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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

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

Ceiling

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.

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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 $-1.7 \color{red}{≤} -1$.

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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 $\lceil x \rceil.$

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