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Irrationality isn't hard, for any of them Q&a for people studying math at any level and professionals in related fields There are natural numbers with arbitrarily long strings of $1's$ or strings of $0's$, for example
And, similarly, there are primes with such strings (there are infinitely many primes that start with any fixed sequence). Not only is it not rational, it's not algebraic Count how many number are there from $1$ to $n$ who have all $10$ digits in it at least once
Can we have a generalized method to solve this problem?
If i have a number x = 0123456789101112, what number type should i make it I thought maybe long or double might work since their ranges are quite large, but when i compile the program, it says the. Answers using stl and boost will be welcomed. Do you have any partial progress yourself
Something you've tried that didn't work I want to know the shortest and most elegant way to prove it This is an infinite string of digits But this string, when we try to interpret it in decimals, does not encode a number, i.e., an element of $ {\mathbb n}$ or $ {\mathbb r}$
Natural numbers, when written in decimals, appear as finite strings, and have a last digit
Noninteger real numbers, when written in decimals, require a decimal point, after which an infinity of digits may appear. 323k subscribers in the learnmath community Questions, no matter how basic…
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