# Check my solution for discontinuous function

1. Dec 6, 2013

### Government\$

1. The problem statement, all variables and given/known data
Let f: R -> R and defined with f(x)={ x, if x $\in$ Q or x^2 if $\in$ R\Q}
a) Prove that function is discontinuous at x=2;
b) Find all points for R in which function is continuous;

3. The attempt at a solution
As far as i know there are infinitely many irrationals but more importantly i can find infinitely many irrational numbers around any number. If that's the case then:

a) In a i need to show that function is discontinuous, so i need to find ε>0 such that for all δ>0, and for some x, i have | x - 2| < δ and | f(x) - 2 | > ε

So if i let ε= 1/2, i can find for any δ, no matter how small, a number x that is irrational. When number is irrational f(x)=x^2. Thus if x>2 and irrational, f(x) is going to be a bit above 4 for values close to 2 and then | f(x) - 2 | approximately 2 which is greater then ε= 1/2, and this works for any δ. Thus it is discontinuous at x=2.

b) If instead of x=2 i let x=q where, q is $\in$ Q, i can use the same proces as in a to show that it is discontinuous at any q, where q is in Q. Also if i let x be irrational i can do same thing again and show that function is not continuous at any irrational number. So this means that it is not continuous at any point .

2. Dec 6, 2013

### brmath

what about the continuity at x = 0? and x = 1?