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对根式相加为无理数的研究

2022-08-26 16:10 作者:奥博格沙特  | 我要投稿

(1)设a%2Cb%20%5Cin%20Q%5Csqrt%7Ba%7D%2C%20%5Csqrt%7Bb%7D%20%5Cnotin%20Q,则%5Csqrt%7Ba%7D%20%2B%20%5Csqrt%7Bb%7D为无理数.

如果不严格证明,该命题是“显然”的,但我相信看这篇文章的朋友都希望得到一个严格的证明.

思路:反证法. 通过变形化为A%5Ctimes%20B%3DC(其中A%20%5Cin%20Q%2CB%20%5Cnotin%20Q%2C%20C%20%5Cin%20Q)的形式,再由有理数运算封闭性得出矛盾.

证明:

假设%5Csqrt%7Ba%7D%2B%5Csqrt%7Bb%7D%3Dqq%5Cin%20Q.

(%5Csqrt%7Ba%7D-q)%5E2%3D(%5Csqrt%7Bb%7D)%5E2

%5Cimplies%20a-2%5Csqrt%7Ba%7Dq%2Bq%5E2%3Db

%5Cimplies%20a-b%2Bq%5E2%3D2%5Csqrt%7Ba%7Dq

q%3D0,则a%3Db%3D0,与%5Csqrt%7Ba%7D%2C%5Csqrt%7Bb%7D%20%5Cnotin%20Q矛盾.

q%20%5Cneq%200

%5Cimplies%20%5Csqrt%7Ba%7D%3D%5Cfrac%7Ba-b%2Bq%5E2%7D%7B2q%7D%20

%5Csqrt%7Ba%7D%20%5Cnotin%20Q%5Cfrac%7Ba-b%2Bq%5E2%7D%7B2q%7D%20%5Cin%20Q,矛盾!

%E2%88%B4%5Csqrt%7Ba%7D%2B%5Csqrt%7Bb%7D为无理数.


(2)设a%2Cb%20%5Cin%20Q%5Csqrt%5B3%5D%7Ba%7D%2C%20%5Csqrt%5B3%5D%7Bb%7D%20%5Cnotin%20Qa%2Bb%20%5Cneq%200,则%5Csqrt%5B3%5D%7Ba%7D%2B%5Csqrt%5B3%5D%7Bb%7D为无理数.

思路:与(1)类似.

证明:

假设%5Csqrt%5B3%5D%7Ba%7D%2B%5Csqrt%5B3%5D%7Bb%7D%3Dqq%20%5Cin%20Q.

%5Cimplies%20a%2Bb%2B3%5Csqrt%5B3%5D%7Bab%7D(%5Csqrt%5B3%5D%7Ba%7D%2B%5Csqrt%5B3%5D%7Bb%7D)%3Dq%5E3(将等式两边三次方)

%5Cimplies%20a%2Bb%2B3q%5Csqrt%5B3%5D%7Bab%7D%3Dq%5E3

%E2%88%B5a%2Bb%20%5Cneq%200

%E2%88%B4q%20%5Cneq%200

%5Cimplies%20%5Csqrt%5B3%5D%7Bab%7D%20%3D%20%5Cfrac%7Bq%5E3-a-b%7D%7B3q%7D

%5Csqrt%5B3%5D%7Bab%7D%20%5Cnotin%20Q

%5Cfrac%7Bq%5E3%20-%20a%20-%20b%7D%7B3q%7D%20%5Cin%20Q,矛盾!

%E2%88%B4%5Csqrt%5B3%5D%7Ba%7D%2B%5Csqrt%5B3%5D%7Bb%7D%20%5Cnotin%20Q

%5Csqrt%5B3%5D%7Bab%7D%20%5Cin%20Q

x%20%3D%20%5Csqrt%5B3%5D%7Ba%7D%2C%20y%20%3D%20%5Csqrt%5B3%5D%7Bb%7D

x%2Cy%20%5Cnotin%20Q%2C%20x%2By%20%5Cin%20Q%2C%20xy%20%5Cin%20Q

%5Cimplies%20x-y%20%5Cnotin%20Q

(否则2x%3D(x%2By)%2B(x-y)%20%5Cin%20Q,矛盾!)

