Moving between bases

Question

Is there any kind of method to move between bases?

I know that there are methods to convert between base x to base y, but I'm looking for something general which won't limit me. Is it possible? Maybe there is some kind of equation.

What is limiting you in the usual way to go from one basis to another? — Tobias Kildetoft, Feb 25 '13 at 19:18
You said you know there are methods. I assumed that meant you actually knew some of those methods. — Tobias Kildetoft, Feb 25 '13 at 19:20
Yes there are some methods to move for example from binary base to hex-decimal. — Imri Persiado, Feb 25 '13 at 19:21
(number).toString(base) for bases up to $36$, or you can adapt this JavaScript code to work with bases above $36$. — , Feb 25 '13 at 20:02
We just answered this today at http://math.stackexchange.com/questions/313954/number-base-conversion/313962#313962 — MJD, Feb 25 '13 at 20:51

Micah · Accepted Answer · 2013-02-25T20:10:01.940

There are at least two different algorithms you might want to use for converting a number $n$ from base $x$ to base $y$, depending on whether you'd rather do your arithmetic in base $x$ or base $y$.

If you're more comfortable doing arithmetic in your starting base:

Write $n=ay+b$ where $0 \leq b < y$ (that is, do "division with remainder" on $n$). Then $b$ will be the rightmost digit of your new representation of $n$, and you repeat the process on $a$ to get all the other digits. Stop when $a$ is less than $y$ (at which point $a$ will be the leftmost digit of your number).

For example, say we want to write the base-ten number $23504$ in base $7$. Then:

$23504$ divided by $7$ is $3357$ with a remainder of $5$, so the rightmost digit is $5$.
$3357$ divided by $7$ is $479$ with a remainder of $4$, so the next-to-rightmost digit is $4$.
$479$ divided by $7$ is $68$ with a remainder of $3$, so the next digit is a $3$.
$68$ divided by $7$ is $9$ with a remainder of $5$, so the next digit is a $5$.
$9$ divided by $7$ is $1$ with a remainder of $2$, so the two leftmost digits are $12$.

Putting all this together, we have $23504_{\mathrm{ten}}=125345_7$.

If you're more comfortable doing arithmetic in your ending base:

Start at the left end of your number and keep a running total. As long as you still have more digits, multiply your running total by $x$ and add the next digit to it. When you run out of digits, your running total is the converted number.

For example, say we want to write the base-$7$ number $125345_7$ in base ten. Then:

We haven't added anything yet, so our running total is $0$. Multiply by $7$ and add the leftmost digit $(1)$ to get $1$.
Our running total is $1$. Multiply by $7$ and add the next digit $(2)$ to get $9$.
Our running total is $9$. Multiply by $7$ and add the next digit $(5)$ to get $68$.
Our running total is $68$. Multiply by $7$ and add the next digit $(3)$ to get $479$.
Our running total is $479$. Multiply by $7$ and add the next digit $(4)$ to get $3357$.
Our running total is $3357$. Multiply by $7$ and add the next digit $(5)$ to get $23504$. Since we're now all out of digits, this is the base-$10$ representation of the number.

So again this demonstrates that $125345_7=23504_{\mathrm{ten}}$ (but starting from the other end).

Finally, maybe you're not comfortable doing arithmetic in either base. If there's some base $z$ you are comfortable doing arithmetic in, that's OK! Just use the second algorithm to convert from base $x$ to base $z$, and the first one to convert from base $z$ to base $y$.

score 1 · Answer 2 · answered Feb 25 '13 at 19:29

To change bases you use the division algorithm: for an integer $n$ (in any base) and new base $y$, there exists a quotient $q$ and remainder $r$ such that $$n = qy + r, \qquad 0\leq r < y.$$ You can compute $q$ and $r$ by performing long division of $n$ by $y$ in base $x$.

The remainder $r$ then becomes the least significant digit (ones digit) of the number $n$ in base $y$. Repeat the process, using $q$ for $n$, to find the next digits.

Moving between bases

2 Answers2