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(a) Define F: ℤ → ℤ by the rule  F(n) = 2 − 3n,  for each integer n. (i) Prove that F is one-to-one. Proof: Suppose  n1  and  n2  are any integers, such that  F(n1) = F(n2).  Substituting from the definition of F gives that  2 − 3n1 =         .  Solving this equation for  n1  and simplifying the result gives that  n1 =         .  Therefore, F is one-to-one. (ii) Provide a counterexample to show that F is not onto. Let  m =         .  For this value of m, the only number n with the property that  F(N) = m  is not an integer. Thus F is not onto. (b) Define G: ℝ → ℝ by the rule  G(x) = 2 − 3x  for each real number x. Prove that G is onto. Proof that G is onto: Let y be any real number. (Scratch work: On a separate piece of paper, solve the equation  y = 2 − 3x for x.  Enter the result - an expression in y - in the box below.) x =        To finish the proof, we need to show (1) that x is a real number, and (2) that  G(x) = y. Now sums, products, and differences of real numbers are real numbers, and quotients of real numbers with nonzero denominators are also real numbers. Therefore, x is a real number. In addition, according to the formula that defines G, when G is applied to x, x is multiplied by 3 and the result is subtracted from 2. When the expression for x (using the variable y) is multiplied by 3, the result is         . And when the result is subtracted from 2, we obtain         . Thus  G(x) = y. Hence, there exists a number x such that x is a real number and  G(x) = y.  Therefore, G is onto.

(a) Define F: ℤ → ℤ by the rule  F(n) = 2 − 3n,  for each integer n. (i) Prove that F is one-to-one. Proof: Suppose  n1  and  n2  are any integers, such that  F(n1) = F(n2).  Substituting from the definition of F gives that  2 − 3n1 =         .  Solving this equation for  n1  and simplifying the result gives that  n1 =         .  Therefore, F is one-to-one. (ii) Provide a counterexample to show that F is not onto. Let  m =         .  For this value of m, the only number n with the property that  F(N) = m  is not an integer. Thus F is not onto. (b) Define G: ℝ → ℝ by the rule  G(x) = 2 − 3x  for each real number x. Prove that G is onto. Proof that G is onto: Let y be any real number. (Scratch work: On a separate piece of paper, solve the equation  y = 2 − 3x for x.  Enter the result - an expression in y - in the box below.) x =        To finish the proof, we need to show (1) that x is a real number, and (2) that  G(x) = y. Now sums, products, and differences of real numbers are real numbers, and quotients of real numbers with nonzero denominators are also real numbers. Therefore, x is a real number. In addition, according to the formula that defines G, when G is applied to x, x is multiplied by 3 and the result is subtracted from 2. When the expression for x (using the variable y) is multiplied by 3, the result is         . And when the result is subtracted from 2, we obtain         . Thus  G(x) = y. Hence, there exists a number x such that x is a real number and  G(x) = y.  Therefore, G is onto.

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter4: Polynomial And Rational Functions
Section4.5: Rational Functions
Problem 54E
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(a)
Define F: ℤ → ℤ by the rule 
F(n) = 2 − 3n,
 for each integer n.
(i)
Prove that F is one-to-one.
Proof: Suppose 
n1
 and 
n2
 are any integers, such that 
F(n1) = F(n2).
 Substituting from the definition of F gives that 
2 − 3n1 = 
 
 
 
 .
 Solving this equation for 
n1
 and simplifying the result gives that 
n1 = 
 
 
 
 .
 Therefore, F is one-to-one.
(ii)
Provide a counterexample to show that F is not onto.
Let 
m = 
 
 
 
 .
 For this value of m, the only number n with the property that 
F(N) = m
 is not an integer. Thus F is not onto.
(b)
Define G: ℝ → ℝ by the rule 
G(x) = 2 − 3x
 for each real number x. Prove that G is onto.
Proof that G is onto: Let y be any real number.
(Scratch work: On a separate piece of paper, solve the equation 
y = 2 − 3x for x.
 Enter the result - an expression in y - in the box below.)
x = 
 
 
 
To finish the proof, we need to show (1) that x is a real number, and (2) that 
G(x) = y.
Now sums, products, and differences of real numbers are real numbers, and quotients of real numbers with nonzero denominators are also real numbers. Therefore, x is a real number.
In addition, according to the formula that defines G, when G is applied to x, x is multiplied by 3 and the result is subtracted from 2.
When the expression for x (using the variable y) is multiplied by 3, the result is 
 
 
 
 . And when the result is subtracted from 2, we obtain 
 
 
 
 . Thus 
G(x) = y.
Hence, there exists a number x such that x is a real number and 
G(x) = y.
 Therefore, G is onto.

  

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