会的帮个忙……

函数f:R到R单调增，满足f(x+y)=f(x)+f(y)，f(1)=1
求证：f(x)=x

主要要无理数范围的证明

I assume you have proved it in the rational number field, and I ll continue your proof.

for any real number x,

and an arbitrarily small postive epsilon,

from the density of rational field, we know
there exist two rationals p and q, such that  x - epsilon < p < x < q < x + epsilon

according to what u have proved, f(p) = p and f(q) = q

the monotony of f =>  f(p) < f(x) < f(q)
or p < f(x) < q

thus x - epsilon < f(x) < x + epsilon

from the arbitrariness from epsilon, we know f(x) = x #

ls牛b！！

easy for me

..............................=  =!

我数盲，哎！！！

前天貌似看到楼上了，嘿嘿