# How can this function be continuous at all points?

I have an exercise in which I need to determine which functions are continuous for all points

Note: $\lfloor x\rfloor$ is floor of $x$.

Function I need to consider is $f(x)= \lfloor x\rfloor+\sqrt{x- \lfloor x\rfloor}$

How can I prove that this function is continuous at integer values if neither $\lfloor x\rfloor$ nor $\sqrt{x- \lfloor x\rfloor}$ are continuous at integer values?

I can determine it visually from graph and intuitively when considering values close to integers from both sides, but how can I rigorously prove it?

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I've updated my answer quite significantly with a graphical and fully analytical/limit based justification. I also explained why it is continuous. I would please ask that you choose to accept one of the answers here (even if not mine) as surely one of these fine folk here deserve the recognition. – TheGreatDuck Apr 23 at 4:40

Hint: For any $n\in \mathbb{N}$, we have $$\lim_{x\to n-} \lfloor x\rfloor=n-1$$ and $$\lim_{x\to n+} \lfloor x\rfloor=n$$

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This isn't really an answer... – TheGreatDuck Apr 23 at 16:20
But it's an incomplete answer. It should be made a comment or finished. – TheGreatDuck Apr 23 at 23:26
It doesn't matter what meta says. See 4. d. of the site terms of service "(d) knowingly post any false, inaccurate or incomplete material". Your post is "incomplete". Now, I personally like hints however stack exchange is shooting them down, so sadly, your post has to change. Sorry. :( – TheGreatDuck Apr 24 at 6:57
@TheGreatDuck, my post is not an incomplete answer, it is a complete hint. If you don't like this, feel free to downvote. – vadim123 Apr 24 at 23:39

It's simple to see that it suffices to prove the continuity of $f$ on the integer. So let $n\in\Bbb Z$ then:

$$\lim_{x\to n^+}f(x)=\lim_{x\to n^+}n+\sqrt{x-n}=n$$and$$\lim_{x\to n^-}f(x)=\lim_{x\to n^-}n-1+\sqrt{x-n+1}=n-1+1=n$$ so $$\lim_{x\to n^+}f(x)=\lim_{x\to n^-}f(x)=n=f(n)$$ hence $f$ is continuous at $n$ and we conclude.

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$1-{\bf 1}_{\Bbb Q}$ and ${\bf 1}_{\Bbb Q}$ are everyhere discontinuous, but their sum is $1$, which is continuous.

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This isn't really an answer. – vadim123 Mar 16 '14 at 20:11
@vadim123 I think it is quite relevant. – Pedro Tamaroff Mar 16 '14 at 20:16

This is easy to consider if we use the modulo identity:

$$f(x)= \lfloor x\rfloor+\sqrt{x\mod 1}$$

("mod" AKA "%" AKA "modulo" is the remainder when a is divided by b)

This means that the sqrt term loops the period from [0,1] of the regular sqrt(x) function infinitely. This is because x % 1 changes back to 0 every time it hits 1 or a multiple of 1. The right hand side of a curve formed by one of these periods is 1 and the left hand side is 0. This means that to force it to become continuous, one would shift the graph up by one every one units in order to counteract a periodic jump of -1 units. That is exactly what the [x] term does.

This can proved further by showing that the sum of all the jumps over the graph of the second term equals the negative of floor. This is done by calculating the portion of the second term's graph that consists of absolutely nothing but jumps and level segments known as the jump series or jump component of that expression. With this we will need the following equation:

$$[x>0]\sum_{n=0}^{JC(x)} \lim_{a\to JI(n)^+}f(a)-\lim_{a\to JI(n)^-}f(a)+[x≦0]\sum_{n=0}^{JC(x)}\lim_{a\to JI(n)^-}f(a)-\lim_{a\to JI(n)^+}f(a)$$

JC (jump counter) is a function that says how many jump locations are from 0 to x. In all cases, a floor function of a monotonic function /is/ the JC function. It's like how one does partial derivatives. You look for an instance of some floor function and calculate the partial jump series with respect to that /type/ of floor function. floor(x) was what it was before we converted to modulo. In the same respect JI (jump iterator) is in all monotonic cases the inverse of the inside of the JC as it says where the n'th jump is. So, it is just x. Of course, f is the function comprised of only the second term. You may notice that this is a piecewise function. Don't be too concerned about that. It's to distinguish which way a jump goes as it's a jump as we move away from the origin rather than the usual sense of finding values as one moves along the x-axis. Don't question it too much. It's just how it is.

Now we plug everything into it:

$$[x>0]\sum_{n=0}^{\lfloor x \rfloor} \lim_{a\to n^+}\sqrt{x-\lfloor x \rfloor}-\lim_{a\to n^-}\sqrt{x-\lfloor x \rfloor}+[x≦0]\sum_{n=0}^{\lfloor x \rfloor}\lim_{a\to n^-}\sqrt{x-\lfloor x \rfloor}-\lim_{a\to n^+}\sqrt{x-\lfloor x \rfloor}$$

We know that n will be an integer value always by simple inspection. An important limit is that the left hand limit of floor at an integer point is n, and the right hand limit is n-1. Everything else collapses as they are continuous functions and limit theorems support this:

$$[x>0]\sum_{n=0}^{\lfloor x \rfloor} \sqrt{n-(n)}-\sqrt{n-(n-1)}+[x≦0]\sum_{n=0}^{\lfloor x \rfloor}\sqrt{n-(n-1)}-\sqrt{n-(n)}$$

Now we complete some algebra:

$$[x>0]\sum_{n=0}^{\lfloor x \rfloor} \sqrt{0}-\sqrt{1}+[x≦0]\sum_{n=0}^{\lfloor x \rfloor}\sqrt{1}-\sqrt{0}$$

$$[x>0]\sum_{n=0}^{\lfloor x \rfloor} -1 +[x≦0]\sum_{n=0}^{\lfloor x \rfloor}1$$

$$[x>0] (-\lfloor x \rfloor) +[x≦0]\sum_{n=0}^{\lfloor x \rfloor}1$$

Now, it would appear that this test has failed, right? After all, the right term should be $\lfloor x \rfloor$? WRONG! Look carefully at the piecewise notation. On the range that term is non-zero, the indexes to the summation are negative. A little unknown and discrete fact is that summations give negative values when counting down. This means our result will have a negative sum:

$$[x>0] (-\lfloor x \rfloor) +[x≦0](-\lfloor x \rfloor)$$

$$([x>0]+[x≦0])(-\lfloor x \rfloor) = -\lfloor x \rfloor$$

The jump component of the second term is $-\lfloor x \rfloor$; therefore, the $\lfloor x \rfloor$ will cancel out the jump component resulting in a fully continuous function on all defined* points.

Alternately, I could have done this with the whole function and shown it was zero, but I wanted to also demonstrate how to /make/ a function continuous, and this does it nicely.

*They're all defined.

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I think your first paragraph does make sense (at least I can understand it). What you are trying to say is that $\lim_{x \to n^{+}}\sqrt{x - [x]} = 0$ and $\lim_{x \to n^{-}}\sqrt{x - [x]} = 1$ and we need a term which counters this and that is $[x]$ so that the left and right limits of $f(x)$ are same as $x \to n$. But frankly speaking your detailed explanation after that does not appear to make any sense and I don't think it was necessary. – Paramanand Singh Jul 5 at 15:01
@ParamanandSingh That is your opinion. I was attempting to be very thorough. Not all questions of this nature have visibly apparent answers. And by that I mean questions asking to find the piecewise constant function that counteracts jump discontinuity. – TheGreatDuck Jul 5 at 20:48