i can't solve it, hoping you can help.
let $f(x)=2x-1+(x-a)^2D(x)$ -- D(x) is dirichlet function.
a)prove that it exists for every x.
b)let $x_0\neq 1$, and let $ f(x) = \begin{cases} 2x-1 & x \notin \mathbb{Q} \\ x^2 \in \mathbb{Q} \end{cases} $. prove that f(x) is not continuous using what you proved above (a). (given hint: assume negatively that f(x) is continuous in $x_0$ and then contradict it using the rules of arithmetics of continuous function).
for a) i tried to represent it as a addition of two functions: let y(x)=2x-1 and $z(x)=(x-a)^2D(x)$, so after proving that y(x) is only continuous for $X_0=1$, i tried to show that f(x)=f(y + z) and then to that it cannot be continuous, but without no luck. for b) i did not prove a, but i think that using it again as an addition of functions can allow me that $x_0$ is continuous and not continuous at the same time, which means a contradiction. though i do not know how to write it mathmatically.
hoping you can help.
The question (a) is to me one so much startling. The existence is immediated. For any $x$, the function $f(x)$ is well defined, because it is not more that an addition, a product, of perfectly defined real functions.
As for (b) have in account that if you approach to a $x_{0}$ for the rational ones: $$\lim_{x\to x_{0}, x\in \mathbb{Q}} f(x)=x_{0}^{2}$$ and if you approach for the irrational ones: $$\lim_{x\to x_{0}, x\in \mathbb{I}} f(x)=2x_{0}-1$$ Those two values only coincide if $x_{0}=1$. In another case the limit does not exist and there is therefore not continuity.