# $\alpha\in R$ then if $k\alpha\in R$ then $k=1,2,-1,-2,1/2,-1/2$

Let $R$ be a root system. Suppose $\alpha\in R$ and if for some $k \in \Bbb{R}$ we have $k\alpha\in R$ then how to prove $k=1,2,-1,-2,1/2,-1/2$? I just know $\frac{2(\alpha,\beta)}{<\alpha,\alpha>}$ will be an integer could any one help me why this term is bounded by $2$ above?

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You know that the quantities $m_1 = 2\frac{\langle k\alpha,\alpha\rangle}{\langle k\alpha, k\alpha\rangle} = \frac{2}{k}$ and $m_2 = 2\frac{\langle \alpha,k\alpha \rangle}{\langle \alpha,\alpha \rangle}= 2k$ are always integers. Now write $2k = n$ for some integer $n$ so that $\frac{4}{n}$ is an integer. The only integers $n$ for which $\frac{4}{n}$ is an integer is
$$n = \pm 1, \pm 2,\pm 4$$
so that $k= \pm 1, \pm 2, \pm \frac{1}{2}$.
You mean that $\frac{2(\alpha,\beta)}{(\alpha,\alpha)}$ is integer. Now up to reversing the roles of $\alpha$ and $\beta$ in $k\alpha=\beta$, you can assume that $|k|\leq1$, and you then immediately get what you want.