app.crowdmark.com C Find the first step in the following "Proof" that is incorrect, and explain in a short paragraph why the reasoning is incorrect. You should have seen in other mathematics courses (such as MATH 1200 or MATH 1090, for example) that a proof consists of a sequence of steps, each one preceding from the previous step by the rules of mathematical logic. This question is asking you to identify the first step in the proof that an incorrect use of logic has been applied and to explain why this is the case. Statement: All functions are continuous. Proof: This will be proved by contradiction, so suppose not. Then there is a function f that is not continuous. This means that there is some real number w such that f is not continuous at w. The first thing we will do is to show that the function - f is also not continuous at w. Why we are doing this should become clear at the end of the argument. To see that - f is not continuous at w, use the definition of continuity at a point (Definition 2.4.1) and the fact that f is not continuous at w to conclude that lim,->w f(x) # f(w). Using the definition of a limit (Definition 2.2.1) it is then possible to find e > 0 such that for all 8 > 0 there is w* such that 0 c. But 1(- D) (w) - (-D(w')1 = 1 - f(w) + f(w')1 = 1f(w) - f(w*) 1 > e thus establishing that the same e shows that limx->w -f(x) # -f(w). Hence -f is not continuous at w. In the text it is shown in Theorem 2.4.2 (1) on page 84 that if f and g are continuous at a point c then the function f + g is also continuous at c. It has already been shown that f and - f are not continuous at w. By Theorem 2.4.2 it therefore follows that f + (-f) is not continuous at w. (1) But f + (-n)(x) = f(x) - f(x) = 0 for all x; in other words, f + (-f) is the constant 0 function. By Example 5 on page 67 of the text we know that limx->w 0 = 0. It follows that the zero function is also continuous at w. This is a contradiction to (1) and, hence, it has been shown that the original assumption Time left Hide that there is a function f that is not continuous leads to a contradiction. Therefore it has been shown that 1:08:00 all functions are continuous. MacBook Pro Q Search or enter website name