Submitted by K Sengupta
Rating: 4.5000 (2 votes)
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We shall prove an even stronger result. RESULT: If a, b and c are three rational numbers so that a < b < c, and 1/a, 1/b and 1/c are in Arithmetic Progression, then (a+b) can never be equal to c. PROOF: Since 1/a, 1/b and 1/c are in A.P., it follows that 1/b - 1/a = 1/c - 1/b giving, b = 2*a*c/(a+c). Accordingly, x + (2xz/(x+z) = z, or, x^2 + 2xz - z^2 = 0 or, x = z( 1 +/- sqrt(2)). so that, for any rational number c, it follows that a is always irrational. This is a contradiction. Hence, the proof. ALTERNATE PROOF (based upon comments by Richard and goFish): 1/b - 1/a = 1/c - 1/b , where a, b and c are rational numbers. or, 2ac = bc + ab Substituting c = a+b and simpliifying, we obtain: 2*(a^2) = b^2, giving: b/a = sqrt(2)--------------(#) Consequently, LHS of (#) is a rational number while RHS of (#) is an irrational number. This is a contradiction. Hence the proof.

	Subject	Author	Date
	AAAAAAAAHHHHHH	phi	2006-03-09 04:58:18
	re: One Approach	goFish	2006-02-21 16:04:01
	One Approach	Richard	2006-02-21 11:22:39