灯下 登录
数学 / 几何原本 / Proposition VII.1

第7卷命题 1 · 互质数的判定定理

Two unequal numbers being set out, and the less being continually subtracted in turn from the greater, if the number which is left never measures the one before it until an unit is left, the original numbers will be prime to one another.

设有两个不相等的数,从较大的数中反复减去较小的数,如果余数始终不能量尽前一个数,直到剩下单位1,则这两个数互质。

A B C D E F G H
fig-1

本页以“互质数的判定定理”整体图解辅助阅读;点、线、角、圆索引已按命题文字和证明步骤校订,可与证明和问答联动。

线

正文图形由校订坐标生成;点、线、角、圆可与证明和问答联动。

分步证明Step-by-step proof
1 / 4
  1. For, the less of two unequal numbers AB, CD being continually subtracted from the greater, let the number which is left never measure the one before it until an unit is left; I say that AB, CD are prime to one another, that is, that an unit alone measures AB, CD. For, if AB, CD are not prime to one another, some number will measure them. Let a number measure them, and let it be E; let CD, measuring BF, leave FA less than itself, let AF, measuring DG, leave GC less than itself, and let GC, measuring FH, leave an unit HA.

    设两不等数AB、CD,从较大者AB中反复减去较小者CD,余数FA小于CD。

  2. Since, then, E measures CD, and CD measures BF, therefore E also measures BF. But it also measures the whole BA; therefore it will also measure the remainder AF.

    再从CD中减去FA得余数GC,从FA中减去GC得余数HA为单位1。

  3. But AF measures DG; therefore E also measures DG. But it also measures the whole DC therefore it will also measure the remainder CG.

    假设存在数E能量尽AB和CD,则E能量尽BF(因CD量尽BF),从而E也量尽AF。

  4. But CG measures FH; therefore E also measures FH. But it also measures the whole FA; therefore it will also measure the remainder, the unit AH, though it is a number: which is impossible.

    同理,E量尽DG、CG、FH,最终量尽单位1,但数不能量尽单位1,矛盾。故AB、CD互质。

《几何原本》第七至九卷的整数论证,可和丢番图不定方程线 diophantus-arithmetica/overview、数论史长卷 math-meta/topic-number-theory-history 对读。