6. Show that 𝑓𝑓: ℝ → ℝ, 𝑓𝑓(𝑥𝑥) = 3𝑥+1/5
is a one-to-one correspondence
To show that a function is a one-to-one correspondence, we need to prove that it is both injective (one-to-one) and surjective (onto).
1. Injective (One-to-one):
If f(x) = f(y) for any x and y in the domain, then x = y.
Let's assume f(x) = f(y):
3x + 1/5 = 3y + 1/5
3x = 3y
x = y
This shows that the function is injective, as any two inputs that map to the same output must be equal.
2. Surjective (Onto):
For every y in the range, there exists at least one x in the domain such that f(x) = y.
Let's consider any y in the range ℝ. We need to find an x in the domain ℝ such that f(x) = y.
3x + 1/5 = y
3x = y - 1/5
x = (y - 1/5)/3
We can see that for any y in ℝ, there exists an x in ℝ that satisfies the equation. Therefore, the function is surjective.
Since the function is both injective and surjective, it can be concluded that 𝑓𝑓: ℝ → ℝ, 𝑓𝑓(𝑥𝑥) = 3𝑥+1/5 is a one-to-one correspondence.
To show that a function is a one-to-one correspondence, we need to prove that it is both injective (one-to-one) and surjective (onto).
1. Injectivity (one-to-one):
Let's assume that 𝑥₁ and 𝑥₂ are two arbitrary real numbers in the domain ℝ such that 𝑓(𝑥₁) = 𝑓(𝑥₂). This means that 3𝑥₁ + 1/5 = 3𝑥₂ + 1/5.
Now, let's solve for 𝑥₁ and 𝑥₂ to see if they are equal:
3𝑥₁ = 3𝑥₂
𝑥₁ = 𝑥₂
Since 𝑥₁ and 𝑥₂ are equal, we have shown that if 𝑓(𝑥₁) = 𝑓(𝑥₂), then 𝑥₁ = 𝑥₂. Therefore, the function 𝑓(𝑥) = 3𝑥 + 1/5 is injective (one-to-one).
2. Surjectivity (onto):
To show that the function is onto, we need to prove that for every real number 𝑦 in the codomain ℝ, there exists a real number 𝑥 in the domain ℝ such that 𝑓(𝑥) = 𝑦.
Let's consider any 𝑦 in ℝ. We need to find an 𝑥 such that 𝑓(𝑥) = 𝑦.
We have 𝑓(𝑥) = 3𝑥 + 1/5
Setting 𝑦 = 3𝑥 + 1/5, we can solve for 𝑥:
3𝑥 + 1/5 = 𝑦
3𝑥 = 𝑦 - 1/5
𝑥 = (𝑦 - 1/5)/3
Therefore, for any 𝑦 in ℝ, we have found an 𝑥 = (𝑦 - 1/5)/3 such that 𝑓(𝑥) = 𝑦. This shows that the function 𝑓(𝑥) = 3𝑥 + 1/5 is surjective (onto).
Since the function 𝑓(𝑥) = 3𝑥 + 1/5 is both injective (one-to-one) and surjective (onto), it is a one-to-one correspondence.
To show that a function 𝑓: ℝ → ℝ is a one-to-one correspondence, we need to demonstrate both injectivity and surjectivity.
Injectivity means that for every pair of distinct elements 𝑥 and 𝑦 in the domain, their images under 𝑓 will be distinct. In other words, if 𝑓(𝑥) = 𝑓(𝑦), then 𝑥 = 𝑦.
To prove this, let's assume 𝑓(𝑥) = 𝑓(𝑦) and show that 𝑥 = 𝑦. Plugging in the equation for 𝑓(𝑥) and 𝑓(𝑦), we get:
3𝑥 + 1/5 = 3𝑦 + 1/5
Notice that both sides of the equation are equal. By subtracting 1/5 from both sides, we get:
3𝑥 = 3𝑦
Finally, by dividing both sides by 3, we obtain:
𝑥 = 𝑦
This demonstrates that if 𝑓(𝑥) = 𝑓(𝑦), then 𝑥 = 𝑦, proving that 𝑓 is injective.
Surjectivity means that for every element 𝑦 in the codomain, there exists an element 𝑥 in the domain such that 𝑓(𝑥) = 𝑦. In other words, 𝑓 maps the entire domain to the entire codomain.
In our case, the codomain is ℝ, which represents the set of real numbers. Given any 𝑦 in ℝ, we can find a corresponding 𝑥 by solving the equation 𝑓(𝑥) = 𝑦.
Let's solve the equation 3𝑥 + 1/5 = 𝑦 for 𝑥. By subtracting 1/5 from both sides and dividing both sides by 3, we get:
3𝑥 = 𝑦 - 1/5
𝑥 = (𝑦 - 1/5)/3
So, for any 𝑦 in ℝ, we can find a corresponding 𝑥 given by (𝑦 - 1/5)/3. This shows that 𝑓 is surjective.
Since 𝑓 is both injective and surjective, 𝑓 is a one-to-one correspondence.