Logical Proofs floor and ceiling functions

[tylerperri]

. Consider the two functions that are defined for any x ∈ R


math.floor(x) = the largest integer less than or equal to x,
math.ceiling(x) = the smallest integer greater than or equal to x.

expressing these definitions using predicate logic... y = math.floor(x) means that y ∈ Z ∧ y ≤ x ∧ (∀z ∈ Z, z ≤ x ⇒ z ≤ y)
andy = math.celing(x) means that y ∈ Z ∧ y ≥ x ∧ (∀z ∈ Z, z ≥ x ⇒ z ≥ y)

(a) Consider the statement:∀x ∈ R, math.ceiling(-x) = −math.floor(x).If you think that the statement is True, prove it.if you think it is False, prove that itsnegation is True.

(b) Consider the statement:∀x ∈ R, ∀n ∈ N, math.floor(n · x) = n · math.floor(x).If you think that the statement is True, prove it. And if you think it is False, prove that itsnegation is True.

 

[Deleted member 4993]

. Consider the two functions that are defined for any x ∈ R


math.floor(x) = the largest integer less than or equal to x,
math.ceiling(x) = the smallest integer greater than or equal to x.

expressing these definitions using predicate logic... y = math.floor(x) means that y ∈ Z ∧ y ≤ x ∧ (∀z ∈ Z, z ≤ x ⇒ z ≤ y)
andy = math.celing(x) means that y ∈ Z ∧ y ≥ x ∧ (∀z ∈ Z, z ≥ x ⇒ z ≥ y)

(a) Consider the statement:∀x ∈ R, math.ceiling(-x) = −math.floor(x).If you think that the statement is True, prove it.if you think it is False, prove that itsnegation is True.

(b) Consider the statement:∀x ∈ R, ∀n ∈ N, math.floor(n · x) = n · math.floor(x).If you think that the statement is True, prove it. And if you think it is False, prove that its negation is True.

What are your thoughts?

Please share your work with us ...even if you know it is wrong.

If you are stuck at the beginning tell us and we'll start with the definitions.

You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:

http://www.freemathhelp.com/forum/announcement.php?f=33

 

[tylerperri]

sorry I forgot to post thoughts.. but yes i am pretty much stuck at the beginning.

 

[pka]

. Consider the two functions that are defined for any x ∈ R
math.floor(x) = the largest integer less than or equal to x,
math.ceiling(x) = the smallest integer greater than or equal to x.

(a) Consider the statement:∀x ∈ R, math.ceiling(-x) = −math.floor(x).If you think that the statement is True, prove it.if you think it is False, prove that itsnegation is True.

(b) Consider the statement:∀x ∈ R, ∀n ∈ N, math.floor(n · x) = n · math.floor(x).If you think that the statement is True, prove it. And if you think it is False, prove that itsnegation is True.

You must be aware of this link, These proofs is as if we are chasing "our one tails."

I suggest that you use these definitions:

\(\displaystyle \begin{align*}\left\lfloor x \right\rfloor &\le x < \left\lfloor x \right\rfloor + 1\quad\text{ floor fun}\text{ction} \\\left\lceil x \right\rceil - 1 &< x \le \left\lceil x \right\rceil\quad\text{ ceiling func}\text{tion} \end{align*}\)

Now turn this around:

If \(\displaystyle n\in\mathbb{Z}\text{ and }n\le x<n+1\text{ then }n=\left\lfloor x \right\rfloor.\) AND If \(\displaystyle m\in\mathbb{Z}\text{ and }m-1< x\le m\text{ then }m=\left\lceil x \right\rceil.\)

\(\displaystyle \begin{align*}\left\lfloor -x \right\rfloor &\le -x < \left\lfloor -x \right\rfloor + 1 \\-\left\lfloor -x \right\rfloor &\ge x > -\left\lfloor -x \right\rfloor - 1\end{align*}\)

By the above \(\displaystyle -\left\lfloor -x \right\rfloor=\left\lceil x \right\rceil.\)

You need to work on these.