# Thread: Working in base 12...

1. ## Working in base 12...

To work in base 12, we must invent new symbols for digits 10 and 11. Use t for ten and e for eleven. For example, 2x5=t 12
a Write out 3, 5 and 8 times table in base 12, up to e times.
b In the times tables in a, some have answers with the last digit 0. Which tables in base 10 [up to 9 times] have answers with the last digit 0?
c Explain why some tables have answers with last digit 0 and others do not, in base 10 [up to 9 times] and in base 10 [up to e times]
d Which tables in base 12 [up to e times] will have no answers with last digit 0?

2. Hello, Bartimaeus!

I assume you are somewhat familiar with base-12 notation.

To work in base-12, we must invent new symbols for digits 10 and 11.
Use $\displaystyle t$ for ten and $\displaystyle e$ for eleven. For example, $\displaystyle 2 \times 5 = t_{12}$

(a) Write out 3-, 5- and 8-times table in base-12, up to $\displaystyle e$-times.
Code:
  x |  1 |  2 |  3 |  4 |  5 |  6 |  7 |  8 |  9 |  t |  e |
-- + -- + -- + -- + -- + -- + -- + -- + -- + -- + -- + -- +
3 |  3 |  6 |  9 | 10 | 13 | 16 | 19 | 20 | 23 | 26 | 29 |
-- + -- + -- + -- + -- + -- + -- + -- + -- + -- + -- + -- +
5 |  5 |  t | 13 | 18 | 21 | 26 | 2e | 34 | 39 | 42 | 47 |
-- + -- + -- + -- + -- + -- + -- + -- + -- + -- + -- + -- +
8 |  8 | 14 | 20 | 28 | 34 | 40 | 48 | 54 | 60 | 68 | 74 |
-- + -- + -- + -- + -- + -- + -- + -- + -- + -- + -- + -- +
(b) In the times tables in (a), some have answers with the last digit 0.
Which tables in base-10 [up to 9-times] have answers with the last digit 0?

In base-10, the rows that have answers ending in 0 are:
. . 2-times, 4-times, 6-times, 8-times, and 5-times.

(c) Explain why some tables have answers with last digit 0 and others do not
in base-10 [up to 9-times] and in base-12 [up to e times]

For base-ten: $\displaystyle 10 = 2\cdot5$
Any row which shares a common factor with 10 has answers ending in 0.
Any row which is relative prime to 10 has no answers ending in 0.

For base=twelve: $\displaystyle 12 = 2^2\cdot3$
Any row which shares a common factor with 12 has answers ending is 0.
. . They are: 2-times, 3-times, 4-times, 6-times, 8-times, 9-times, and $\displaystyle t$-times

(d) Which tables in base-12 [up to $\displaystyle e$-times] will have no answers with last digit 0?

Any row which is relatively prime to 12 has no answers ending in 0.
. . They are: 5-times, 7-times, and $\displaystyle e$-times.

3. Thankyou Soroban. You are a Super member.
I am a little familiar with writing in bases, but never in any higher than base ten. Is there, I suppose you could call it, formula, table or a trick in working in converting on base into another?

4. Hello again, Bartimaeus!

There is a procedure for changing a base-ten number to another base.

Example: write $\displaystyle 389$ in base-5.

Step 1: Divide $\displaystyle 389$ by $\displaystyle 5$ and note the remainder:
. . . . . . $\displaystyle 389 \div 5\:=\:77\quad rem.\boxed{4}$

Step 2: Divide the quotient by 5 and note the remainder:
. . . . . . .$\displaystyle 77 \div 5 \,=\,15\quad rem.\boxed{2}$

Repeat Step 2 until a zero quotient each reached.
. . . . . . . $\displaystyle 15 \div 5 \,=\,3\quad rem.\boxed{0}$
. . . . . . . . $\displaystyle 3 \div 5\,=\,0\quad rem.\boxed{3}\;\;\uparrow$

Now read up the remainders: .$\displaystyle 3024_5$

The division can be written like this:
Code:
    5 ) 3 8 9
-------
5 ) 7 7   4
-----
5 ) 1 5   2
-----
5 ) 3   0
---
0   3 ↑

5. Now, suppose you wanted to change this base 5 to base 10.

$\displaystyle 3024_{5}$

Starting from the far right multiply each digit by successive powers of 5(starting with 0):

$\displaystyle 4*5^{0}=4$
$\displaystyle 2*5^{1}=10$
$\displaystyle 0*5^{2}=0$
$\displaystyle 3*5^{3}=375$

6. Nice explanation, Cody!

But has anyone ever seen this method?
[The explanation/derivation is long, but the method is very fast.]

