[a0;a1,a2,a3,・・・]とするとき、

[a0;a1,a2,a3,・・・]/n=[a0/n;na1,a2/n,na3,・・・]、a0,a2,a4,a6,・・・がｎで割り切れるとき

n[a0;a1,a2,a3,・・・]=[na0;a1/n,na2,a3/n,・・・]、a1,a3,a5,a7,・・・がｎで割り切れるとき

が成り立つのは容易に理解できるであろう

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[a0;a1,a2,a3,・・・]=a0+1/(a1+1/(a2+1/(a3+・・・)))

n[a0;a1,a2,a3,・・・]=na0+n/(a1+1/(a2+1/(a3+・・・)))

=na0+1/(a1/n+1/(na2+n/(a3+・・・)))

=na0+1/(a1/n+1/(na2+1/(a3/n+・・・)))

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[a0;a1,a2,a3,・・・]/n=a0/n+1/n(a1+1/(a2+1/(a3+・・・)))

=a0/n+1/(na1+n/(a2+1/(a3+・・・)))

=a0/n+1/(na1+1/(a2/n+1/n(a3+・・・)))

=a0+1/(na1+1/(a2/n+1/(na3+・・・)))

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