Do Transcendental Numbers Guarantee Certain Sequences?

Oct 2015
128
0
Antarctica
One of my friends made that argument that, just because a number repeats with no pattern (with certain transcendental numbers being a good example of this) then any given series of digits is guaranteed to appear somewhere within the number. For example:

"Since Pi goes on forever and has no pattern, then somewhere in that string of digits lies the binary code for the Microsoft Word program."

I find it difficult to either accept or reject this statement. Obviously it would take an unreasonable amount of digits for us to actually find something useful like that, but is that assumption even true in the first place? Just because a number is transcendental and with no pattern, does that guarantee that any given string of numbers will appear somewhere in the decimal approximation, regardless of how many digits it takes to find it? Or is this just a naive assumption that isn't always true? Furthermore, how can we know this with certainty?
 
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v8archie

Math Team
Dec 2013
7,713
2,682
Colombia
The number
0.090090009000090000090000009...​
is transcendental (I think) but clearly omits most finite strings of digits.

In fact we can prove that there are transcendental numbers that contain only the numerals 0 and 9 in their decimal expansion. There are uncountably many binary sequences - by Cantor's diagonal argument. Replace every 1 with a 9 and precede the sequence by "0." to create a decimal between zero and 1. Since the algebraic numbers are countable, not all of the numbers we created can be algebraic.
 
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Oct 2015
128
0
Antarctica
I'm sorry for such a late reply. Thanks for the explanation!