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• Algebraic and transcendental formal numbers

We will now give some examples of algebraic formal numbers. Let us start with the particular case of a rational number. We wish to compute the inverse of 1011 in F(2). First we observe that 1011 * 10111 = 10000001 . On the other hand we also have

11 * 0.111111....[1].... = 1.111111....[1].... + 0.111111....[1].... = 1

By changing T into T7 in the above formula, we obtain

1 = 10000001*0.000000100000010000001....[0000001]....

We recall that the blocks between square brackets are for ever repeated. Consequently we can write

1011-1 = 10111 * 10000001-1 = 10111 * 0.000000100000010000001....[0000001]....

which leads to

1011-1 = 0.001011100101110010111....[0010111]....

This expansion is periodic in agreement with the property stated above since 1011-1 is a rational number. Notice that, in analogy with the case of real numbers, the periodicity results from the fact that every formal integer over a finite field divides Tl ( Tm -1) for some positive integers l and m.

Let us now look at some algebraic irrational numbers. In F(2), the equation X2 = 1.1 has no solution, in other words the square root of 1.1 does not exist in F(2). Nevertheless the square root of 1.1 does exist in F(3) and an approximation is 1.21121.... .
Again in F(2), the equation X3 = 1.1 has a unique solution, in other words the cubic root of 1.1 does exist and an approximation is 1.11100001.... .

At last let us consider the following equation

X2 + 10 * X + 1 = 0

If we work in F(2), this can be written

X = 0.1 + 0.1 * X2

which leads to

X = 0.1 + 0.1 * (0.1 + 0.1 * X2)2 =0.1 + 0.1 * (0.01 + 0.01 * X4) = 0.101 + 0.001 * X4

Repeating the same arguments, we would obtain

X = 0.1010001 + 0.0000001 * X8

By iteration we see that the above equation has a unique root X2 in F(2) and we have

X2 = 0.101000100000001000.....

where the sequence of the coefficients of X2 is such that a-i is 1 if i is equal to 2k - 1 and 0 otherwise. It could be shown that the same equation has a unique solution in F(p) for all prime numbers p . We give an approximation for the solutions X3 and X5 in F(3) and F(5) respectively.

X3 = 0.102020102000000...  and   X5 = 0.1040200040302010......

The sequence of the coefficients for X2 is very regular. Even though the sequences of the coefficients for X3 and for X5 do not appear to be as regular, it can be shown that those sequences are in a certain sense not completely at random. Indeed mathematicians have defined a special class of sequences taking only a finite number of values. These sequences, called automatic sequences, are in a way a generalization of ultimately periodic sequences. A sequence is automatic if and only if it is the sequence of the coefficients of an algebraic formal number over a finite field. Consequently it is possible to decide whether a given formal number is algebraic or transcendental. Let us consider in F(p) the following number

X = 0.0110101000101000101000100000101....

where the sequence of the coefficients of X is such that a-i is 1 if i is a prime number and 0 otherwise. Then it can be proved that this number X is transcendental in F(p) for all prime numbers p. It is highly probable that the real number represented by the same expansion, considered as its decimal expansion, is also transcendental but the proof of that is yet out of reach.

 

 

 

EPILOGUE

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Bordeaux 09-02-2002
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