# continuous? monotone? one to one?

• Apr 3rd 2011, 08:02 AM
tn11631
continuous? monotone? one to one?
If I=[0,1] and f: I->R is defined by

f(x)= {x if x is rational, 1-x if x is irrational.

What point of I is f continuous and why?

Also, is f monotone and/or is f one-to-one on I?
• Apr 3rd 2011, 08:12 AM
Plato
Quote:

Originally Posted by tn11631
If I=[0,1] and f: I->R is defined by
f(x)= {x if x is rational, 1-x if x is irrational.
What point of I is f continuous and why?
Also, is f monotone and/or is f one-to-one on I?

$\displaystyle f(x) = \left\{ {\begin{array}{rl}{x,} & {\text{if x is rational}} \\ {1 - x,} & {\text{otherwise}} \\ \end{array} } \right.$

What is $\displaystyle f(0.5)=~?$ What about $\displaystyle \displaystyle \lim _{x \to 0.5} f(x)~?$
• Apr 3rd 2011, 08:32 AM
tn11631
Quote:

Originally Posted by Plato
$\displaystyle f(x) = \left\{ {\begin{array}{rl}{x,} & {\text{if x is rational}} \\ {1 - x,} & {\text{otherwise}} \\ \end{array} } \right.$

What is $\displaystyle f(0.5)=~?$ What about $\displaystyle \displaystyle \lim _{x \to 0.5} f(x)~?$

Oo sorry I should have drawn the graph first :/ but now would f(x) be monotone. I wanna say no because as soon as x switches from rationals to irrationals the graph goes from decreasing to increasing and vice versa.
• Apr 3rd 2011, 08:48 AM
Plato
Quote:

Originally Posted by tn11631
Oo sorry I should have drawn the graph first :/ but now would f(x) be monotone. I wanna say no because as soon as x switches from rationals to irrationals the graph goes from decreasing to increasing and vice versa.

It is increasing on the rationals and decreasing on the irrational.
So is it monotone?
• Apr 3rd 2011, 09:03 AM
tn11631
Quote:

Originally Posted by Plato
It is increasing on the rationals and decreasing on the irrational.
So is it monotone?

Well for montone doesn't it have to be either sticktly increasing or stricktly decreasing..in this case it is but f(x) as a whole switches between the rationals and the irrationals so no its not monotone? I'm not sure, maybe I'm not understanding this correctly?
• Apr 3rd 2011, 09:22 AM
Plato
That is correct. It is not monotone.
Consider $\displaystyle f(0.3),\;f\left( {\frac{\pi } {{10}}} \right)\;\& \,f\left( {\frac{\pi } {{11}}} \right)$
• Apr 3rd 2011, 09:38 AM
tn11631
Quote:

Originally Posted by Plato
That is correct. It is not monotone.
Consider $\displaystyle f(0.3),\;f\left( {\frac{\pi } {{10}}} \right)\;\& \,f\left( {\frac{\pi } {{11}}} \right)$

Yay! Sorry I'm just excited that I managed to pull something out of my brain :) however what about I being one-to-one? I had to look up this definiton becaus e I couldn't remember it but it says that a 1-1 is one that assumes each value in its image exactly once. I wanna say that it is NOT one-to-one but I don't know why and I feel like I'm wrong :/
• Apr 3rd 2011, 09:54 AM
Plato
You want to show that if $\displaystyle f(a)=f(b)$ then $\displaystyle a=b$.
That is easy if both $\displaystyle a~\&~b$ are rational or both irrational.

BUT, what if one is rational and the other irrational?
• Apr 3rd 2011, 11:33 AM
tn11631
Quote:

Originally Posted by Plato
You want to show that if $\displaystyle f(a)=f(b)$ then $\displaystyle a=b$.
That is easy if both $\displaystyle a~\&~b$ are rational or both irrational.

BUT, what if one is rational and the other irrational?

So then no its not one-to-one. That's what I was thinking with the whole rational irrational part but I wasn't sure how to explain it.
• Apr 3rd 2011, 12:25 PM
Plato
Quote:

Originally Posted by tn11631
So then no its not one-to-one. That's what I was thinking with the whole rational irrational part but I wasn't sure how to explain it.

No that does not follow.
If a is rational and b is irrational could it happen that $\displaystyle f(a)=f(b)~?$
• Apr 4th 2011, 07:35 AM
tn11631
Quote:

Originally Posted by Plato
No that does not follow.
If a is rational and b is irrational could it happen that $\displaystyle f(a)=f(b)~?$

Oh no now I'm confused. So based on what you said it's making me think yes. However, when i look at the picture of the graph it cannot be that for every point and I thought that was the point of one-to-one. ...also could you possibly explain why for the first part where f is continuous at .5 I thought isaw it but i don't understand it.
• Apr 4th 2011, 07:51 AM
Plato
Quote:

Originally Posted by tn11631
Oh no now I'm confused. So based on what you said it's making me think yes. However, when i look at the picture of the graph it cannot be that for every point and I thought that was the point of one-to-one. ...also could you possibly explain why for the first part where f is continuous at .5 I thought isaw it but i don't understand it.

If $\displaystyle f(a)=f(b)$ both are rational then $\displaystyle a=b$.
If both are irrational then $\displaystyle 1-a=1-b$ or again $\displaystyle a=b$.

If a is rational and b is irrational then that means $\displaystyle a=1-b$ or $\displaystyle b=1-a$.
But $\displaystyle 1-a$ is rational. Thus that case is impossible.
The function is one-to-one.
• Apr 4th 2011, 08:26 AM
Hartlw
continuous at .5

f(.5+d)=.5+d, d rational
f(.5+d)=1-(.5+d) = .5-d, d irrational

l f(.5+d)-.5 l = ldl < e if ldl < e