MAT-314 Relations and Functions

MAT-314 Relations and Functions

• 1.
  Abstract Algebra, Saracino SecondEdition Section 7 Relations and Functions
• 2.
  Lehman College, Departmentof Mathematics Relations (Definitions) (1 of 2) Let 𝐴 and 𝐵 be two nonempty sets, then a relation (or binary relation) ℛ from 𝐴 into 𝐵 is a set of ordered pairs (𝑥, 𝑦), where 𝑥 ∈ 𝐴 and 𝑦 ∈ 𝐵. Recall the Cartesian Product of 𝐴 and 𝐵, defined as the set: 𝐴 × 𝐵 = 𝑎, 𝑏 𝑎 ∈ 𝐴, 𝑏 ∈ 𝐵 What is the difference between the two definitions? Answer: The cartesian product 𝐴 × 𝐵 is the set of all ordered pairs, while the relation ℛ is a subset of 𝐴 × 𝐵. For example, let 𝐴 = ℤ and let 𝐵 = ℤ, then the set ℛ1 = 1, 2 , 2, 5 , 3, −4 , 5, 2 is a relation from ℤ into ℤ, but it is not all of ℤ × ℤ.
• 3.
  Lehman College, Departmentof Mathematics Relations (Definitions) (2 of 2) The set ℛ2 = 𝑥, 𝑦 ∈ ℝ × ℝ | 𝑥2 + 𝑦2 = 1 is a relation from ℝ into ℝ (the unit circle centered at the origin). It is not the entire cartesian plane ℝ2 = ℝ × ℝ. Let 𝐴 and 𝐵 be two nonempty sets and let ℛ be a relation from 𝐴 into 𝐵. The domain of ℛ is the set 𝑥 ∈ 𝐴 | ∃ 𝑦 ∈ 𝐵, such that 𝑥, 𝑦 ∈ 𝑅 That is, the set of all first elements of ordered pairs in ℛ. The image or range of ℛ is the set 𝑦 ∈ 𝐵 | ∃ 𝑥 ∈ 𝐴, such that 𝑥, 𝑦 ∈ 𝑅
• 4.
  Lehman College, Departmentof Mathematics Relations (Examples) Example 1: State the domain and image (range) of the relation ℛ1 = 1, 2 , 2, 5 , 3, −4 , 5, 2 Answer: Domain: 1, 2, 3, 5 ; Range: −4, 2, 5 . Example 2: State the domain and image (range) of the relation ℛ2 = 𝑥, 𝑦 ∈ ℝ × ℝ | 𝑥2 + 𝑦2 = 1 Answer: Domain: −1, 1 ; Range: −1, 1 . Example 3: State the domain and image (range) of the relation ℛ3 = 1, 7 , 3, −2 , 1, 5 , 2, 2 Answer: Domain: 1, 2, 3 ; Range: −2, 2, 5, 7 . Example 4: State the domain and image (range) of the relation ℛ4 = 𝑥, 𝑦 ∈ ℝ × ℝ | 𝑦 = 𝑥2 + 1 Answer: Domain: (−∞, ∞); Range: [1, ∞).
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  Lehman College, Departmentof Mathematics Inverse Relations (Definitions) (1 of 2) Let ℛ be a relation from set 𝐴 into set 𝐵. The inverse of ℛ, denoted ℛ−1 , is the relation from 𝐵 into 𝐴 consisting of those ordered pairs 𝑦, 𝑥 which, when reversed, belong to ℛ. ℛ−1 = 𝑦, 𝑥 ∈ 𝐵 × 𝐴 𝑥, 𝑦 ∈ ℛ Example 1: State the inverse 𝑅1 −1 of the relation ℛ1 = 1, 2 , 2, 5 , 3, −4 , 5, 2 Answer: ℛ1 −1 = 2, 1 , 5, 2 , −4, 3 , 2, 5 . Example 2: State the inverse ℛ5 −1 of the relation ℛ5 = 1, 2 , 3, 3 , 4, 5 , 5, 1 Answer: ℛ5 −1 = 2, 1 , 3, 3 , 5, 4 , 1, 5 .
• 6.
