, 1 \, 0 \, \ldots) + (1 \, 0 \, 0 \, \ldots) = (1 \, 1 \, 0 \ldots)$.

At last back to the base question. In this case our base for the above defined vector space would be infinite (it's an interesting question to figure out what possible bases one could choose). And the for the canonical base $$ (1 \, 0 \, 0 \, 0 \, \ldots ) \qquad (0 \, 1 \, 0 \, 0 \, \ldots) \qquad (0 \, 0 \, 1 \, 0 \, \ldots) \qquad \ldots $$ The above map would identify the n-th entry in the tuple with the n-th number (reading from the right) of the representation of the number in base5 notation. The above map (as pointed out in other answer) however does not work well with the usual addition as we know it since for example: $$ (0 \, 4 \, 0 \, \ldots ) + (0 \, 1 \, 0 \, \ldots) = (0 \, 0 \, 0 \, \ldots) $$ Where in Base5 one would get $40+10=100$ (in base10 $20+5=25$)

Note that this can be applied analogously to every other base$n$ where $n$ is prime.

Although I was not able to give you a conclusive answer this might have given you another perspective at the question you have asked.

(Edit: This is similar to the answer given by Lee Mosher in the sense that the entries in the tuple here represent the scalar used for each power of the 'base number')