The number of discontinuity points of the function $f(x) = (x \bmod 2)^2 + (x \bmod 4)$ in the interval $0 < x < 9$ is?

Question

For any real number x and any positive integer $n$, we can uniquely write $x = mn + r$, where $m$ is an integer (positive, negative or zero) and $0 ≤ r < n.$ With this notation we define $x \bmod n = r.$ The number of discontinuity points of the function $f(x) = (x \bmod 2)^2 + (x \bmod 4)$ in the interval $0 < x < 9$ is?

How should I approach this problem?

Answer 1

$f(x)$ is continuous at $a$ iff $\lim_{x\to a-} f(x) = \lim_{x\to a+} f(x)$.

If for some natural $n$ and $\delta$, $n > \delta > 0$ then $nk+\delta \mod n = \delta \mod n = 0$ but $nk- \delta \mod n=-\delta \mod n = n-\delta$. So $\lim_{x\to nk^+}x\mod n = \lim_{x\to nk^+} x- nk = \lim_{\delta \to 0} \delta = 0$ but $\lim_{x\to nk^-} x\mod n = \lim_{x\to nk^-}x - (nk - n) = \lim+{\delta \to 0} n-\delta = n$.

But for any $a$ so that $nk < a < nk + n$ and $\delta < \min(a-nk, nk+n -a)$ we have $a\pm \delta \mod n = \delta$ so $\lim_{x\to a^+} x\mod n = \lim_{x\to a^-} x\mod n = a-n$.

So $x \mod n$ is continuous at $a$ if and only if $a$ is not a multiple of $n$.

$f(x) = (x\mod 2)^2 + (x\mod 4)$ will be continuous at all $x$ that are not an even integer.

If $x$ is an even integer we must consider if $x$ is a multiple of $4$ or not.

If $2 > \delta > 0$ then

$f(4k + \delta) = (4k + \delta\mod 2)^2 + (4k + \delta \mod 2) = \delta^2 + \delta \to 0$.

While $f(4k-\delta) = (4k - \delta \mod 2)^2 + (4k -\delta \mod 2) = (2-\delta)^2 + (4-\delta) \to 4+4 = 8$.

$0 \ne 8$ so $f$ is not continuous at $4k$.

$f(4k +2+ \delta) = (4k + 2+\delta\mod 2)^2 + (4k + 2+\delta \mod 2) = \delta^2 + (2 + \delta) \to 2$.

While $f(4k +2-\delta) = (4k +2 - \delta \mod 2)^2 + (4k +2 -\delta \mod 2) = (2-\delta)^2 + (2-\delta) \to 4+2 = 6$.

$2 \ne 6$ so $f$ is not continuous at $4k$.

So $f$ is not continuous at even integers.

Note: $g(x) = (x\mod 2)^2 - (x\mod 4)$ will have a different answer.

What will it be?

Answer 2

If two functions are continuous at $c$, then their sum is continuous at $c$ as well. Similarly for the product.

So you need to find where $g(x)=x\bmod n$ is not continuous. The answer is easy: at the integral multiples of $n$.

Thus the points you need to look at are $2,4,6,8$.

Let's look also at $h(x)=(x\bmod n)^2$. Can you prove that it is not continuous at the integral multiples of $n$? Just consider the limits from the left and from the right.

Can $f$ be continuous at $2$ or $6$? No, because $x\bmod4$ is continuous there, while $(x\bmod 2)^2$ isn't.

Thus we just need to look at $4$ (the answer is the same at $8$). Take a number slightly less than $4$; then $x\bmod 2$ is “large” and similarly $x\bmod 4$. On the other hand, if the number is slightly larger than $4$, then $x\bmod 2$ and $x\bmod 4$ are “small”.

Make the argument rigorous.

Answer 3

The graph of the function $f(x)=x\mod2$ defined over the real is formed by the diagonals (with positive slope) of the squares $[2n,2n+2]\times[0,2]$ then its discontinuities are done for the values $x=2n$. Similarly the discontinuities of the function $g(x)=x\mod4$ are presented for the values $x=4n$. It follows that, because of $4=2\times2$ the discontinuities of the function $$F(x)=(f(x))^2+g(x)$$ are done in the even numbers, so in our case, for $x=2,4,6,8$.