设 f(a)=af(a)=af(a)=a,f(b)=bf(b)=bf(b)=b。在 (i) 中令 x=a,y=bx=a,y=bx=a,y=b:f(ab)=f(af(b))=bf(a)=abf(ab)=f(af(b))=bf(a)=abf(ab)=f(af(b))=bf(a)=ab,故 ababab 也是不动点。令 x=1/a,y=ax=1/a, y=ax=1/a,y=a:1=f(1)=f(1af(a))=af(1a)1=f(1)=f(\tfrac1a f(a))=a f(\tfrac1a)1=f(1)=f(a1f(a))=af(a1),得 f(1/a)=1/af(1/a)=1/af(1/a)=1/a:不动点的倒数仍是不动点。