$$\bex \forall\ m\in \bbZ^+\ra \sqrt{m}\in (\bbR\bs \bbQ)\cup \bbZ^+. \eex$$
转载于:https://www.cnblogs.com/zhangzujin/p/4107672.html
$$\bex \forall\ m\in \bbZ^+\ra \sqrt{m}\in (\bbR\bs \bbQ)\cup \bbZ^+. \eex$$
转载于:https://www.cnblogs.com/zhangzujin/p/4107672.html