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Let $f$ be the following function from $\mathbb{R} \rightarrow \mathbb{Z}: f(x) = m$ where m is the unique integer such that $m \le x < m+1$. In other words, $f(x)$ is equal to the largest integer that is less than or equals to x. For instance, $f(2.3) = 2$, $f(1) = 1$, $f(-1.37) = -2$. Prove f is an increasing function.

I know that the definition of increasing function is "if for any $x,y \in A$ such that $x \le y$, we have $f(x) \le f(y)$." I have this unorganized ideas how to prove this problem, and I'm not sure how to proceed.

Let $x \le y$. Set $f(x) = m$ and $f(y) = n$ where $m \le x < m+1$ and $n \le y < n+1$. Then, we have $f(y)-f(x) = n - m$. Since $y \ge x$, we have $n - m \le y - x < n + 1 - m -1 \Rightarrow n - m \le y - x < n - m$. Hence, $f(y)-f(x) = n - m = y - x$.

I don't know how to connect this idea to show that $f(x) \le f(y)$. How do I go to next part? Do I even have the right idea?

EDIT

the definition of increasing function is "if for any $x,y \in A$ such that $x < y$, we have $f(x) \le f(y)$."

Here's my proof with the right definition.

Let $x < y$. Set $f(x) = m$ and $f(y) = n$ where $m \le x < m+1$ and $n \le y < n+1$. Then, we have $f(y)-f(x) = n - m$. Since $y > x$, we have $n - m \le y - x < n + 1 - m -1 \Rightarrow n - m \le y - x < n - m$. Hence, $f(y)-f(x) = n - m = y - x$. Since $y > x$, we have $f(y)-f(x) > 0$. Therefore, $f$ is an increasing function.

I don't know if this does make sense..

so0jong
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  • ahhh I noticed that there was a typo in my lecture slide with the definition. It should be "$x<y$" not $x \le y$.. – so0jong Mar 17 '21 at 23:53
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    It's not always true that $n-m \leq y-x$. If $x=\frac{3}{4}, y=\frac{5}{4}$, then $n-m=1$ and $y-x=\frac{1}{2}$. – aschepler Mar 17 '21 at 23:58
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    Your assertion that $n - m = y - x$ says $y = x$ is an integer, which need not be true. Rather than try to find your mistake, I suggest you draw a picture of the graph of $f$ and think about cases for the values of $y$ and $x$ depending on how far apart they are and whether they are themselves integers. – Ethan Bolker Mar 17 '21 at 23:58
  • I meant to say $n - m = y - x$ because $n - m \le y - x < n - m$. I thought $n - m \le y - x < n - m$ might imply that $y - x = n - m$. – so0jong Mar 18 '21 at 00:10

1 Answers1

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$m \leq x \leq y <n+1$. But $n$ and $m$ are integers so $m <n+1$ implies $m \leq n$.

Kavi Rama Murthy
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