[proofplan] The proof is a direct computation from the definition of the coroot pairing. Multiplying the two Cartan integers cancels the root lengths in exactly the same way as the formula for the cosine of the angle between two nonzero vectors. The crystallographic axiom gives integrality, while non-proportionality forces $\cos^2\theta < 1$, so the product is an integer in the interval $[0,4)$. [/proofplan] [step:Multiply the two coroot pairings and identify the cosine factor] Since $\alpha,\beta \in \Phi$, both roots are nonzero, so $(\alpha,\alpha)>0$ and $(\beta,\beta)>0$. By the definition of the coroots, \begin{align*} \alpha^\vee = \frac{2\alpha}{(\alpha,\alpha)} \qquad\text{and}\qquad \beta^\vee = \frac{2\beta}{(\beta,\beta)}. \end{align*} Therefore the two coroot pairings are \begin{align*} \langle \beta,\alpha^\vee\rangle &= \frac{2(\beta,\alpha)}{(\alpha,\alpha)}, \\ \langle \alpha,\beta^\vee\rangle &= \frac{2(\alpha,\beta)}{(\beta,\beta)}. \end{align*} Multiplying these identities and using the symmetry $(\beta,\alpha)=(\alpha,\beta)$ of the inner product gives \begin{align*} \langle \beta,\alpha^\vee\rangle \langle \alpha,\beta^\vee\rangle &= \frac{2(\beta,\alpha)}{(\alpha,\alpha)} \frac{2(\alpha,\beta)}{(\beta,\beta)} \\ &= \frac{4(\alpha,\beta)^2}{(\alpha,\alpha)(\beta,\beta)}. \end{align*} By the definition of $\theta$, \begin{align*} \cos^2\theta = \frac{(\alpha,\beta)^2}{(\alpha,\alpha)(\beta,\beta)}. \end{align*} Substituting this expression into the preceding identity yields \begin{align*} \langle \beta,\alpha^\vee\rangle \langle \alpha,\beta^\vee\rangle = 4\cos^2\theta. \end{align*} [/step] [step:Use crystallographic integrality and non-proportionality to restrict the product] Because $\Phi$ is crystallographic, the Cartan integers satisfy \begin{align*} \langle \delta,\gamma^\vee\rangle \in \mathbb{Z} \end{align*} for all $\gamma,\delta \in \Phi$. Applying this with $(\gamma,\delta)=(\alpha,\beta)$ and $(\gamma,\delta)=(\beta,\alpha)$ gives \begin{align*} \langle \beta,\alpha^\vee\rangle \in \mathbb{Z} \qquad\text{and}\qquad \langle \alpha,\beta^\vee\rangle \in \mathbb{Z}. \end{align*} Hence their product is an integer. Since $\theta \in [0,\pi]$, we have $0 \leq \cos^2\theta \leq 1$. If $\cos^2\theta = 1$, then equality holds in the [Cauchy-Schwarz inequality](/theorems/432) for the nonzero vectors $\alpha$ and $\beta$, so $\alpha$ and $\beta$ are linearly dependent. That would make them proportional, contrary to the hypothesis. Thus \begin{align*} 0 \leq \cos^2\theta < 1. \end{align*} Using the identity already proved, \begin{align*} 0 \leq \langle \beta,\alpha^\vee\rangle \langle \alpha,\beta^\vee\rangle = 4\cos^2\theta < 4. \end{align*} The product is therefore an integer in the interval $[0,4)$, and the only such integers are \begin{align*} 0,1,2,3. \end{align*} This proves \begin{align*} \langle \beta,\alpha^\vee\rangle \langle \alpha,\beta^\vee\rangle \in \{0,1,2,3\}. \end{align*} [/step]