Why or why not?
Thank you
Thank you
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No, it can't.
The difference between terms, call it d, is either rational or irrational.
Case 1. d is rational
If the arithemetic progression has one rational term t, then the other terms are t + nd for some n. These are all rational.
If the arithmetic series has one irrational term, u, then the other terms are u + nd for some n, and these will all be irrational. (Proof: If you have one rational term, say u + nd = a/b, then u = a/b -nd, which is rational, contradicting the assumptions that u is irrational.)
Case 2, d is irrational
If the arithmetic series has one rational term, a/b + nd, then the others are a/b +md. If one of these is rational for n not equal to m, then you can solve for d and show that it's rational, which is a contradiction So none of the others can be rational--the series can have at most one rational term. E.g., first term = 0, nth term = nsqrt2.
The difference between terms, call it d, is either rational or irrational.
Case 1. d is rational
If the arithemetic progression has one rational term t, then the other terms are t + nd for some n. These are all rational.
If the arithmetic series has one irrational term, u, then the other terms are u + nd for some n, and these will all be irrational. (Proof: If you have one rational term, say u + nd = a/b, then u = a/b -nd, which is rational, contradicting the assumptions that u is irrational.)
Case 2, d is irrational
If the arithmetic series has one rational term, a/b + nd, then the others are a/b +md. If one of these is rational for n not equal to m, then you can solve for d and show that it's rational, which is a contradiction So none of the others can be rational--the series can have at most one rational term. E.g., first term = 0, nth term = nsqrt2.