Example: .Evaluate $\displaystyle f(x) = 2x^3 - 5x^2 + 3x - 1$ at $\displaystyle x = 3$

We have: .$\displaystyle f(x)\:=\:2x^3 - 5x^2 + 2x - 1$

Factor $\displaystyle x$ from the first three terms: .$\displaystyle f(x)\:=\2x^2 - 5x + 3)\cdot x - 1$

Factor $\displaystyle x$ from the first two terms: .$\displaystyle f(x)\:=\[2x-5]\cdot x + 3)\cdot x - 1$

Then: .$\displaystyle f(3)\:=\[2\cdot3 - 5]\cdot3 + 3)\cdot3 - 1$

Now "read" the steps we will take: .$\displaystyle \begin{array}{cccccc}\text{2 times 3}\\ \text{minus 5} \\ \text{times 3} \\ \text{plus 3} \\ \text{times 3} \\ \text{minus 1}\end{array}$ . $\displaystyle \begin{array}{cccccc}6\\ 1\\3\\6\\18\\\boxed{17}\end{array}$

We could have done this mentally ... no squaring or cubing involved.

But if all that factoring is required, it's not much of a shortcut, is it?

Look again at the factored polynomial: .$\displaystyle f(x)\:=\:\left([2x - 5]\[\cdot x + 3\right)\cdot x - 1$

"Drop" the grouping symbols: .$\displaystyle 2x - 5\cdot x + 3\cdot x - 1 \;= \;2x - 5x + 3x - 1$

This is the original polynomial with the exponents removed . . .

If we are given: .$\displaystyle g(x)\;=\;3x^4 - 7x^3 + 4x^2 - 9x + 8$
. . we can (mentally) drop the exponents: .$\displaystyle 3\cdot x - 7\cdot x + 4\cdot x - 9\cdot x + 8$
. . and we have the factored form (minus the grouping symbols).

For $\displaystyle g(2)$: 3 times 2, minus 7, times 2, plus 4, times 2, minus 9, times 2, plus 8

Important: We must complete each addition/subtraction before the next multiplication.

On our calculator, it would be:
. . $\displaystyle 3\;\boxed{\times}\;2\;\boxed{-}\;7\;\boxed{=}\;\;\boxed{\times}\;2\;\boxed{+}\;4 \;\boxed{=}\;\;\boxed{\times}\;2\;\boxed{-}\;9\;\boxed{=}\;\;\boxed{\times}\;2\;\boxed{+}\;8 \;\boxed{=}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Back to base-five . . .

Example: Convert $\displaystyle 3024_5$ to base-ten.

The value is: .$\displaystyle 3\cdot5^3 + 0\cdot5^2 + 2\cdot5 + 4$ .and we have a "polynomial in 5".

So we have: .$\displaystyle 3\cdot5 + 0\cdot 5 + 2\cdot 5 + 4\quad\Rightarrow\quad\begin{array}{cccccc}\text{3 times 5} \\ \text{plus 0}\\ \text{times 5}\\ \text{plus 2}\\ \text{times 5}\\ \text{plus 4}\end{array}$ $\displaystyle \begin{array}{cccccc}15\\15\\75\\77\\385\\\boxed{3 89}\end{array}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

This procedure is even more spectacular in base-two.

Example: Write $\displaystyle 101011_2$ in base-ten.

