Solution: First we divide $\mathbb{N}_{0}$ to $A_i=\{x|2^i=<x<2^{i+1}\}$ So we will colour each $A_i$ with 3 color,named p,q,r so i) is sastisfied $A_1$ color p,we will prove that if $A_i$ is colored for all $i=1,k$,then we can color $A_{k+1}$ Assume that $A_{k+1}$ can not be colored,so there are $x_1>=y_1>z_1$ of color p,$x_2>=y_2>z_2$ of color q,$x_3>=y_3>z_3$ of color r sastisfied ii),so $x_1,x_2,x_3$ belong to $A_{k+1}$ Case 1:$k$ is even:$k=2t$ Then we get $2^{k+1}+y_i>z_i>=2^k+y_i$=>$2^{t+1}>z_i>2^t$(contradiction) Case 2:$k$ is odd:$k=2t+1$ In the same way we get $2^{t+3/2}>z_i>2^{t+1/2}$,there are 3 numbers by Dirichlet theorem,2 of them lie in same $A_t$ or $A_{t+1}$,contradiction,done