When we first encounter irrational numbers, we usually define them as decimal numbers with infinite number of decimals that do not repeat in a periodic fashion. The canonical example one sees is the number:(adsbygoogle = window.adsbygoogle || []).push({});

[tex]

\beta = 0.1010010001\ldots

[/tex]

i.e. there is one more zero between each consecutive pair of ones.

Another way of representing this number is by a series. Let us find the positions of the ones in this representation. Let us denote by [itex]n_{k}[/itex] the decimal position of thek-th one. Obviously, [itex]n_{1} = 1[/itex].

The number of zeros between the second and first one is 1.

The number of zeros between the third and second one is 2.

...

The number of zeros between the (k+ 1)-st and thek-th one isk.

However, the number of decimal places between a figure in then-th and a figure in them-th place is equal to [itex]n - m - 1[/itex] (not counting the end decimals themselves). So, we can write the recursion.

[tex]

n_{k+1} - n_{k} - 1 = k, \ k \ge 1

[/tex]

[tex]

n_{k+1} - n_{k} = k + 1, \ k \ge 1

[/tex]

[tex]

n_{k} - n_{k - 1} = k, \ k \ge 2

[/tex]

which, together with the initial condition has the solution:

[tex]

n_{k} = \sum_{j=1}^{k} {j} = \frac{k(k + 1)}{2}.

[/tex]

As a sanity check, we confirm that this formula gives 1, 3, 6, 10 as the decimal places where the first, second, third and fourth one occur, respectively. The value of a one in then-th decimal place is:

[tex]

\frac{1}{10^{n}}

[/tex]

Combining these results, we see that we can write [itex]\beta[/itex] as:

[tex]

\beta = \sum_{k = 1}^{\infty} {\frac{1}{10^{\frac{k (k+1)}{2}}}}

[/tex]

The question is: Is this number algebraic or transcendental?

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# Is this number algebraic or transcendental?

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