We have: .$\displaystyle 1\times2 + 0\times2 + 1\times 2 +0\times2 + 1\times2 + 1$
. . . . . . . . . . .$\displaystyle 2\quad\,2\quad\;4\quad5\;\;\;10\;\;\;10\;\;\;20\; \;\;21\;\;42\;\;\:\boxed{43}$

You have the answer while others are discovering that it begins with $\displaystyle 2^5.$

7. Originally Posted by Soroban
But has anyone ever seen this method?
It is Horner's Rule.

And reading that I see "The Horner scheme is often used to convert between different positional numeral systems — in which case x is the base of the number system, and the ai coefficients are the digits of the base-x representation of a given number." And I didn't know that connection. So thanks, Soroban!

8. So, I have a question...
Is it possible to write a given number x in a given base y where y is not a integer i.e. write 1220 is base e. The problem I'm having is that to use the traditional method for converting bases with remainders, your remainder is a decimal. Also the practicality seems to run down when you realize that you have e (2.71828182846...) numbers in your system.
Dan

9. Originally Posted by dan
So, I have a question...
Is it possible to write a given number x in a given base y where y is not a integer i.e. write 1220 is base e. The problem I'm having is that to use the traditional method for converting bases with remainders, your remainder is a decimal. Also the practicality seems to run down when you realize that you have e (2.71828182846...) numbers in your system.
Dan
No, for example you cannot express,
$\displaystyle \pi = a_0+a_1e+a_2e^2+...+a_ne^n$
Because, they are algebraically independent.
Meaning there is not polynomial with rational coefficients that when evaluavated at $\displaystyle e$ gives $\displaystyle \pi$.

10. Originally Posted by dan
So, I have a question...
Is it possible to write a given number x in a given base y where y is not a integer i.e. write 1220 is base e. The problem I'm having is that to use the traditional method for converting bases with remainders, your remainder is a decimal. Also the practicality seems to run down when you realize that you have e (2.71828182846...) numbers in your system.
Dan
One system: Golden Ratio Base.

I think this is dealt with in general in volume 2 of Don Knuth's TAOCP "SemiNumerical Algorithms", but my copy is at work so I can't check it now.

RonL

11. Originally Posted by ThePerfectHacker
No, for example you cannot express,
$\displaystyle \pi = a_0+a_1e+a_2e^2+...+a_ne^n$
Because, they are algebraically independent.
Meaning there is not polynomial with rational coefficients that when evaluavated at $\displaystyle e$ gives $\displaystyle \pi$.
I believe those are called 'transcendental numbers'. Pi and e are classic examples.

12. Originally Posted by ThePerfectHacker
No, for example you cannot express,
$\displaystyle \pi = a_0+a_1e+a_2e^2+...+a_ne^n$
Because, they are algebraically independent.
Meaning there is not polynomial with rational coefficients that when evaluavated at $\displaystyle e$ gives $\displaystyle \pi$.
Have I missed something? Why would you expect the base $\displaystyle e$ representation of $\displaystyle \pi$ to terminate?

RonL

13. Originally Posted by CaptainBlack
Have I missed something? Why would you expect the base $\displaystyle e$ representation of $\displaystyle \pi$ to terminate?

RonL
No I said you cannot express it.

14. Originally Posted by ThePerfectHacker
No I said you cannot express it.
To express $\displaystyle \pi$ in base $\displaystyle e$ we would like a representation in the form:

$\displaystyle \pi = \sum_{r=n_0}^{\infty} a_r\ e^{-r}$

where $\displaystyle n_0$, may be negative.

In fact does the greedy algorithm not give a construction with all the $\displaystyle a_i$s are either $\displaystyle 0,\ 1$ or $\displaystyle 2$?

RonL

15. Originally Posted by CaptainBlack
To express $\displaystyle \pi$ in base $\displaystyle e$ we would like a representation in the form:

$\displaystyle \pi = \sum_{r=n_0}^{\infty} a_r\ e^{-r}$

where $\displaystyle n_0$, may be negative.

In fact does the greedy algorithm not give a construction with all the $\displaystyle a_i$s are either $\displaystyle 0,\ 1$ or $\displaystyle 2$?

RonL
It appears that in base $\displaystyle e$ one representation of $\displaystyle \pi$ is:

$\displaystyle \pi \approx 10.101002020..._{\mbox{base e}}$

Is this unique? I think so, but haven't checked.

RonL

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