Exercise 1.2 · Q3: The greatest integer function

Basics you need

Solution

1. $[x]$ is well defined for every real $x$

   The intervals $[n, n+1)$⁠, one for each integer $n$⁠, are

   $$\dots,\ [-2,-1),\ [-1,0),\ [0,1),\ [1,2),\ [2,3),\ \dots$$

   They follow on from each other without gaps or overlaps, so every real $x$ lies in exactly one of them, say $[n, n+1)$⁠. Then $n \le x$⁠, so $n$ is an integer not exceeding $x$⁠; and every larger integer is at least $n+1 > x$⁠. So $n$ is the greatest integer $\le x$⁠, and $[x] = n$⁠. Thus $f$ gives each real number exactly one output, and that output is an integer.

   We need two facts from this step. "Exactly one output" confirms $f$ is a function from $\R$ to $\R$⁠. "The output is always an integer" is the key to the onto part. Writing $[x] = n \iff n \le x < n+1$ also gives a reliable way to compute values, including for negative numbers, where guessing often goes wrong.
   The greatest integer $\le x$ is the $n$ with $n\le x<n+1$ (Basic 1).

2. Not one-one: $f(1) = f(1.5)$

   Compute both values with the rule from Step 1:

   $$\begin{aligned} 1 \le 1 < 2 \ &\Longrightarrow\ f(1) = [1] = 1,\\ 1 \le 1.5 < 2 \ &\Longrightarrow\ f(1.5) = [1.5] = 1. \end{aligned}$$

   So $f(1) = f(1.5)$ although $1 \ne 1.5$⁠. Two different inputs have the same image, so $f$ is not one-one.

   One-one claims that equal outputs always come from equal inputs. A single pair that breaks this is a complete disproof. The pair is not special: every $x$ in $[1, 2)$ has $[x] = 1$⁠, so infinitely many inputs share the output $1$⁠. On the graph this is the flat step at height 1.
   One counterexample disproves a "for all" statement (Basic 4). Basics: how to prove and disprove

3. Not onto: $0.5$ is never an output

   The number $0.5$ belongs to the codomain $\R$⁠. By Step 1, $f(x) = [x]$ is an integer for every $x\in\R$⁠. Since $0 < 0.5 < 1$⁠, the number $0.5$ is not an integer, so

   $$f(x) \ne 0.5 \qquad \text{for every } x\in\R.$$

   No element of the domain is mapped to $0.5$⁠, so $f$ is not onto.

   To disprove onto we must choose a value in the codomain and show that no input reaches it. Checking a few inputs would not be enough; the argument "every output is an integer, and $0.5$ is not" covers all real inputs at once. Any non-integer real number ($\sqrt2$⁠, $-3.7$⁠, $\pi$⁠, …) would serve equally well. The range of $f$ is exactly $\Z$⁠, which is much smaller than the codomain $\R$⁠.
   Onto means every element of the codomain is an image (Basic 3). Definition 6, and the Remark after it: $f$ is onto if and only if its range equals the codomain. Basics: one-one and onto

4. Conclude, and note what the codomain does

   $f$ is not one-one (Step 2) and not onto (Step 3), so the greatest integer function $f : \R\to\R$ is neither one-one nor onto.

   If the codomain were $\Z$ instead of $\R$⁠, the function would become onto: for every integer $m$⁠, $m \le m < m+1$⁠, so $[m] = m$⁠, and $m$ is the image of itself. It would still not be one-one, because $[1] = [1.5]$ does not depend on the codomain.

   Onto is a statement about the codomain, so changing the codomain can change the answer. One-one depends only on which inputs share outputs, so changing the codomain cannot repair it. This is why the question has to say $f : \R\to\R$⁠, and why both sets matter.

Every fact used, and where it comes from