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数学 / 几何原本 / Proposition X.40

第10卷命题 40 · 有理中矩面无理线

If two straight lines incommensurable in square which make the sum of the squares on them medial, but the rectangle contained by them rational, be added together, the whole straight line is irrational; and let it be called the side of a rational plus a medial area.

若两线段平方不可通约,其平方和为中项面,而所成矩形为有理面,则两线段之和为无理线段,称为有理中矩面边。

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线

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分步证明Step-by-step proof
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  1. For let two straight lines AB, BC incommensurable in square, and fulfilling the given conditions [X. 34 ], be added together; I say that AC is irrational.

    设两线段AB、BC平方不可通约,且满足给定条件,相加得AC。

  2. For, since the sum of the squares on AB, BC is medial, while twice the rectangle AB, BC is rational, therefore the sum of the squares on AB, BC is incommensurable with twice the rectangle AB, BC; so that the square on AC is also incommensurable with twice the rectangle AB, BC.

    因AB、BC平方和为无理面,而二倍矩形AB·BC为有理面,故平方和与二倍矩形不可通约。

  3. [X. 16 ] But twice the rectangle AB, BC is rational; therefore the square on AC is irrational.

    由X.16,AC上的正方形与二倍矩形AB·BC亦不可通约。

  4. Therefore AC is irrational.

    二倍矩形为有理面,故AC上的正方形为无理面,因此AC为无理线段。