Algebraic Proofs for the Trachtenberg System

We will have a look at the algebraic proofs for the Trachtenberg System and see why the methods Jakow Trachtenberg used in his system work. In the book the algebraic proofs for only a few parts of the system were shown but here we will try to include as much of the system as we can.

One part of the system we won't be showing a proof for is the fast method for long division because the method is not exact and you have to make adjustments to your results as you go.

We won't be going into much detail on explaining the proofs as if this part interests you then you will most likely understand what is going on. If not you can always contact me or leave a comment below.

Expanded Notation

The first thing we will do is to separate the numbers out into individual digits using expanded notation.
For example taking the number $4637$ , we have 4 thousands, six hundreds, 3 tens and 7 units (ones) making up this number so we can write it out like so:

$4,637 = 4 \times 1,000 + 6 \times 100 + 3 \times 10 + 7 \times 1$

This is the case for any four digit number and so we can represent that by generalizing the number and replacing each digit with a letter.

$abcd = a \times 1000 + b \times 100 + c \times 10 + d \times 1$

The $a$ , $b$ , $c$ and $d$ each can represent any digit from $0$ to $9$ .
In the case of the number 4637 then:

$\begin{equation*} \begin{split} a &= 4\\ b &= 6\\ c &= 3\\ d &= 7\\ \end{split} \end{equation*}$

We could replace $abcd$ with any four digit number and the formula would be correct.

Extra Zeros

We can add a zero in front of a number and it has no affect on the number, the value stays the same.

$4637 = 04637$

or more generally, using our letters:

$0abcd = 0 \times 10,000 + a \times 1,000 + b \times 100 + c \times 10 + d \times 1$

We can also include adding zero times any unit to our number without changing its value. Because any value multiplied by zero is always zero.

$0abcd = 0 \times 10,000 + a \times 1,000 + b \times 100 + c \times 10 + d \times 1 + 0 \times 1$

Adding the $0 \times 1$ on the end is the same as adding zero to the number, it has no effect. However, this will be used later in the proofs. This zero on the end does not change the value but will serve as a placeholder in later formulas.

In the following algebraic formulas we will use $N$ to represent our general number:

$\begin{equation*} \begin{split} N &= abcd\\ N &= 0abcd \end{split} \end{equation*}$

For consistency for the remainder of the algebra we will use the dot notation for multiplication so instead of:

$abcd = 0 \times 10,000 + a \times 1,000 + b \times 100 + c \times 10 + d \times 1$

We will use:

$abcd = 0\cdot10,000 + a\cdot1,000 + b\cdot100 + c\cdot10 + d\cdot1$

We will start with the easier proofs then work our way to the more complicated ones.

Numbers of any length

Taking the expanded form of our four digit number we can write it in power notation.

$a\cdot1,000 + b\cdot100 + c\cdot10 + d\cdot1 = a\cdot 10^3 + b\cdot 10^2 + c\cdot 10^1 + d\cdot 10^0$

Each letter in our number is multiplied by a power of 10 and we can represent all of them with $10^n$ . We can now replace the letters in our number with new letters like this:

$\begin{equation*} \begin{split} a &= a_3\\ b &= a_2\\ c &= a_1\\ d &= a_0 \end{split} \end{equation*}$

Where each subscript in our new letters matches the power the letter was multiplied by, this means we can write.

$a\cdot 10^3 + b\cdot 10^2 + c\cdot 10^1 + d\cdot 10^0$

$a_3\cdot 10^3 + a_2\cdot 10^2 + a_1\cdot 10^1 + a_0\cdot 10^0 = \sum_{n=0}^{n=3}a_{n} \cdot 10^{n}$

Now instead of just having a four digit number we can replace this with a $k$ digit number, where $k$ may be any number. We can now represent our general number $N$ in this format.