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7 Fill in the blanks Consider the incomplete proof below: Theorem: Let f: A→A be any function from a set to itself, and define g: A →A by g(x) = f(f(x)). Then f is injective if and only if g is injective. Proof: First, we will prove that assuming that Let x, y € A be arbitrary, and suppose g(x) = g(y), or in other words f(f(x)) = f(f(y)). By the assumption we made, this means that f(x) = f(y), and by using the same assumption again, we get x = y. Second, we will prove that suppose f(x) = f(y). Then assuming that , so g(x) = g(y); by assumption, this means that x = y. Let x, y € A be arbitrary, and Which part of the theorem is being proved in the first paragraph, and which part is being proved in the second paragraph? In the second paragraph, explain how we can go from f(x) = f(y) to g(x) = g(y).

7 Fill in the blanks Consider the incomplete proof below: Theorem: Let f: A→A be any function from a set to itself, and define g: A →A by g(x) = f(f(x)). Then f is injective if and only if g is injective. Proof: First, we will prove that assuming that Let x, y € A be arbitrary, and suppose g(x) = g(y), or in other words f(f(x)) = f(f(y)). By the assumption we made, this means that f(x) = f(y), and by using the same assumption again, we get x = y. Second, we will prove that suppose f(x) = f(y). Then assuming that , so g(x) = g(y); by assumption, this means that x = y. Let x, y € A be arbitrary, and Which part of the theorem is being proved in the first paragraph, and which part is being proved in the second paragraph? In the second paragraph, explain how we can go from f(x) = f(y) to g(x) = g(y).

Calculus For The Life Sciences
Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter6: Applications Of The Derivative
Section6.CR: Chapter 6 Review
Problem 2CR
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7 Fill in the blanks Consider the incomplete proof below: Theorem: Let f: A→A be any function from a set to itself, and define g: A →A by g(x) = f(f(x)). Then f is injective if and only if g is injective. Proof: First, we will prove that assuming that Let x, y € A be arbitrary, and suppose g(x) = g(y), or in other words f(f(x)) = f(f(y)). By the assumption we made, this means that f(x) = f(y), and by using the same assumption again, we get x = y. Second, we will prove that suppose f(x) = f(y). Then assuming that , so g(x) = g(y); by assumption, this means that x = y. . Let x, y € A be arbitrary, and Which part of the theorem is being proved in the first paragraph, and which part is being proved in the second paragraph? In the second paragraph, explain how we can go from f(x) = f(y) to g(x) = g(y).
Transcribed Image Text:7 Fill in the blanks Consider the incomplete proof below: Theorem: Let f: A→A be any function from a set to itself, and define g: A →A by g(x) = f(f(x)). Then f is injective if and only if g is injective. Proof: First, we will prove that assuming that Let x, y € A be arbitrary, and suppose g(x) = g(y), or in other words f(f(x)) = f(f(y)). By the assumption we made, this means that f(x) = f(y), and by using the same assumption again, we get x = y. Second, we will prove that suppose f(x) = f(y). Then assuming that , so g(x) = g(y); by assumption, this means that x = y. . Let x, y € A be arbitrary, and Which part of the theorem is being proved in the first paragraph, and which part is being proved in the second paragraph? In the second paragraph, explain how we can go from f(x) = f(y) to g(x) = g(y).
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