6. Show that 𝑓: ℝ → ℝ, 𝑓(𝑥) = 3𝑥+1/5 is a one-to-one correspondence
To show that a function is a one-to-one correspondence, we need to demonstrate that it is both injective (one-to-one) and surjective (onto).
1. Injective: Suppose 𝑓(𝑥₁) = 𝑓(𝑥₂) for some 𝑥₁, 𝑥₂ ∈ ℝ. Then we have:
3𝑥₁ + 1/5 = 3𝑥₂ + 1/5
Subtracting 1/5 from both sides:
3𝑥₁ = 3𝑥₂
Dividing both sides by 3:
𝑥₁ = 𝑥₂
Thus, 𝑓(𝑥) = 3𝑥 + 1/5 is injective.
2. Surjective: To show that 𝑓(𝑥) = 3𝑥 + 1/5 is onto, we need to demonstrate that for every 𝑦 ∈ ℝ, there exists 𝑥 ∈ ℝ such that 𝑓(𝑥) = 𝑦. Let 𝑦 be any real number. We can solve the equation 𝑓(𝑥) = 𝑦 for 𝑥 as follows:
3𝑥 + 1/5 = 𝑦
Subtracting 1/5 from both sides:
3𝑥 = 𝑦 - 1/5
Dividing both sides by 3:
𝑥 = (𝑦 - 1/5)/3
Thus, for any 𝑦 ∈ ℝ, we can find 𝑥 = (𝑦 - 1/5)/3 such 𝑓(𝑥) = 𝑦.
Since 𝑓(𝑥) = 3𝑥 + 1/5 is both injective and surjective, it is a one-to-one correspondence.
To show that the function 𝑓(𝑥) = 3𝑥 + 1/5 is a one-to-one correspondence, we need to prove two things:
1. 𝑓(𝑥) is injective (one-to-one)
2. 𝑓(𝑥) is surjective (onto)
Let's start with the injectivity:
1. Injectivity (One-to-One):
To show that 𝑓(𝑥) is injective, we need to prove that 𝑓(𝑥₁) = 𝑓(𝑥₂) implies 𝑥₁ = 𝑥₂ for any 𝑥₁, 𝑥₂ ∈ ℝ.
Let's assume 𝑓(𝑥₁) = 𝑓(𝑥₂):
Then, we have:
3𝑥₁ + 1/5 = 3𝑥₂ + 1/5
Subtracting 1/5 from both sides:
3𝑥₁ = 3𝑥₂
Dividing by 3:
𝑥₁ = 𝑥₂
This shows that 𝑓(𝑥) is injective.
2. Surjectivity (Onto):
To show that 𝑓(𝑥) is surjective, we need to prove that for every 𝑦 in the codomain (ℝ), there exist 𝑥 in the domain (ℝ) such that 𝑓(𝑥) = 𝑦.
Let 𝑦 be any real number in the codomain.
We want to find 𝑥 such that 𝑓(𝑥) = 𝑦:
𝑓(𝑥) = 3𝑥 + 1/5
Setting 𝑦 = 3𝑥 + 1/5, we can solve the equation for 𝑥:
3𝑥 + 1/5 = 𝑦
Subtracting 1/5 from both sides:
3𝑥 = 𝑦 - 1/5
Dividing by 3:
𝑥 = (𝑦 - 1/5)/3
Since the equation 𝑥 = (𝑦 - 1/5)/3 holds for any 𝑦 in the codomain (ℝ), 𝑓(𝑥) is surjective.
Therefore, we have shown that 𝑓(𝑥) = 3𝑥 + 1/5 is both injective (one-to-one) and surjective (onto), which proves that it is a one-to-one correspondence.
To show that the function 𝑓: ℝ → ℝ, 𝑓(𝑥) = 3𝑥 + 1/5 is a one-to-one correspondence, we need to prove that the function is both injective (one-to-one) and surjective (onto).
First, let's prove that the function 𝑓 is injective. A function is injective if distinct elements in the domain map to distinct elements in the codomain.
To show this, let 𝑦₁ and 𝑦₂ be two distinct elements in the codomain ℝ. We need to prove that 𝑓(𝑥₁) ≠ 𝑓(𝑥₂) for any 𝑥₁ and 𝑥₂ in the domain ℝ.
Suppose 𝑦₁ = 𝑓(𝑥₁) = 3𝑥₁ + 1/5 and 𝑦₂ = 𝑓(𝑥₂) = 3𝑥₂ + 1/5. To prove that 𝑓 is injective, we need to show that if 𝑦₁ ≠ 𝑦₂, then 𝑥₁ ≠ 𝑥₂.
So, if 𝑦₁ ≠ 𝑦₂, then we have 3𝑥₁ + 1/5 ≠ 3𝑥₂ + 1/5. By subtracting 1/5 from both sides, we get 3𝑥₁ ≠ 3𝑥₂. Dividing both sides by 3, we have 𝑥₁ ≠ 𝑥₂.
Therefore, when 𝑦₁ ≠ 𝑦₂, 𝑥₁ ≠ 𝑥₂, which establishes that the function 𝑓 is injective.
Next, let's prove that the function 𝑓 is surjective. A function is surjective if every element 𝑦 in the codomain ℝ has a corresponding element 𝑥 in the domain ℝ such that 𝑦 = 𝑓(𝑥).
Let 𝑦 be an arbitrary element in the codomain ℝ. We need to find an 𝑥 such that 𝑦 = 𝑓(𝑥).
So, let's solve the equation 𝑦 = 3𝑥 + 1/5 for 𝑥. Rearranging the equation, we have 𝑥 = (𝑦 - 1/5)/3.
Now, we have found an 𝑥 such that 𝑦 = 𝑓(𝑥). Therefore, the function 𝑓 is surjective.
Since 𝑓 is both injective and surjective, it is a one-to-one correspondence.
Therefore, we have shown that the function 𝑓: ℝ → ℝ, 𝑓(𝑥) = 3𝑥 + 1/5 is a one-to-one correspondence.