如何证明函数可逆？含整数到非负整数分段函数的可逆性求证

Hey there! Let's tackle your two questions about function invertibility clearly and step by step.

A function is invertible if and only if it's a bijection—meaning it's both injective (one-to-one) and surjective (onto). Here's how to approach this:

• Prove injectivity (one-to-one):
  A function $f$ is injective if whenever $f(a) = f(b)$, it must follow that $a = b$. The standard method is to assume $f(a) = f(b)$ and then algebraically derive that $a = b$. Alternatively, you can show distinct inputs always map to distinct outputs ($a \neq b \implies f(a) \neq f(b)$), but the first approach is usually simpler.

• Prove surjectivity (onto):
  A function $f$ is surjective if every element in the codomain has at least one corresponding element in the domain that maps to it. For any $y$ in the codomain, you need to find an $x$ in the domain such that $f(x) = y$—typically by solving for $x$ in terms of $y$ and verifying $x$ is valid for the domain.

• Construct the inverse function directly:
  If you can define a function $f^{-1}$ that satisfies two conditions:

  1. For every $x$ in $f$'s domain, $f^{-1}(f(x)) = x$ (the inverse undoes $f$), and
  2. For every $y$ in $f$'s codomain, $f(f^{-1}(y)) = y$ (applying $f$ after the inverse gives back the original $y$),
     then $f$ is invertible by definition, since the inverse function exists.

First, let's restate the function definition for clarity:
$$f(x) = \begin{cases} 2x, & \mbox{if } x\mbox{≥0} \ -(2x+1), & \mbox{if } x\mbox{<0} \end{cases}$$
Where $\mathbb{Z}$ is the set of all integers, and $\mathbb{N_0}$ is the set of non-negative integers ($0, 1, 2, ...$).

We can prove this is invertible either by showing it's a bijection, or by constructing its inverse function. Let's cover both methods for thoroughness.

Method 1: Construct the inverse function #

Looking at how $f$ maps inputs to outputs:

• Non-negative integers $x \geq 0$ map to even non-negative integers: $0 \to 0$, $1 \to 2$, $2 \to 4$, etc.
• Negative integers $x < 0$ map to odd positive integers: $-1 \to 1$, $-2 \to 3$, $-3 \to 5$, etc.

Since every element in $\mathbb{N_0}$ is either even or odd, we can define the inverse function $f^{-1}: \mathbb{N_0} \mapsto \mathbb{Z}$ as:
$$f^{-1}(y) = \begin{cases} \frac{y}{2}, & \mbox{if } y\mbox{ is even} \ -\frac{y+1}{2}, & \mbox{if } y\mbox{ is odd} \end{cases}$$

Now verify this inverse works:

1. Check $f^{-1}(f(x)) = x$ for all $x \in \mathbb{Z}$:

   • If $x \geq 0$: $f(x) = 2x$ (even), so $f^{-1}(f(x)) = \frac{2x}{2} = x$.
   • If $x < 0$: Let $x = -k$ where $k > 0$. Then $f(x) = -(2(-k)+1) = 2k-1$ (odd). So $f^{-1}(f(x)) = -\frac{(2k-1)+1}{2} = -\frac{2k}{2} = -k = x$.

2. Check $f(f^{-1}(y)) = y$ for all $y \in \mathbb{N_0}$:

   • If $y$ is even: Let $y = 2k$ where $k \geq 0$. Then $f^{-1}(y) = k$, so $f(f^{-1}(y)) = 2k = y$.
   • If $y$ is odd: Let $y = 2k+1$ where $k \geq 0$. Then $f^{-1}(y) = -\frac{(2k+1)+1}{2} = -(k+1)$, so $f(f^{-1}(y)) = -(2(-(k+1))+1) = -(-2k-2+1) = -(-2k-1) = 2k+1 = y$.

Since both conditions hold, $f^{-1}$ is indeed the inverse of $f$, so $f$ is invertible.

Method 2: Prove $f$ is a bijection #

Step 1: Prove injectivity

Assume $f(a) = f(b)$. We have three cases:

• If $f(a)$ is even: Then $a \geq 0$ and $b \geq 0$, so $2a = 2b \implies a = b$.
• If $f(a)$ is odd: Then $a < 0$ and $b < 0$, so $-(2a+1) = -(2b+1) \implies 2a+1 = 2b+1 \implies a = b$.
• It's impossible for $f(a)$ to be even and $f(b)$ to be odd (since even ≠ odd), so the only valid cases lead to $a = b$. Thus $f$ is injective.

Step 2: Prove surjectivity

Take any $y \in \mathbb{N_0}$:

• If $y$ is even: Let $x = \frac{y}{2}$. Since $y$ is even and non-negative, $x$ is a non-negative integer (so $x \in \mathbb{Z}$), and $f(x) = 2x = y$.
• If $y$ is odd: Let $x = -\frac{y+1}{2}$. Since $y$ is odd, $y+1$ is even, so $\frac{y+1}{2}$ is a positive integer, making $x$ a negative integer (so $x \in \mathbb{Z}$). Then $f(x) = -(2x+1) = -(2(-\frac{y+1}{2})+1) = -(-y-1+1) = y$.

Every $y$ has a corresponding $x$, so $f$ is surjective.

Since $f$ is both injective and surjective, it's a bijection—hence invertible.

内容的提问来源于stack exchange，提问作者user529503