[proofplan]
We use the defining model of relative algebraic K-theory: $K(R,I)$ is the homotopy fibre of the map $K(R)\to K(R/I)$ induced by the quotient homomorphism. This gives a fibre sequence of spectra, and the standard long exact sequence of homotopy groups for a fibre sequence supplies the desired low-degree segment. The only work is to identify each homotopy group in that segment with the corresponding absolute or relative K-group and to name the connecting map.
[/proofplan]
[step:Form the quotient-induced fibre sequence]
Let
\begin{align*}
\pi:R\to R/I
\end{align*}
denote the quotient homomorphism. Applying the spectrum-valued algebraic K-theory functor gives a map of spectra
\begin{align*}
K(\pi):K(R)\to K(R/I).
\end{align*}
By the convention in the statement, the relative K-theory spectrum $K(R,I)$ is defined as the homotopy fibre of $K(\pi)$. Therefore there is a homotopy fibre sequence of spectra
\begin{align*}
K(R,I)\to K(R)\to K(R/I).
\end{align*}
[/step]
[step:Apply the long exact homotopy sequence in degrees one and zero]
We use the standard long exact sequence of homotopy groups associated to a fibre sequence of spectra. Applied to
\begin{align*}
K(R,I)\to K(R)\to K(R/I),
\end{align*}
it gives the exact segment
\begin{align*}
\pi_1K(R,I)\to \pi_1K(R)\to \pi_1K(R/I)\xrightarrow{\partial}\pi_0K(R,I)\to \pi_0K(R)\to \pi_0K(R/I).
\end{align*}
The map $\partial$ is, by definition, the connecting homomorphism in this long exact sequence.
[guided]
The reason the proof is formal is that relative K-theory has been defined by a homotopy fibre. A fibre sequence of spectra
\begin{align*}
F\to E\to B
\end{align*}
has an associated long exact sequence of homotopy groups. In the present situation the spectrum $F$ is $K(R,I)$, the spectrum $E$ is $K(R)$, and the spectrum $B$ is $K(R/I)$. Thus the long exact sequence gives, in the relevant degrees,
\begin{align*}
\pi_1K(R,I)\to \pi_1K(R)\to \pi_1K(R/I)\xrightarrow{\partial}\pi_0K(R,I)\to \pi_0K(R)\to \pi_0K(R/I).
\end{align*}
This is an exact sequence of groups because homotopy groups of spectra are groups in all degrees, including degree $0$. The displayed map $\partial$ is not an additional construction: it is precisely the boundary, or connecting, homomorphism supplied by the long exact sequence of the fibre sequence. Exactness means that at each displayed term, the image of the incoming homomorphism is the kernel of the outgoing homomorphism.
[/guided]
[/step]
[step:Identify the homotopy groups with low algebraic K-groups]
By the definition of absolute algebraic K-groups in low degrees,
\begin{align*}
\pi_1K(R)=K_1(R)
\end{align*}
and
\begin{align*}
\pi_0K(R)=K_0(R).
\end{align*}
The same identifications applied to the [quotient ring](/page/Quotient%20Ring) $R/I$ give
\begin{align*}
\pi_1K(R/I)=K_1(R/I)
\end{align*}
and
\begin{align*}
\pi_0K(R/I)=K_0(R/I).
\end{align*}
Finally, by the definition of relative K-groups from the relative spectrum $K(R,I)$,
\begin{align*}
\pi_1K(R,I)=K_1(R,I)
\end{align*}
and
\begin{align*}
\pi_0K(R,I)=K_0(R,I).
\end{align*}
Substituting these identifications into the exact segment from the previous step gives
\begin{align*}
K_1(R,I) \longrightarrow K_1(R) \longrightarrow K_1(R/I) \xrightarrow{\partial} K_0(R,I) \longrightarrow K_0(R) \longrightarrow K_0(R/I).
\end{align*}
This is exactly the claimed low-degree relative exact sequence.
[/step]
