Post by Anders D. Nygaard Post by J. J. Lodder
Brouwer's questions are what they are
precisely because Turing machines can't answer them,
What are Brouwer's questions? I can easily find Hilbert's
(they are usually known as "problems", though), but not his.
A Brouwerian question is a question
that can in principle be settled by computation,
but not in practice.
It is easy to invent infinitely many of them,
so if one happens to get settled, just take another one.
Obsolete and no longer needed,
for the halting problem for Turing machines
does just the same for you in a more general way.
In this context Brouwerian questions can be seen
as illustrative examples for the undecidability of the halting problem.
Brouwer used them to construct other counter-examples,
such as a number that is neither smaller than zero, nor equal to zero,
nor greater than zero.
PS Brouwer's original question was:
Are there seven consecutive zeroes in \pi?
This prompted Hilbert's reply: "God knows!"
This question seemed preposterously difficult to settle in 1925,
but nowadays we know that the answer is yes.
(which is not surprising, assuming normality of \pi)
A later question, also due to Brouwer is:
We define a number B constructively as follows:
The first decimal will be zero,
unless the first decimal of \pi equals 7, otherwise it will be 1
The second decimal will be zero, unless the second and third decimals of
\pi both equal 7, else 1
And so on, with the nth decimal equal to zero,
unless decimals n to 2n-1 of \pi are all equal to 7.
We know nowadays that B begins with 10 trillion zeroes,
but is B equal to zero?
In words: is there an n,
such that there are n consecutive zeros in \pi,
starting at decimal position n.
Best guess is that we will never know.
Not even Marvin can help us out.