无奇完全数的证明(乙)(黄振东) 作者:黄振东, 单位:利川市 龙船调 编辑部. 摘要:设奇玩全数n=q^a*a^2n*b^2m,奇数数幂的约数和与奇数数幂的比,除5^2a为有限小数外,其余都为无限小数,无限小数的乘积不为整数,所以:n不为完全数. 关键词:奇数,完全数,约束和,比,有限小数,无限小数. abstract: a strange to play for the entire q antity n = q ^ a * a * b ^ ^ 2 n 2 m, odd n mber power s bm ltiple and with an odd n mber of power than, except 5 ^ 2 a for finite decimal, the rest is for infinite decimals, the prod ct of infinite decimals is not as an integer, so: n is not perfect. key words: odd n mber, perfect n mber, constrained s m, ratio, finite decimal, infinite decimal. 1定理:无奇完全数. 2证明: 2,1设奇玩全数n=q^a*a^2n*b^2m 2,2奇数数幂的约数和与奇数数幂的比,除5^2a为有限小数外,其余都为无限小数,无限小数的乘积不为整数,所以:n不为完全数.无奇完全数.证毕!