Prove Continuous Function f at 1/√2

[rachelro]

if f : (0, 1]--> R is given by f(x) = 0 if x is irrational, and f(x) = 1/(m+n) if x = m/n in (0, 1] in lowest terms for integers m and n. How can i prove that this function is continuous at 1/√2?

 

[shreyakmath]

At x=1/sqrt(2),
LHL=0 and RHL=0 becoz x+ and x- are both irrational here.
Also, Lt(x--1/sqrt(2))=0
x being irrational.

Hence f(x) is continuous at x=1/sqrt(2)

 

[Vargo]

Essentially, you need to use the fact that if p/q is close to sqrt(2), then q must be large. The closer you want it to be, the larger q must be. Therefore, the value of the function for p/q sufficiently close to sqrt(2) must be small. Perhaps you could use something like this: Given Q, let e_Q be the infimum of | (p/q) - sqrt(2)| for q less than or equal to Q. Then e_Q is nonincreasing, is never 0, and tends to zero as Q tends to infinity. You could use that in your proof.

 

[DonAntonio]

At x=1/sqrt(2),
LHL=0 and RHL=0 becoz x+ and x- are both irrational here.
Also, Lt(x--1/sqrt(2))=0
x being irrational.

Hence f(x) is continuous at x=1/sqrt(2)

Perhaps it' only me but I really can't understand whatever it is you're trying to argue above. Besides this, I think it'd be better

for all if you write down your piece with LaTeX.

DonAntonio

 

[HallsofIvy]

At x=1/sqrt(2),
LHL=0 and RHL=0 becoz x+ and x- are both irrational here.

What do you mean by "x+" and "x-"??

Also, Lt(x--1/sqrt(2))=0
x being irrational.

What is "Lt"?

Hence f(x) is continuous at x=1/sqrt(2)

 

[DonAntonio]

if f : (0, 1]--> R is given by f(x) = 0 if x is irrational, and f(x) = 1/(m+n) if x = m/n in (0, 1] in lowest terms for integers m and n. How can i prove that this function is continuous at 1/√2?

Clearly, if \,\{y_n\}\, is an irrational seq. s.t. \,\displaystyle{y_n\to\frac{1}{\sqrt{2}}}\, , then \,\displaystyle{0=f(y_n)=f\left(\frac{1}{\sqrt{2}}\right)=0}

OTOH, if \,\displaystyle{\left\{x_n=\frac{a_n}{b_n}\right\}}\, is a rational seq. s.t. \,\displaystyle{x_n\to\frac{1}{\sqrt{2}}}\, , with \,(a_n,b_n)=1\,\,\forall n\, , then

Lemma: If \,\displaystyle{\left\{x_n=\frac{a_n}{b_n}\right\}}\, is a rational seq. that converges to an irrational number , then \,b_n\to \infty

Proof: Exercise (try contradiction and check what happens when an integer seq. converges...)

DonAntonio

 

[rachelro]

Thanks guys for the replies.

shreyakmath - your argument is not valid as 0- is not defined.
DonAntonio - it is a good way to proceed, but I've come to a simpler solution (may not be 100 correct).

Here is the solution:

We can show that f(x) is continious at x=1/√2 using the ε-δ definition. Let ε>0 and set δ=ε. we first note that |f(x)|≤|x| for all x. Indeed, if x is irrational, f(x)=0, and if x is rational f(x)=1/(m+n); x=m/n.

Thus, for 0<|x|<δ=ε, we have:

|f(x)-f(1/√2)|=|f(x)|≤|x|<ε.

this will then concludes the proof!

 

[DonAntonio]

Thanks guys for the replies.

shreyakmath - your argument is not valid as 0- is not defined.
DonAntonio - it is a good way to proceed, but I've come to a simpler solution (may not be 100 correct).

Here is the solution:

We can show that f(x) is continious at x=1/√2 using the ε-δ definition. Let ε>0 and set δ=ε. we first note that |f(x)|≤|x| for all x. Indeed, if x is irrational, f(x)=0, and if x is rational f(x)=1/(m+n); x=m/n.

Thus, for 0<|x|<δ=ε, we have:

|f(x)-f(1/√2)|=|f(x)|≤|x|<ε.

this will then concludes the proof!

¡Muy bien! ¡ Bravo ! Ausgezeichnet ! מצויין ! Excellent ! I really loved it.

DonAntonio

 

[Vargo]

I believe there is an error. Your proof demonstrates that if x is close to 0 then f(x) is close to f(1/sqrt(2)).

You need to show that if x is close to 1/sqrt(2), then f(x) is close to 1/sqrt(2).