No; the only condition for qualyfing as an irrational number is that the same pattern of digits doesn't repeat over and over again, as it does with a rational number. For example, 8/7 is a rational number; the decimal expansion is 1.142857 142857 142857 ... As you see, the same pattern of digits repeats over and over. The number may start with different digits, but if after a while the same pattern repeats again and again, the number is rational.
The following number is irrational: 0.101001000100001000001 ... The pattern doesn't repeat, because a zero is added every time. And, in this example, the decimal expansion doesn't contain any digits other than 0 and 1.
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Yes. Every irrational number has a non-terminating, non-repeating decimal representation.
Every irrational number can be represented by a non-terminating non-repeating decimal. Rounding this decimal representation to a suitable degree will provide a suitable approximation.
All real numbers have a decimal representation. Rational numbers have decimal representations that terminate or repeat infinitely. Irrational numbers have decimal representations that are non-terminating and non-repeating.
Every rational number has a decimal expansion that either terminates (like 42.23517) or repeats (like 26.1447676767676...)Pi's decimal expansion neither terminates nor repeatsHence, Pi cannot be rational.If we could prove the first two statements, this would constitute a proof that Pi is irrational, but most people cannot provide proof of either. Most proofs on this issue are quite technical, but I'm hoping to return to this question with a suitable answer soon.
Yes, but most of them (the irrational numbers) will have infinitely long, non-repeating representations.