%E2%88%B5x%5E3-y%5E3%3D(x-y)(x%5E2%2Bxy%2By%5E2)%3D(x-y)%5B(x%2By)%5E2-xy%5D

x%5E3-y%5E3%3Da-b%20%5Cin%20Q

x-y%20%5Cnotin%20Q

(x%2By)%5E2-xy%20%5Cin%20Q

%E2%88%B4(x%2By)%5E2-xy%3D0,即x%5E2%2Bxy%2By%5E2%3D0

x为主元,

%5CDelta%20%3Dy%5E2-4y%5E2%3D-3y%5E2%5Cgeq%200

%5Cimplies%20y%3D0

%5Cimplies%20x%3D0

这与x%2Cy%20%5Cnotin%20Q矛盾.

%E2%88%B4x%2By%5Cnotin%20Q

综上,%5Csqrt%5B3%5D%7Ba%7D%2B%5Csqrt%5B3%5D%7Bb%7D%20%5Cnotin%20Q

遗憾的是,该方法虽然巧妙,但很难在次数和项数上做推广.


(3)将(2)作为引理,可得:

h(x)%20%5Cin%20Z%5Bx%5Dh(%5Csqrt%5B3%5D%7Bt%7D)%3D0t%20%5Cin%20Q%5Csqrt%5B3%5D%7Bt%7D%20%5Cnotin%20Q),则h(x)有因式x%5E3-t.

证明:设h(x)%3D(x%5E3-t)q(x)%2Br(x)deg r%5Cleq%202

x%3D%5Csqrt%5B3%5D%7Bt%7D%20%5Cimplies%20r(%5Csqrt%5B3%5D%7Bt%7D)%3D0

r(x)%3Dax%5E2%2Bbx%2Bca%2Cb%2Cc%20%5Cin%20Q

a(%5Csqrt%5B3%5D%7Bt%7D)%5E2%2Bb(%5Csqrt%5B3%5D%7Bt%7D)%2Bc%3D0

a%2C%20b%20%5Cneq%200

a%5E3t%5E2%2Bb%5E3t%20%5Cneq%200

(否则-%5Cfrac%7Bb%7D%7Ba%7D%3D%5Csqrt%5B3%5D%7Bt%7D,由有理数运算封闭性可知矛盾)

由(2)中的结论可知:

%5Csqrt%5B3%5D%7Ba%5E3t%5E2%7D%2B%5Csqrt%5B3%5D%7Bb%5E3t%7D%20%5Cnotin%20Q

a(%5Csqrt%5B3%5D%7Bt%7D)%5E2%2Bb(%5Csqrt%5B3%5D%7Bt%7D)%20%5Cnotin%20Q

%5Cimplies%200%3Da(%5Csqrt%5B3%5D%7Bt%7D)%5E2%2Bb(%5Csqrt%5B3%5D%7Bt%7D)%2Bc%20%5Cnotin%20Q

矛盾!

a%2Cb中恰有一数为0

显然也有a(%5Csqrt%5B3%5D%7Bt%7D)%5E2%2Bb(%5Csqrt%5B3%5D%7Bt%7D)%20%5Cnotin%20Q

矛盾!

%E2%88%B4a%3Db%3D0

%5Cimplies%20c%3D0

%5Cimplies%20r(x)%3D0

%E2%88%B4h(x)%3D(x%5E3-t)q(x)

Q.E.D.

注:x%5E3-t称为%5Csqrt%5B3%5D%7Bt%7D的极小多项式.

事实上,证明(3)是我研究(2)的原因.


本文中的证明仅为个人方法,如有雷同,纯属巧合.

如果读者有更好的方法,或者发现问题,欢迎指出!

对根式相加为无理数的研究的评论 (共 条)

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