- #1
kidsmoker
- 88
- 0
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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