  Lehman College, Departmentof Mathematics Functions (1 of 5) Let 𝐴 and 𝐵 be two nonempty sets, a function 𝑓 from 𝐴 into 𝐵 is a relation from 𝐴 into 𝐵 such that each element of the domain of 𝑓 is assigned one and only one element of the image of 𝑓. Which of the following relations are functions? 1. ℛ1 = 1, 2 , 2, 5 , 3, −4 , 5, 2 2. ℛ2 = 𝑥, 𝑦 ∈ ℝ × ℝ | 𝑥2 + 𝑦2 = 1 3. ℛ3 = 1, 7 , 3, −2 , 1, 5 , 2, 2 4. ℛ4 = 𝑥, 𝑦 ∈ ℝ × ℝ | 𝑦 = 𝑥2 + 1 5. ℛ5 = 1, 2 , 3, 3 , 4, 5 , 5, 1
• 7.
  Lehman College, Departmentof Mathematics Functions (2 of 5) 1. ℛ1 = 1, 2 , 2, 5 , 3, −4 , 5, 2 2. ℛ2 = 𝑥, 𝑦 ∈ ℝ × ℝ | 𝑥2 + 𝑦2 = 1 3. ℛ3 = 1, 7 , 3, −2 , 1, 5 , 2, 2 4. ℛ4 = 𝑥, 𝑦 ∈ ℝ × ℝ | 𝑦 = 𝑥2 + 1 5. ℛ5 = 1, 2 , 3, 3 , 4, 5 , 5, 1 Answers: ℛ1, ℛ4 and ℛ5 are functions. ℛ3 is not a function, since 1 7 ∈ ℛ3 and 1, 5 ∈ ℛ3 ℛ2 is not a function, since 0, 1 ∈ ℛ2 and 0, −1 ∈ ℛ2 (this is called the vertical line test in precalculus)
• 8.
  Lehman College, Departmentof Mathematics Functions (3 of 5) Let A and B be two nonempty sets and let 𝑓 be a function from 𝐴 into 𝐵. We will write 𝑓: 𝐴 → 𝐵 and say 𝑓 is a function from 𝐴 to 𝐵, or 𝑓 maps 𝐴 into 𝐵. If 𝑓 maps an element 𝑥 ∈ 𝐴 to an element 𝑦 ∈ 𝐵. That is, 𝑥, 𝑦 ∈ 𝑓, we write 𝑓 𝑥 = 𝑦. We say 𝑓 is onto 𝐵 if for every 𝑦 ∈ 𝐵 there exists 𝑥 ∈ 𝐴, such that 𝑓 𝑥 = 𝑦. Example 1: Consider the function 𝑓: ℝ → ℝ, such that 𝑓 𝑥 = 𝑥2 . Is 𝑓 onto ℝ? Answer: No, 𝑓 is not onto ℝ, since −1 is not in the image of 𝑓. That is, the equation 𝑥2 = −1 has no solution in ℝ.
• 9.
  Lehman College, Departmentof Mathematics Functions (4 of 5) Let 𝐴 and 𝐵 be sets and let 𝑓: 𝐴 → 𝐵 be a function. We say 𝑓 is one-to-one if for two different elements 𝑥1, 𝑥2 ∈ 𝐴 (𝑥1 ≠ 𝑥2) we have 𝑓 𝑥1 ≠ 𝑓 𝑥2 . Which of the following functions are one-to-one? ℛ1 = 1, 2 , 2, 5 , 3, −4 , 5, 2 ℛ5 = 1, 2 , 3, 3 , 4, 5 , 5, 1 Answer: ℛ1 is not one-to-one, but ℛ5 is one-to-one.  ℛ1 is not one-to-one since 1, 2 ∈ ℛ1 and 5, 2 ∈ ℛ2, so 1 ≠ 5, but 𝑓 1 = 𝑓 5 .  ℛ5 is one-to-one since the second element in each pair does not repeat (this is called the horizontal line test in precalculus).
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  Lehman College, Departmentof Mathematics Functions (5 of 5) (Contrapositive definition) Let 𝐴 and 𝐵 be sets and let 𝑓: 𝐴 → 𝐵 be a function. We say 𝑓 is one-to-one if for 𝑥1, 𝑥2 ∈ 𝐴, 𝑓 𝑥1 = 𝑓 𝑥2 implies 𝑥1 = 𝑥2. Example 1: Consider the function 𝑓: ℝ → ℝ, such that 𝑓 𝑥 = 𝑥. Is 𝑓 one-to-one? Answer: Yes, 𝑓 is one-to-one, since if 𝑥1 ≠ 𝑥2 then 𝑓 𝑥1 = 𝑥1 and 𝑓 𝑥2 = 𝑥2, so 𝑓 𝑥1 ≠ 𝑓 𝑥2 . Example 2: Consider the function 𝑓: ℝ → ℝ, such that 𝑓 𝑥 = 𝑥2 . Is 𝑓 one-to-one? Answer: No, 𝑓 is not one-to-one, since −1 ≠ 1, but 𝑓 −1 = 1 and 𝑓 1 = 1, so 𝑓 −1 = 𝑓 1 .
