- #1

kidsmoker

- 88

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Hi.

In the book I'm reading it gives the function

f(x) = 0, if x is irrational

f(x) = 1/q, if x=p/q in lowest terms.

It says this is continuous at all irrational x. This i can understand i think, because you can show that f(x) tends to zero, as x tends to a, for all a. For this you just take epsilon to be the distance from a to the closest rational number. It can be seen from the graph:

http://img80.imageshack.us/img80/7246/26351357js6.jpg

Then they give the function

f(x) = 0, if x is irrational

f(x) = x, if x is rational.

But they say this is discontinuous at all values of x other than x=0. I don't really understand why this is the case. How come you can't use the same argument as above to show that f(x) tends to zero, as x tends to a, for all a again?

Thanks!

In the book I'm reading it gives the function

f(x) = 0, if x is irrational

f(x) = 1/q, if x=p/q in lowest terms.

It says this is continuous at all irrational x. This i can understand i think, because you can show that f(x) tends to zero, as x tends to a, for all a. For this you just take epsilon to be the distance from a to the closest rational number. It can be seen from the graph:

http://img80.imageshack.us/img80/7246/26351357js6.jpg

Then they give the function

f(x) = 0, if x is irrational

f(x) = x, if x is rational.

But they say this is discontinuous at all values of x other than x=0. I don't really understand why this is the case. How come you can't use the same argument as above to show that f(x) tends to zero, as x tends to a, for all a again?

Thanks!

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