Can an arithmetic progression have infinitely many rational and infinitely many irrational terms
Favorites|Homepage
Subscriptions | sitemap
HOME > > Can an arithmetic progression have infinitely many rational and infinitely many irrational terms

Can an arithmetic progression have infinitely many rational and infinitely many irrational terms

[From: ] [author: ] [Date: 12-01-02] [Hit: ]
is either rational or irrational.Case 1.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,......
Why or why not?

Thank you

-
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.
1
keywords: and,an,irrational,Can,arithmetic,infinitely,terms,have,many,rational,progression,Can an arithmetic progression have infinitely many rational and infinitely many irrational terms
New
Hot
© 2008-2010 http://www.science-mathematics.com . Program by zplan cms. Theme by wukong .