Question

設(shè)f(x)對一切自然數(shù)有定義,且①f(x)是整數(shù);②f(2)=2;③f(m·n)=f(m)·f(n)對一切自然數(shù)成立;④當(dāng)m＞n時,有f(m)＞f(n),試證:f(n)=n.

Answer 1

證明:(1)由于2=f(2)=f(1·2)=f(1)·f(2)=2f(1),∴f(1)=1,即n=1時,命題成立.

(2)設(shè)n≤k時,有f(k)=k.

當(dāng)n=k+1時,若k+1為偶數(shù),則k+1=2i(i∈N且i≤k),

∴f(k+1)=f(2i)=f(2)·f(i)=2i=k+1;若k+1為奇數(shù),則k+2為偶數(shù),即k+2=2(i+1)(i∈N且i+1≤k).

∴f(k+2)=f［2(i+1)］=f(2)·f(i+1)=2(i+1)=k+2.

由于k＜k+1＜k+2,

∴f(k)＜f(k+1)＜f(k+2)且f(n)為整數(shù),故f(k+1)=k+1,即當(dāng)n=k+1時結(jié)論成立.

由(1)(2),知對于n∈N都有f(n)=n.