GR0568 Question 20

[bonaluram]

Hello, can anybody help about the following question?

Let f be the function defined on the real line by
f(x) is
x/2 if x is rational
x/3 if x is irrational

If D is the set of points of discontinuity of f, then D is the

(A) empty set
(B) set of rational numbers
(C) set of irrational numbers
(D) set of nonzero real numbers
(E) set of real numbers

The answers is D.
Thanks for your help.

[origin415]

For every point x of R, you can make a convergent sequence entirely in the rationals and entirely in the irrationals, as they are both dense in R.

Take $$a_n$$ to be a sequence which converges to x in the rationals, $$b_n$$ no be a sequence which converges in the irrationals, then f is continuous iff
$$\lim f(a_n) = \lim f(b_n)$$
$$\frac{x}{2} = \frac{x}{3}$$
This only happens at x = 0, so the set of discontinuities is everything else.

[joey]

We can think about this qualitatively, too. Each rational has an irrational very close to it and vice versa, so if x/2 differs significantly from x/3 near x, then f is jumping all over the place on each neighborhood of x. In that case, f is clearly discontinuous at x. This holds whenever x≠0. On the other hand, both x/2 and x/3 approach f(0)=0 as x approaches 0, so f is continuous at x=0.

Missing the continuity at x=0 is probably the most common error here--I made it when I took this test. Reading all the answer choices even when you immediately see the one you think is correct can lessen the probability of this error. You're likely to think, "Oh maybe zero is important", and then quickly see that the situation is different there.

[bonaluram]

thank you so much guys, both answers helped me a lot.