Question: This question arose while I was tutoring a grade 11 student on the subjects of polynomial division and the Remainder Theorem, and I could not provide a satisfactory answer to it. Presumably, $f(x)$ is a polynomial. I am aware we can write $f(x)$ in the form
$$
f(x) = d(x)q(x) + r(x)
$$
representing its division by $d(x)$, where $q(x)$ is the quotient of the division and $r(x)$ is the remainder, with deg $r(x) < $ deg $d(x)$. But for
$$
f(x) = (x+5)q(x) + (x+3),
$$ assuming $r(x) = (x+3)$, the strict inequality does not hold. I tried analogously shifting the question to integers, where, for example, we can write 27 upon the division of 6 as
$$
27 = (6)(4) + 3
$$
and we would then have the smallest multiple of 6 greater than 27 is one more than the quotient, so 5. But I am not sure how to extend this to the question posed.

Answer: 
> 
> Let $r$ be a rational.
> 
> 
> 

we have $\;f(r)=r$

and $\;\; \forall n>0$,

$$f(r+\frac{\sqrt{2}}{n})=r+\frac{\sqrt{2}}{n}+1$$

thus

$$\lim\_{n\to+\infty}f(r+\frac{\sqrt{2}}{n})=r+1\neq f(r)$$

which means that $\;f\;$ is not continuous at $x=r$.

> 
> Let $s$ be an irrational.
> 
> 
> 

we have $f(s)=s+1$ and

$$s=\lim\_{n\to+\infty}\frac{\lfloor 10^n s \rfloor}{10^n}$$

but

$$\lim\_{n\to+\infty}f(\frac{\lfloor 10^n s \rfloor}{10^n})=\lim\_{n\to+\infty}\frac{\lfloor 10^n s \rfloor}{10^n}$$

$=s\neq f(s)$ .

$f$ is not continuous at $x=s$.
Regardless of delta we can find a y within delta of x that is more than distance 1 from x (there are rationals and irrationals arbitrarily close to x). But then there exists no delta corresponding to epsilon less than 1.