At 12:20 PM 1/19/1997, Sir Robin of Locksley wrote:

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Is it possible to prove that number 0.1234567891011121314151617181920... iz irrational?

Most definately. All you need to do is prove that the set of this number is uncountable, ergo is irrational. If you have friend who have done math in real analysis they can explain more.

The set of this number contains one member - the number. Therefore, this is a countable set. A rational number is a number which can be described as a fraction in which the numerator and the denominator are integers. Jim Choate implied that "infinity" is an integer. This is not correct. The set of integers is said to be infinite. Infiniteness is a property of the set, not a member of the set. All rational numbers are expressible in a decimal form of finite length or in a decimal form which ends in a repeatable string of digits. (Consider the process by which each successive digit is found. It is a fraction of the denominator and some remainder with a zero added. The remainder is less than the denominator multiplied by the base, 10 in this case. When a remainder repeats, the state of the algorithm repeats, and its output will therefore be repeated. Since the remainder is bounded, the algorithm must repeat a state. Note that this holds regardless of the base used.) If our number, 0.1234..., does not repeat then it is not rational. Let's show that it does not repeat. If it repeats, there is a sequence of digits which repeats indefinitely. Such a sequence must be longer than the longest integer. (Because if it was part of an integer, the other parts of the integer would not always be the same and it couldn't repeat.) But, there is no longest integer. Therefore there is no repeating sequence. Therefore our number, 0.1234..., is not a rational number. Math Man