Question 517528
*[tex \sqrt{c}] and *[tex \sqrt{c-1}] are strictly increasing functions of c, and are defined for all *[tex 1 \le c \le 2]. In addition, they are continuous so it follows that *[tex \sqrt{c} + \sqrt{c-1}] is continuous and strictly increasing on [1,2]. At c=1, *[tex \sqrt{c} + \sqrt{c-1}], and at c=2, *[tex \sqrt{c} + \sqrt{c-1}] is larger than 2. By the intermediate value theorem there must be some value c within [1,2] such that *[tex \sqrt{c} + \sqrt{c-1} = 2].


Update: *[tex \sqrt{c}] and *[tex \sqrt{c-1}] are continuous because they satisfy the conditions for continuity: that they exist everywhere (on [1,2]), the limit as  c goes to some number exists, and the limit is equal to the number evaluated at c.


http://en.wikipedia.org/wiki/Continuous_function