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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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I'm pretty sure it's E
What about 0 ?

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I think it's just as discontinuous as every other point
Interesting. I'm going with D. Pathological example, though :)
I think it is D. I am fairly certain of this. If only I could rationalize my opinion!
i support E.
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It is obviously not A since the function is defined at some points. It cannot be B or C since each of those functions contain infinitely many discontinuous points. The only reason I say D is because both functions share 0 as a vertex which means the function is continuous at 0.
The definition of continuity at a point is that there exists a limit as you approach that point, and that the function equals the limit at that point. Both are true for zero. This is an interesting example of a function that is continuous at only one isolated point. Kinda weird sounding, but there you have it.
in theory it's undefined at zero
Not it isn't, zero is rational.
or you're right sorry
that was silly
hmm.. foolish of me :)
you're right, that's weird
Since 0 is rational, f(x) has a removable discontinuity at 0/3. The point (0, 0) is removable but it defined by the other function. It just so happens that the other function defines f(0) as 0/2 = 0 anyway. This is why I think 0 is continuous.
Why do you say it has a removable discontinuity?
If we focus only on f(x) = x/3 if x is irrational, then we would expect 0/3 to be discontinuous since 0 is rational and is not defined.
But that's not the function. The function is piecewisely defined at every single point. It's never undefined because any point you choose is either rational or irrational.
But piecewise functions are also discontinuous. In this function, each subfunction has infinitely many points of discontinuity. Combined, however, we obtain the set of all real numbers. x/2 is discontinuous for every irrational (the removed discontinuities lay at x/3) x/3 is discontinuous for every rational (the removed discontinuities lay at x/2) We would expect, then, that when x = 0, the removable discontinuity at x/3 would be found at x/2. It happens that 0/3 = 0/2. Thus the removed discontinuity is found at the same location.
A removable discontinuity is defined as a point at which a function has a finite limit but the function either does not exist at that point, or it does not equal the limit at that point. Neither of those two things apply here. The function exists at zero, and the limit is zero. It is just good old-fashioned continuous.
by the function does not exist, I really mean the function is not well-defined.
You would not agree that f(x) contains two functions containing infinitely many discontinuous points? y = x/2 if x is rational y = x/3 if x is irrational
I would completely agree that f(x) has infinitely many irrational points. I disagree that it has a removable discontinuity at 0.
I'm sorry, discontinuous, not irrational
Ah, yes. I agree that 0 is continuous. My reasoning was that if we focused only on y = x/3 if x is irrational, that line by itself has a discontinuous point at (0, f(0)). However, since 0 is rational, and f(x) = x/2 when x is rational, then f(0) = 0 anyway. f(x) then is continuous at 0. You already said D didn't you ... we are in agreement :D
My brain probably processed the problem incorrectly but resulted in the correct answer.
I think your consideration was logical, but it is just very particular to this problem because there are infinitely many irrational points interspersed between rationals and vice versa. Generally speaking, a function is continuous at a if lim f(x) as x->a = f(a), which is perfectly valid here :)

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