• 11.
  Lehman College, Departmentof Mathematics Functions (5 of 5) Example 3: Consider the function 𝑔: ℝ → ℝ, such that 𝑔 𝑥 = 𝑥3 . Is 𝑔 one-to-one? Is 𝑔 onto ℝ? Answer: 𝑔 is both one-to-one and onto ℝ. Proof: Let 𝑎3 = 𝑏3 for some 𝑎, 𝑏 ∈ ℝ. If 𝑎 = 0, then 𝑏 = 0, and if 𝑏 = 0, then 𝑎 = 0, so 𝑎 = 𝑏. We will limit ourselves to 𝑎 ≠ 0 and 𝑏 ≠ 0. Since 𝑎3 = 𝑏3 , then 𝑎3 − 𝑏3 = 0, and We will show if 𝑎 ≠ 0 and 𝑏 ≠ 0, then 𝑎2 + 𝑎𝑏 + 𝑏2 > 0. 0 = 𝑎3 − 𝑏3 = (𝑎 − 𝑏)(𝑎2 + 𝑎𝑏 + 𝑏2 ) 𝑎2 + 𝑎𝑏 + 𝑏2 = 2𝑎2 + 2𝑎𝑏 + 2𝑏2 2 = 𝑎 + 𝑏 2 + 𝑎2 + 𝑏2 2 > 0
• 12.
  Lehman College, Departmentof Mathematics Functions (5 of 5) Example 3 (cont’d): Since, if 𝑎3 − 𝑏3 with 𝑎 ≠ 0, 𝑏 ≠ 0 and Then, 𝑎 − 𝑏 = 0, and 𝑎 = 𝑏. So, 𝑔 is one-to-one. To show that 𝑔 is onto ℝ, let 𝑏 ∈ ℝ be arbitrary and suppose 𝑏 = 𝑔(𝑥) for some 𝑥. Then 𝑏 = 𝑥3 , and we have the equation 𝑥3 − 𝑏 = 0. Now, lim 𝑥→∞ (𝑥3 −𝑏) = ∞ and lim 𝑥→−∞ (𝑥3 − 𝑏) = −∞, so since 𝑥3 − 𝑏 is continuous on ℝ, then by the Intermediate Value Theorem, 𝑥3 − 𝑏 = 0 has at least one solution 𝑥 ∈ ℝ. Therefore, 𝑔 is onto ℝ. 0 = 𝑎3 − 𝑏3 = (𝑎 − 𝑏)(𝑎2 + 𝑎𝑏 + 𝑏2 ) 𝑎2 + 𝑎𝑏 + 𝑏2 = 𝑎 + 𝑏 2 + 𝑎2 + 𝑏2 2 > 0
• 13.
  Lehman College, Departmentof Mathematics Inverse Functions (1 of 2) Let 𝐴 and 𝐵 be sets and let 𝑓: 𝐴 → 𝐵 be a function. We say 𝑓 is invertible if the inverse relation 𝑓−1 from 𝐵 to 𝐴 is also a function. We will write 𝑓−1 : 𝐵 → 𝐴. Which of the following functions are invertible? If so, what is the inverse? ℛ1 = 1, 2 , 2, 5 , 3, −4 , 5, 2 ℛ5 = 1, 2 , 3, 3 , 4, 5 , 5, 1 Answer: 𝑅1 is not invertible, but 𝑅5 is invertible. ℛ1 −1 = 2, 1 , 5, 2 , −4, 3 , 2, 5 , so ℛ1 is not inversible since 2, 1 ∈ ℛ1 −1 and 2, 5 ∈ ℛ1 −1 ℛ5 −1 = 2, 1 , 3, 3 , 5, 4 , 1, 5 is a function, so ℛ5 is invertible.
