let x be an irrational number. show that for every k>0 there exist n,m ∈Z such that 0<|nx-m|<k

Question

Use direct proof or contrapositive proof or proof by contradiction

Answer 1

Since x is irrational there exists an integer N for which

N<x<N+1. Divide the interval [N,N+1] into n² parts as
xj=N+j/n² for j=0,1,2,...,n², (x_n²=N+1)
All of the x_j are rational. x_j+1-x_j=1/n². No x_j is x.
There is a j* such that x_j*<x<x_j*+1 That is a rational
which is closer than 1/n² to x. Remember that x_j=N+j/n² = (n²N+j)/n²
0<|x-xj*|<1/n²
0<|x-(n²N+j*)/n²|<1/n²
0<|nx-(n²N+j*)/n|<1/n