This is a fascinating and deep question that touches on the relationship between two different representations of real numbers: their decimal expansions and their continued fraction expansions. The relationship between these is notoriously complex.

The short answer to your question is yes, such an irrational number exists.

In fact, there are uncountably many of them. However, constructing one explicitly is extremely difficult, and the proof of existence relies on advanced methods from number theory and fractal geometry.

Deeper Explanation

Let's break down the properties you're asking for.

1. Decimal Expansion with Digits {0, 1}: Let be the set of real numbers in whose decimal expansion consists only of the digits 0 and 1. This is a classic example of a Cantor set. Its Hausdorff dimension is . This means it's a "small" set in the sense that its Lebesgue measure is 0, but it's still uncountably infinite.

2. Bounded Continued Fraction Entries: Let be the set of irrational numbers in whose continued fraction partial quotients are all bounded by (i.e., for all ). Numbers in this set are called badly approximable numbers. They are "as irrational as possible" in the sense that they cannot be approximated "too well" by rational numbers. The set is also an uncountable set with Lebesgue measure 0.

Your question is: Is the intersection non-empty? (You asked for entries less than 10, so they are bounded by ).

The Intuition and a First Step

Let's try to see what the constraints imply. Let be such a number. The continued fraction algorithm is: (if is not an integer)

Since is formed by digits 0 and 1, . This can't be right. is rational. . Ah, no, the largest number is if you only allow digits after the decimal point. But if you mean any number, the set is not bounded. The question refers to 0. ..., so we are in . The largest number in is . The smallest non-zero number is... well, there is no smallest. But any number starts with .

Let's assume , so its continued fraction starts with . Then . The first partial quotient is . We require . This means , which implies , so .

This gives a crucial first condition: The decimal expansion of the number must begin with 0.1.... It cannot be 0.1 exactly, because that is a rational number () and its continued fraction is , which has a quotient of 10, which is not less than 10. So must be strictly greater than . This means its decimal expansion must be of the form 0.1... followed by at least one more non-zero digit (a '1').

For example, let's take your number, . . So . This is allowed. The next step is . So . The next step is . So . And so on. This matches the start of the continued fraction you provided.

The problem is that as you define more digits of , you might be forced to create a very large partial quotient at some point. A generic (random) sequence of 0s and 1s would, with probability 1, produce unbounded partial quotients. So a special, non-random sequence is needed.

The Decisive Theorem

This problem is a specific case of a more general question that has been resolved. A key result in this area is from a paper by Burger et al. (2010), although related results exist from other authors as well.

Theorem (paraphrased from Burger et al.): Let be an integer base, and let be a set of allowed digits with . Let be the set of real numbers whose base- expansions use only digits from . Then for any integer , the set of numbers in that are badly approximable with bound is non-empty and has positive Hausdorff dimension.

Let's apply this to your problem:

• The base is .
• The set of allowed digits is . We have , so the theorem applies.
• You want the continued fraction entries to be less than 10, which means they are bounded by .
• The condition in the theorem is . Here, this means , which is true.

Therefore, the theorem directly implies that the set of numbers with only digits 0 and 1 and continued fraction entries bounded by 9 is not only non-empty but is an uncountable set with positive Hausdorff dimension.

Why is this true? A Heuristic Argument

The proof of the theorem is quite technical, but a heuristic argument based on Hausdorff dimension gives a strong indication of why it should be true.

• As mentioned, .
• The Hausdorff dimension of the set of badly approximable numbers is known to be . A more precise result by Hensley gives . For , this is .

There is a principle in fractal geometry that if you have two sets and on the real line, and , then their intersection is "likely" to be non-empty (under certain conditions on how the sets are constructed).

In our case, .

This sum being greater than 1 strongly suggests that the two sets are not "thin" enough to be able to avoid each other, and their intersection should be non-empty. The formal proof essentially makes this intuition rigorous.

Conclusion

Yes, there exists an irrational number whose decimal expansion contains only the digits 0 and 1 and whose continued fraction entries are all less than 10. The existence is guaranteed by powerful theorems at the intersection of Diophantine approximation and fractal geometry, but finding or writing down such a number explicitly remains a formidable challenge.