Discrete math - The ceiling of a real number x, denoted by$⌈𝑥⌉$, is the unique integer that satisfies the inequality

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.

• – Shaun Apr 22 at 6:21
• – Shaun Apr 22 at 6:22
• 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 $$f(x)=\lfloor f(x) \rfloor = \lfloor x \rfloor.$$ – Dbchatto67 Apr 22 at 6:35
• The above argument proves the uniqueness of $\lceil x \rceil.$ – Dbchatto67 Apr 22 at 6:41
• Sorry in the above argument $\lfloor f(x) \rfloor$ should be replaced by $\lceil f(x) \rceil$ – 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 $$x So $$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 $$\lceil x \rceil.$$