I have a discrete math question below with a solution written by my teacher. I'm really lost as to what answers I'm trying to find exactly. I don't understand how the teacher got 1 and -1 for the first two values, and I also don't understand how and why the last line turns $⌈(𝟔.𝟑 − ⌈𝟏.𝟕⌉)⌉$ into $⌈𝟔.𝟑 − 𝟐⌉ = ⌈𝟒.𝟑⌉ = 𝟓$

I would appreciate it if someone could walk me through the solution.

enter image description here

  • $\begingroup$ Here's what I see. $\endgroup$ – Shaun Apr 22 at 6:21
  • $\begingroup$ Here's a MathJax tutorial :) $\endgroup$ – Shaun Apr 22 at 6:22
  • 1
    $\begingroup$ Let there be an integer valued function $f$ such that $$f(x) -1 < x \leq f(x),\ \forall x \in \Bbb R.$$ Then we have $$x \leq f(x) < x+1.$$ This shows that $$f(x)=\lfloor f(x) \rfloor = \lfloor x \rfloor.$$ $\endgroup$ – Dbchatto67 Apr 22 at 6:35
  • $\begingroup$ The above argument proves the uniqueness of $\lceil x \rceil.$ $\endgroup$ – Dbchatto67 Apr 22 at 6:41
  • $\begingroup$ Sorry in the above argument $\lfloor f(x) \rfloor$ should be replaced by $\lceil f(x) \rceil$ $\endgroup$ – Dbchatto67 Apr 22 at 6:46

A drawing of the function $f(x)=\lceil x\rceil$ can help:


Taking for granted what the drawing says it's very easy to check any of the substitutions (e.g. clearly $\lceil 2.4\rceil=\lceil 2.15\rceil=\lceil 2.836\rceil=3$ simply reading it)

You can too check the truth of the substitutions right into the inequalities. E. g. $\lceil1.7\rceil=2$ satisfies $2-1<1.7\leq2$

Finally, the very name helps: for a number some integer is its ceiling.


You just need to know that $\lceil{x}\rceil$ is the smallest integer that is greater than or equal to $x$.

Let $x = 0.1$.

Is $0 ≥ x$? No.

Is $1 ≥ x$? Yes.

Is $2 ≥ x$? Yes. However, it is not the smallest integer as $1$ also satisfies this condition.

Try this procedure with $x = -1.7$ and note that $-1.7 \color{red}{≤} -1$.


Let there be an integer valued function $f$ such that $$f(x)-1 < x \leq f(x),\ \forall x \in \Bbb R.$$ Then we have $$x \leq f(x) < x+1,\ \forall x \in \Bbb R.$$ Now two cases may arise which are

$(1)$ $x \in \Bbb Z.$

$(2)$ $x \notin \Bbb Z.$

If $x \in \Bbb Z$ then since $f$ is an integer valued function with $f(x) \in [x,x+1)$ it follows that $f(x) = x.$ But then we have $f(x) = \lceil f(x) \rceil = \lceil x \rceil.$

Now if $x \notin \Bbb Z$ then $\exists$ a unique integer $n$ with $x<n<x+1.$ So $n$ is the least integer just exceeding $x.$ Therefore $$\lceil x \rceil = n.\ \ \ \ (1)$$ On the other hand since $f$ is integer valued with $f(x) \in [x,x+1)$ and the only integer in the interval $[x,x+1)$ is $n$ so it also follows that $$f(x)=n.\ \ \ \ (2)$$ Combining $(1)$ and $(2)$ it follows that $$f(x) = \lceil x \rceil$$ as required.

This proves the uniqueness of the ceiling function $\lceil x \rceil.$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.