What is $10$ in base $11$ and what is $11$ in base $11$?

Question

I am not understanding how to write numbers in bases. For example, $10=10\times11^0$ so it should be $10$ in base $11$. But also $11=0\times11^0+1\times11^1$ so $11$ is $10$ in base $11$. So both $10$ and $11$ are $10$ in base $11$?

Answer 1

In base 10, we have 10 symbols for digits. In base 2, we have 2. Likewise, in base 11, we must have 11 - but 0, 1, 2, ..., 9 only makes ten. So we have to introduce a new symbol (we generally use A) to be that eleventh digit. "A" represents a value of 10 in a single digit.

So in base 11, 10 will be denoted A (because it is A 1's and zero 11's). 11 will be denoted 10 (because it is one 11 and zero 1's). Similarly, 21 would be 1A (one 11 and A 1's).

Answer 2

For an integer base $b \ge 2$ and nonnegative integer $n$, we can express $n$ in base $b$ using the notation $$n = (d_m, d_{m-1}, \ldots, d_1, d_0)_b = \sum_{k=0}^m d_k b^k,$$ where $$d_k \in \{0, 1, \ldots, b-1\}$$ for all $k$; moreover, such a representation is unique. That is to say, writing $n$ in base $b$ amounts to finding a sequence $\{d_k\}_{k \ge 0}$ of nonnegative integers strictly less than $b$ for which the above sum equals $n$.

For example, let us choose $n = 32597$, $b = 23$. Then the largest power of $23$ that is less than or equal to $n$ is $23^3 = 12167$, and the the quotient of $n$ divided by $23^3$ is $2$. This tells us $d_3 = 2$, so $$n = 2 \cdot 23^3 + d_2 \cdot 23^2 + d_1 \cdot 23^1 + d_0 \cdot 23^0.$$ To solve for the subsequent terms, we simply subtract and proceed with the remainder: $$n - 2 \cdot 23^3 = 8263,$$ and the largest power of $23$ that is less than $8263$ is $23^2 = 529$, and $\lfloor 8263 / 529 \rfloor = 15$, so $d_2 = 15$. Repeating the process until the end, we obtain $$32597 = 2 \cdot 23^3 + 15 \cdot 23^2 + 14 \cdot 23^1 + 6 \cdot 23^0,$$ and we can write this more compactly as $$n = (2, 15, 14, 6)_{23}.$$

This was a fairly detailed example. At this point, it's worth noting that, for convenience, we can think of such representations as comprising an infinite sequence of digits $d_k$, and that if $n$ is finite, then there exists some $m$ for which $d_k = 0$ for all $k > m$; so for example, $$(2,15,14,6)_{23} = (\ldots, 0, 2, 15, 14, 6)_{23}$$ or if we wrote the digits in "reverse"--that is, in order of increasing significance--we might say $$\{d_k\}_{k\ge 0} = (6, 14, 15, 2, 0, 0, 0, \ldots).$$ This affords us the convenience of not having to specify $m$, and that in some sense, the base-$b$ representation of any nonnegative integer is an infinite sequence of nonnegative integers less than $b$.

That said, how would we answer the original question? Well, in base $b = 11$, the number $n = 10$ is less than $b$, so we simply write $$10 = (10)_{23} = 10 \cdot 11^0.$$ Thus, the sequence corresponding to such $n$ and $b$ is $$\{d_k\}_{k\ge 0} = (10,0,0,0,\ldots).$$ Note we do not need to resort to using letters or other symbols.

What is $11$ in base $11$? Well, this should be clear now: $$11 = 1 \cdot 11^1 + 0 \cdot 11^0 = (1,0)_{11} = (0,1,0,0,0,\ldots).$$

Let's try some more. What is $n = 57$ in base $b = 3$? We have $$57 = 2 \cdot 3^3 + 0 \cdot 3^2 + 1 \cdot 3^1 + 0 \cdot 3^0 = (2,0,1,0)_3 = (0,1,0,2,0,0,0,\ldots).$$ How about $300$ in base $100$? This is just $$300 = 3 \cdot 100^1 + 0 \cdot 100^0 = (3,0)_{100} = (0,3,0,0,0,\ldots).$$ By now, you should find the process familiar. Try more examples; just pick an integer $b \ge 2$ and any nonnegative $n$, and work it out yourself.

Answer 3

We can say that $11 = 10\times 1 + 1$ and $10 = 10\times 1 + 0$ without knowing what the base is. The base could be 9 + 4. The question is, what is $10 - 1$?