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  Lehman College, Departmentof Mathematics Inverse Functions (2 of 2) For a function to be invertible we see that it must be one-to-one. Theorem 1. Let 𝐴 and 𝐵 be nonempty sets and let 𝑓: 𝐴 → 𝐵 be a function. Then 𝑓 is invertible if and only if 𝑓 is one-to-one and onto 𝐵. Example 3: Let 𝐺 be a group and let 𝑥 be a fixed element of 𝐺. Define a function 𝑓𝑥: 𝐺 → 𝐺 by 𝑓 𝑎 = 𝑥𝑎 for all 𝑎 ∈ 𝐺. Is 𝑓𝑥 one-to-one? Is 𝑓𝑥 onto 𝐺? Answer: Let 𝑎, 𝑏 ∈ 𝐺, then if 𝑓𝑥 𝑎 = 𝑓𝑥(𝑏), we have 𝑥𝑎 = 𝑥𝑏. By left-cancellation 𝑎 = 𝑏, so 𝑓𝑥 is one-to-one. Let 𝑏 be an arbitrary element of 𝐺. Is there an 𝑎 ∈ 𝐺, such that 𝑏 = 𝑓𝑥(𝑎)? Set 𝑏 = 𝑥𝑎, left-multiplying by 𝑥−1 ∈ 𝐺, we obtain 𝑥−1 𝑏 = 𝑥−1 𝑥 = (𝑥−1 𝑥)𝑎 = 𝑎. Hence, 𝑓𝑥 is onto 𝐺.
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  Lehman College, Departmentof Mathematics Homomorphisms of Groups (1 of 5) Let (𝐺1,∗) and (𝐺2,∘) be two groups with identity elements 𝑒1 and 𝑒2, respectively. Let 𝜑: 𝐺1 → 𝐺2 be a function, such that: Then 𝜑 is called a homomorphism from 𝐺1to 𝐺2. Here, the functions ∗ and ∘ are the group operations in 𝐺1 and 𝐺2, respectively. Let us establish that homomorphisms do exist. Let (𝐺1,∗) and (𝐺2,∘) be two groups with identity elements 𝑒1 and 𝑒2, respectively. Let 𝑓: 𝐺1 → 𝐺2 be given by 𝑓 𝑎 = 𝑒2 for all 𝑎 ∈ 𝐺1. That is, the function 𝑓 sends every element of 𝐺1 to the identity element 𝑒2 of 𝐺2. 𝜑 𝑎 ∗ 𝑏 = 𝜑 𝑎 ∘ 𝜑(𝑏) for all 𝑎, 𝑏 ∈ 𝐺1
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  Lehman College, Departmentof Mathematics Homomorphisms of Groups (2 of 5) To show that 𝑓 is a homomorphism. we need to show that Let 𝑎, 𝑏 ∈ 𝐺1 be arbitrary, then since 𝑓 sends every every element of 𝐺1 to 𝑒2 ∈ 𝐺2. It follows that 𝑓 𝑎 ∘ 𝑓 𝑏 = 𝑒2 ∘ 𝑒2 = 𝑒2 and 𝑓 𝑎 ∗ 𝑏 = 𝑒2, so 𝑓 𝑎 ∗ 𝑏 = 𝑓 𝑎 ∘ 𝜑 𝑏 , and 𝑓 is a homomorphism. 𝑓 is called the trivial homomorphism. Example 4. Conjugation by a fixed element is a group homomorphism. Let 𝐺 be a group and let 𝑥 be a fixed element of 𝐺. Define a function𝑓𝑥: 𝐺 → 𝐺 by 𝑓𝑥 𝑎 = 𝑥𝑎𝑥−1 for all 𝑎 ∈ 𝐺. Then, 𝑓𝑥is a homomorphism from 𝐺 to itself. 𝑓 𝑎 ∗ 𝑏 = 𝑓 𝑎 ∘ 𝑓(𝑏) for all 𝑎, 𝑏 ∈ 𝐺1 𝑓 𝑎 = 𝑓 𝑏 = 𝑓 𝑎 ∗ 𝑏 = 𝑒2
• 17.
