Please explain why the first step of the proof is incorrect.

Find the first step in the following "Proof" that is incorrect, and explain in a short paragraph why the reasoning is incorrect. Statement: The equation 26456+36=64+6236+cos(6) (8) has a positive solution. In other words, there is some positive number 6 satisfying (8}. Proof: By dividing both sides of the Equation (8) by 64 + 62 36 + cos(6) it follows Equation (8) has a solution if 264 562 + 33 64+62 _36+cos(6) 1 (9) does. Let 2 17(9): 26456 +319 (10) 64+6236+cos(6)' From {2.5.4} of the text it follows that cos is a continuous function. By Theorem 2.4.2 (i), (ii) and (iii) it follows that 264 56 + 36 and 64 + 562 + 36 l 005(6) are both continuous functions. By applying Theorem2.4.2 (v) it follows that F is also a continuous function of 6. Then Equation (8} will have a positive solution provided that F(6) : 1 does. This will be established by using Theorem 2.6.1. Begin by noting that if 6 = 1 then 264 562 l 36 = 0 and hence F(1) = 0. While it is a natural idea that that evaluating 264 562 l 36 at 6 = 1 will be useful, finding a value of6 such that 264 562 + 36 > 1 is not that obvious. In such situations it can be useful to evaluate the limit at 6 goes to 00. With this in mind, observe that , 294 592 + 36 _ 6320 64 +62 36+cos(6) 1' 234/64 56/64 + 39/94 1 9330 94/94 + 92/94 _ 39/94 + com/e4 , 2 56-9 + 36/6-3 all{go 1 + 62 363 l (:os(lSl)/64 111119100 2 562 + 36/6-3 I 2. liuzrlg>0 1 l 6'2 36'3 l cos(6)/64 Lete : 1/2. Then there is some K such that |F(6) 2' 3/2 > 1. To summarize, the following facts about F have been shown: . F is continuous at all 6 Z 0 - F(1) : 0 1 such that F((:3) > 1. It then follows from Theorem 2.6.1 that there must be some 6 between 1 and Q such that F(6) = 1