Prove that the function: f ( x ) = { <mtable columnalign="left left"

Question

Prove that the function:

f (x) = {\begin{cases} x, & if x is an irrational number \\ 0 & if x is a rational number \end{cases}

is discontinuous at every irrational number using both the precise definition of a limit and the fact that every nonempty open interval of real numbers contains both irrational and rational numbers.

Answer 1

Let

a

be irrational, so

f (a) = a \neq 0

. Let

ϵ = | a |

and assume there is

δ > 0

such that

| f (x) - f (a) | < ϵ

for all

x

with

| x - a | < δ

. By the existence of rationals

x

with

| x - a | < δ

(for example

x = \frac{1}{n} ⌊ n a ⌋

for

x = \frac{1}{n} ⌊ n a ⌋

with

n > \frac{1}{δ}

) we arrive at a contradiction because for this

x

we have

f (x) = 0

and hence

| f (x) - f (a) | ≮ ϵ

.

Prove that the function: f ( x ) = { <mtable columnalign="left left"

Want to know more about Irrational numbers?

Answers (1)

Not exactly what you’re looking for?

Relevant Questions