1. If f(x) = 5x - 6, prove that f is continuous in its domain.
2. If f(x) = 1 if x is rational and f(x) = 0 if x is irrational, prove that x is not continuous at any point of its domain.

We have already seen from the graph that the first function seems to be continuous while the second one does not. We have to formally prove it, though.

Pick any > 0. Take any sequence { } converging to c. Then there exists an integer N such that

• | - c | < / 5 for n > N

Then

• | f() - (5 c - 6) | = | 5 - 6 - 5 c + 6 | = 5 | - c | <

for n > N. But then the sequence { f() } converges to 5 c - 6, or in other words: if a sequence {} converges to c, then f() converges to f(c). That proves continuity of the first function.

As for the second one: if c is any real number we can find a sequence of rational numbers converging to c, as well as another sequence of irrational numbers also converging to c. But then the sequence { f() } is identically 1, and the sequence { f() } is identically 0. But then f does not have a limit at c, and hence can not be continuous at c either (we have seen this argument - more formally - in a previous example already).