Segal, in the papers "Fredholm complexes" and "Equivariant K-theory", gives the following equivalent definitions of $K$-theory.
For $X$ a compact top. space, let $\mathcal{L}(X) $ be the set of complexes $$0\to E^1\overset{d^1}{\to} E^2\overset{d^2}{\to}\cdots \overset{d^{N-1}}{\to} E^N\to 0$$ where $E_i\to X$ is a vector bundle and the differentials $d_i$ are morphisms of v.bundles.
We say that two complexes $E^\bullet, F^\bullet$ are homotopic, written $E^\bullet \simeq F^\bullet$ if they are restriction of a complex in $\mathcal{L} (X\times[0,1])$ to the boundary $X\times \{0\} $, $X\times \{1\} $.
We say that two complexes $E^\bullet, F^\bullet$ are equivalent, written $E^\bullet \sim F^\bullet$ if exist $V_0^\bullet, V_1^\bullet\in \mathcal{L}(X)$ s.t. $V_i^\bullet$ are both acyclic complexes and $$E^\bullet \oplus {V_0}^\bullet \simeq F^\bullet \oplus {V_1}^\bullet.$$
Segal says that the map $\chi: \frac{\mathcal{L}(X)}{\sim} \to K(X) $ defined by $\chi(E^\bullet) = \sum_i (-1)^{i} E^i$ is a bijection. This does not seem correct to me:
Consider for example $X = \ast$ $$E^\bullet = 0\to \mathbb R\to 0 \to \mathbb R\to 0$$ and $$F^\bullet = 0\to \mathbb R^2\to 0 \to 0 \to 0$$ Then $\chi(E) = \chi (F) = -\mathbb R^2 \in K(\ast)$, however $E^\bullet$ and $F^\bullet$ cannot be equivalent because the equivalence relation (as the homotopy relation) is given in terms of morphisms of complexes hence it must respect the grading of the complex.
Note that the $V_0^\bullet$ and the $V_1^\bullet$ in the definition of "equivalent" need not be the same. So for instance, you could have $$V_0^\bullet=0\to\mathbb{R}\to\mathbb{R}\to 0\to 0$$ and $$V_1^\bullet=0\to 0\to\mathbb{R}\to\mathbb{R}\to 0$$ and then $E^\bullet\oplus V_0^\bullet$ and $F^\bullet\oplus V_1^\bullet$ have the same dimension in each degree, so they could be homotopic. (In fact they are, since each one is homotopic to the complex $0\to\mathbb{R}^2\to\mathbb{R}\to\mathbb{R}\to 0$ with all differentials $0$ by just linearly interpolating the differentials.)