Exercise 1.2 · Q1: $f(x)=1/x$ on the non-zero reals

Basics you need

Solution

1. Check that $f$ maps $\R_*$ into $\R_*$

   Take any $x\in\R_*$⁠, so $x \ne 0$⁠. Then $\frac1x$ is defined, and it is a single real number. It is not zero, because

   $$x \cdot \frac1x = 1 \ne 0,$$

   and if $\frac1x$ were $0$⁠, the product on the left would be $0$⁠. So $f(x) = \frac1x \in \R_*$ for every $x \in \R_*$⁠: $f$ is a genuine function from $\R_*$ to $\R_*$⁠.

   The question states $f : \R_*\to\R_*$⁠, and both halves of the proof depend on it. One-one needs $f(x)$ to be defined for every input (that is why $0$ is excluded from the domain), and onto only makes sense if all outputs lie in the stated codomain (that is why $0$ is excluded from the codomain too: $\frac1x$ is never $0$⁠).
   A product of real numbers is zero only if one of the factors is zero. Since $x\cdot\frac1x = 1$⁠, neither factor can be zero (Basic 4).

2. One-one: equal reciprocals force equal numbers

   Let $x_1, x_2\in\R_*$ be arbitrary and suppose $f(x_1) = f(x_2)$⁠, that is,

   $$\frac{1}{x_1} = \frac{1}{x_2}.$$

   Multiply both sides by $x_1x_2$⁠, which is non-zero because $x_1 \ne 0$ and $x_2 \ne 0$⁠:

   $$\begin{aligned} x_1x_2\cdot\frac{1}{x_1} &= x_1x_2\cdot\frac{1}{x_2}\\ x_2 &= x_1. \end{aligned}$$

   So $f(x_1)=f(x_2)$ implies $x_1 = x_2$⁠, and $f$ is one-one.

   One-one is a statement about every pair of inputs, so we work with arbitrary $x_1, x_2$⁠, not with chosen numbers. Clearing the fractions turns the equation into $x_2 = x_1$ directly. Multiplying by $x_1x_2$ is allowed only because it is not zero; multiplying by zero would turn any equation into $0 = 0$ and prove nothing.
   Definition 5: $f$ is one-one if $f(x_1)=f(x_2) \Rightarrow x_1=x_2$ for all $x_1, x_2$ in the domain (Basic 2). Basics: one-one and onto

3. Onto: every non-zero $y$ is a reciprocal

   Take an arbitrary $y\in\R_*$⁠. We want $x\in\R_*$ with $\frac1x = y$⁠. Solving for $x$ (multiply both sides by $x$⁠, then divide by $y\ne0$⁠) suggests $x = \frac1y$⁠. Now check this candidate:

   • $x = \frac1y$ is defined because $y \ne 0$⁠, and it is non-zero by Step 1, so $x\in\R_*$⁠;
   • $f(x) = \dfrac{1}{1/y} = y$⁠.

   Every $y\in\R_*$ is the image of $\frac1y\in\R_*$⁠, so $f$ is onto. Together with Step 2, $f : \R_*\to\R_*$ is one-one and onto (bijective).

   Onto is again a "for every $y$" claim, so we take a general $y$ and build an input. Solving $f(x) = y$ tells us where to look; the two checks then prove that the candidate is a legal input (it lies in the domain $\R_*$⁠) and that it really works. Here the inverse of $f$ is $f$ itself: applying the reciprocal twice gives back the number you started with.
   Definition 6: $f : X\to Y$ is onto if every $y\in Y$ equals $f(x)$ for some $x\in X$ (Basic 3). Dividing 1 by a fraction flips it: $\frac{1}{1/y} = y$⁠.

4. The second question: same rule, new domain $\N$

   Now consider $g : \N\to\R_*$⁠, $g(x) = \frac1x$⁠. It is a function into $\R_*$⁠: every $n\in\N$ is non-zero, so $\frac1n$ is defined and, by Step 1, non-zero.

   The formula has not changed, but the set of inputs has shrunk from all non-zero reals to $\{1, 2, 3, \dots\}$⁠, while the codomain is still all of $\R_*$⁠. So both properties have to be tested again.

   One-one and onto are properties of the triple (domain, codomain, rule). Shrinking the domain can only remove inputs, so it can never create two inputs with the same output (one-one is safe), but it can leave outputs unreached (onto is at risk). The next two steps confirm exactly this.
   A function is its domain, codomain and rule together (Basic 1). Basics: what a function is

5. $g$ is still one-one

   Let $n_1, n_2\in\N$ with $g(n_1) = g(n_2)$⁠, that is, $\frac{1}{n_1} = \frac{1}{n_2}$⁠. Both are non-zero, so multiplying by $n_1n_2$ exactly as in Step 2 gives $n_2 = n_1$⁠. Hence $g$ is one-one.

   The algebra of Step 2 used only that $x_1$ and $x_2$ are non-zero real numbers. Natural numbers are non-zero real numbers, so the same argument applies word for word. (In general, restricting a one-one function to a smaller domain keeps it one-one.)

6. $g$ is not onto: the value $2$ is never reached

   The number $2$ belongs to the codomain $\R_*$⁠. Suppose some $n\in\N$ had $g(n) = 2$⁠. Then

   $$\frac1n = 2 \ \Longrightarrow\ 1 = 2n \ \Longrightarrow\ n = \frac12.$$

   But $\frac12$ is not a natural number, so no $n\in\N$ has $g(n) = 2$⁠. Therefore $g$ is not onto.

   In fact far more is missed. For $n \in \N$ we have $n \ge 1$⁠, so $0 < \frac1n \le 1$⁠: every output of $g$ lies in $(0, 1]$⁠. Every negative number (such as $-1$⁠) and every number greater than $1$ is missed, and so is a value like $0.3$ inside $(0,1]$⁠, since $\frac1n = 0.3$ would need $n = \frac{10}{3}$⁠.

   To disprove onto we need one concrete value of the codomain and a proof that nothing in the domain maps to it. Solving $\frac1n = 2$ finds the only real candidate, $n=\frac12$ (Step 2 showed the reciprocal is one-one, so there cannot be another). Since that single candidate is not in the domain $\N$⁠, the value 2 is unreached. Compare Step 3: on $\R_*$ the same candidate $\frac12$ was allowed, which is exactly why $f$ was onto.
   To disprove "for every $y$ there is an $x$", produce one $y$ for which there is no $x$ (Basic 3). Basics: how to prove and disprove

7. Answer the question

   The "result" was that the function is one-one and onto. With domain $\N$ and codomain $\R_*$⁠, the function is one-one but not onto. So the result is not true.

   The question asks whether "the result" (both properties together) still holds. One property failing is enough for the answer "No", but a complete answer says which property survives and which fails, with a reason for each.

Every fact used, and where it comes from