Let $A=K[x]$ be the ring of polynomials over the field $K$. Let $m\in A$ with $\deg(m)\ge1$, and let $I=(m)$ the principal ideal generated by $m$. Let $f,g \in A$ so that $\deg(f)<\deg(m)$ and $\deg(g)<\deg(m)$.
Prove that if $f\equiv g\bmod I$ , then $f=g$. Conclude that the congruence classes $\bmod I$ are in bijective correspondence with the $f\in A$ so that $\deg(f)<\deg(m)$.
Finally show who is $R[x] /(x^2+1)$.
My attempt so far: $f+I=g+I$ iff $f-g\in(m)$ iff $m\mid f-g$ iff $f-f'=am$ for some $a\in A$. But $A$ is a Euclidean domain (also a PID and UFD) so $\deg(m)\le \deg(f-g)$ but $\deg(f-g)\lt \deg(m)$ right? I don't know what to do from here but I think that $m$ should be zero. Any help or hints will be very appreciated.
I think you were close :
$$f\equiv g\pmod I\iff f-g=am\;\;\text{and}\;\;\deg(f-g)<\deg m\implies f-g=0$$
and you have the first part.
For the second part: write every element in $\;\Bbb R[x]/\langle x^2+1\rangle\;$, using the above, as
$$\;(ax+b)+I\;,\;\;I:=\langle x^2+1\rangle\;,\;\;a,b\in\Bbb R$$
Observe that, for example, $\;x\cdot x=x^2=-1\pmod I\;$ , and now realize the similarity we have here with the complex numbers...Finally, show $\;\Bbb R[x]/I\cong\Bbb C\;$