It is actually an open problem whether every finite sequence of digits occurs in the decimal expansion of pi (and likewise open for expansions in any other base).
But it is most assuredly not the case that all irrational numbers have every sequence of digits, for instance the number with decimal expansion
.54544544454444544444....
there being n 4’s between the n-th 5 and the (n+1)-st 5 is irrational, but only contains a very restricted set of finite digit sequences — in particular having only 4’s and 5’s as digits, but even more restricted than that: any sequence with a single 5 and any number of 4’s before and after will occur, but once two 5’s occur (necessarily non-adjacent) the first can be preceded by at most one fewer 4’s than occur between the left-most 5 and the next 5, the number of 4’s between successive 5’s must increase by one and the number of 4’s after the right-most 5 must be at most one more than the number of 4’s between the two right-most 5’s.
Actually, like my example, pi is a very tame irrational number: though less obviously so than for my example, its decimal expansion can, in principle, be generated by running an algorithm. Like the rational numbers, the set of real numbers, both rational and irrational, whose decimal (or binary or base b) expansion can be generated by an algorithm is countable, so there are uncountably many “non-algorithmic” irrationals (my favorite being the “oracular number for FORTRAN”, defined by listing all valid FORTRAN programs ordered by total number of symbols and within each number in lexicographic order by the standard ordering on ASCII symbols, the i-th digit right of the binary point is 1 if the i-th program would halt in finite time when run on an machine with infinite memory, and 0 otherwise). That, and its friends, other binary, decimal and base b oracular numbers for various and sundry ways of describing algorithms, are some of the very few non-algorithmic irrationals one can describe at all.
What is this? Gizmo Republic ????
“But it is most assuredly not the case that all irrational numbers have every sequence of digits, for instance the number with decimal expansion”
Yes, I stand corrected, you are right. However, since there are an infinite number of irrational numbers between any two whole numbers, or between any two numbers of any kind, for that matter, then it’s obvious that there are still an infinite amount of irrational numbers that DO contain any possible sequence of numbers. By their very nature, it wouldn’t be anything extraordinary at all to find such a thing to be true about any number you random picked.
Whether you can mathematically prove that or not is another matter, but I think it’s a foregone conclusion logically, which is why mathematicians bother to try to prove it in the first place. It’s intuitively true, so someone must find the proof someday, or so it would seem.