4. For the non-negative real number $x^{4}$ rounded to the nearest integer, denote it as $\langle x\rangle$, that is, when $n$ is a non-negative integer, if
$$
\begin{array}{l}
n-\frac{1}{2} \leqslant x<n+\frac{1}{2}, \\
\text { then }\langle x\rangle=n(\text { for example, }\langle 0\rangle=\langle 0.48\rangle=0,\langle 0.64\rangle= \\
\langle 1.493\rangle=1) .
\end{array}
$$

Given the following conclusions about $\langle x\rangle$:
(1) If $\langle 2 x-1\rangle=3$, then the range of the real number $x$ is $\frac{7}{4} \leqslant x<\frac{4}{9}$;
(2) When $x \geqslant 0, m$ is a non-negative integer, we have
$$
\langle x+m\rangle=m+\langle x\rangle ;
$$
(3) $\langle x+y\rangle=\langle x\rangle+\langle y\rangle$.

Among these, the correct conclusions are $\qquad$ (fill in all correct sequence numbers).
4. (1), (2).

Since $\langle 2 x-1\rangle=3$, we have
$$
3-\frac{1}{2} \leqslant 2 x-1<3+\frac{1}{2} \Rightarrow \frac{7}{4} \leqslant x<\frac{9}{4} \text {. }
$$

Therefore, conclusion (1) is correct.
Let $\langle x\rangle=n$. Then
$n-\frac{1}{2} \leqslant x<n+\frac{1}{2}$ (where $n$ is a non-negative integer).
Thus, $(n+m)-\frac{1}{2} \leqslant x+m<(n+m)+\frac{1}{2}$, and $n+m$ is a non-negative integer.
Therefore, $\langle x+m\rangle=n+m=m+\langle x\rangle$.
Thus, conclusion (2) is correct.
When $x=y=0.7$,
$$
\langle x\rangle=\langle y\rangle=1,\langle x+y\rangle=\langle 1.4\rangle=1 \text {. }
$$

Therefore, $\langle x+y\rangle \neq\langle x\rangle+\langle y\rangle$.
Thus, conclusion (3) is not correct.