  Lehman College, Departmentof Mathematics Homomorphisms of Groups (3 of 5) To show that 𝑓𝑥 is a homomorphism. we need to show that Let 𝑎, 𝑏 ∈ 𝐺 be arbitrary, then. On one hand, On the other hand, It follows that 𝑓𝑥 𝑎𝑏 = 𝑓𝑥 𝑎 𝜑(𝑏), and 𝜑 is a homomorphism from 𝐺 to itself. In the attached homework, we will show that 𝑓𝑥 is one-to-one and onto 𝐺. 𝑓𝑥 𝑎𝑏 = 𝑓𝑥 𝑎 𝑓𝑥(𝑏) for all 𝑎, 𝑏 ∈ 𝐺 𝑓𝑥 𝑎 = 𝑥𝑎𝑥−1 and 𝑓𝑥 𝑏 = 𝑥𝑏𝑥−1 𝑓𝑥 𝑎 𝑓𝑥 𝑏 = 𝑥𝑎𝑥−1 𝑥𝑏𝑥−1 = 𝑥𝑎 𝑥−1 𝑥 𝑏𝑥−1 = 𝑥𝑎𝑏𝑥−1 𝑓𝑥 𝑎𝑏 = 𝑥𝑎𝑏𝑥−1
• 18.
  Lehman College, Departmentof Mathematics Homomorphisms of Groups (4 of 5) One-to-one and onto homomorphisms of groups are called isomorphisms. An Isomorphism of a group 𝐺 onto itself is called an automorphism of 𝐺. Let (𝐺1,∗) and (𝐺2,∘) be two groups with identity elements 𝑒1 and 𝑒2, respectively. Let 𝜑: 𝐺1 → 𝐺2 be a group homomorphism, then 1) 𝜑 𝑒1 = 𝑒2 2) 𝜑 𝑎 −1 = 𝜑(𝑎−1 ) for all 𝑎 ∈ 𝐺1 3) The set 𝜑 𝐺1 = 𝜑 𝑎 ∈ 𝐺2 | 𝑎 ∈ 𝐺1 , called the image of 𝜑, is a subgroup of 𝐺2 (recall range space of a linear operator). 4) The set Ker 𝜑 = 𝑎 ∈ 𝐺1 | 𝜑 𝑎 = 𝑒2 , called the kernel of 𝜑, is a subgroup of 𝐺1 (recall null space of a linear operator).
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  Lehman College, Departmentof Mathematics Homomorphisms of Groups (5 of 5) We will establish the first two results: Now in 𝐺1, 𝑒1 ∗ 𝑒1 = 𝑒1, so But in the group 𝐺2, 𝜑 𝑒1 = 𝜑 𝑒1 ∘ 𝑒2 so, we have By left-cancellation of 𝜑 𝑒1 in 𝐺2, we have 𝑒2 = 𝜑 𝑒1 For any element 𝑎 ∈ 𝐺1, we have 𝑎 ∗ 𝑎−1 = 𝑒1, so But in 𝐺2, 𝜑 𝑎 ∘ 𝜑 𝑎 −1 = 𝑒2, so by uniqueness of inverses in 𝐺2, 𝜑 𝑎 −1 = 𝜑(𝑎−1 ) 𝜑 𝑒1 = 𝜑 𝑒1 ∗ 𝑒1 = 𝜑 𝑒1 ∘ 𝜑 𝑒1 𝜑 𝑒1 ∘ 𝑒2 = 𝜑 𝑒1 ∘ 𝜑 𝑒1 𝑒2 = 𝜑 𝑒1 = 𝜑 𝑎 ∗ 𝑎−1 = 𝜑 𝑎 ∘ 𝜑(𝑎−1 ) 𝜑 𝑒1 = 𝑒2 and 𝜑 𝑎 −1 = 𝜑(𝑎−1 ) for all 𝑎 ∈ 𝐺1
• 20.
  Lehman College, Departmentof Mathematics Homework 7. Part 1 (due 04/04/2020) 7.3 Let 𝑓: ℝ → ℝ be given by 𝑓 𝑥 = 𝑎𝑥 + 𝑏, where 𝑎 and 𝑏 are fixed constants. a) Show that if 𝑎 ≠ 0, then 𝑓 is one-to-one and onto ℝ, so that 𝑓−1 exists. b) Assuming that 𝑎 ≠ 0, find an explicit formula for the inverse function 𝑓−1 . 7.7 Let 𝐺 be a group and let 𝑥 be a fixed element of 𝐺. Define a function 𝑓𝑥: 𝐺 → 𝐺 by 𝑓𝑥 𝑎 = 𝑥𝑎𝑥−1 for each 𝑎 ∈ 𝐺. Is 𝑓𝑥 one-to-one? Is 𝑓𝑥 onto 𝐺? 7. 8 Let 𝐺 be a group. Define a function 𝑓: 𝐺 → 𝐺 by 𝑓 𝑎 = 𝑎−1 for each 𝑎 ∈ 𝐺. Is 𝑓 one-to-one? Is 𝑓 onto 𝐺?