What is the proof for the property ⌊−x⌋ = −⌈x⌉?

  • Thread starter Thread starter twoski
  • Start date Start date
AI Thread Summary
The discussion revolves around proving the properties ⌊−x⌋ = −⌈x⌉ and ⌈−x⌉ = −⌊x⌋. Participants explore the use of the relationship ⌊x⌋ = n, where x = n + m, and consider various cases for m, including whole numbers and rational numbers. There is a consensus that breaking the proof into cases may not be necessary, and a focus on non-integer cases could simplify the proof. A proposed proof using the definitions of the floor and ceiling functions is debated, with some participants questioning the correctness of the final conclusion regarding the signs. The discussion emphasizes the importance of clarity in mathematical proofs and the need to ensure all cases are adequately addressed.
twoski
Messages
177
Reaction score
2

Homework Statement



Prove:

⌊−x⌋ = −⌈x⌉


⌈−x⌉ = −⌊x⌋ .

The Attempt at a Solution



Can i use the property ⌊x⌋ = n, x = n + m where 0 <= m < 1 or do i need to incorporate the negation into this property?

I think there would be 2 cases to this proof.

Case 1: m = 0, the lower bound.

Case 2: m = 1/2.

Case 3: m = 1, the upper bound.

I just don't know how i'd go about proving these cases.
 
Physics news on Phys.org
twoski said:
Can i use the property ⌊x⌋ = n, x = n + m where 0 <= m < 1
Yes, that's a valid relationship regardless of the sign of x.
I think there would be 2 cases to this proof.

Case 1: m = 0, the lower bound.
Case 2: m = 1/2.
That wouldn't be very satisfactory. You really need to prove it for all m in the range, not just special values.
Case 3: m = 1, the upper bound.
m < 1, so just two cases (as you wrote). But it probably isn't necessary to break it into separate cases for m at all. Separate cases for x +ve/-ve might be useful.
 
Oh, I'm not thinking straight today.

Looking at it again, wouldn't there be 2 cases, one where x is a whole number and one where it is a rational number? So in other words, the cases would be m = 0 and 1 > m > 0.

My professor has thus far taught us proofs involving floor/ceiling using the property that any value being floored/ceilinged can be expressed as x = n + m, and we manipulate m to come up with the cases.
 
twoski said:
wouldn't there be 2 cases, one where x is a whole number and one where it is a rational number?
I hope you mean "and one where it is not a whole number"; otherwise you're leaving out the irrationals. Anyway, as I said, I doubt it's useful to break it up that way. Go ahead and try it, but do the non-integer case first; I expect you'll find it handles both.
 
Proof for ⌊−x⌋ = −⌈x⌉

By the definition of the floor function, x – 1 < ⌊x⌋ ≤ x

It follows that −x + 1 > −⌊x⌋ ≥ −x

Next, let ⌈x⌉ = n where x ≤ n < x + 1

We note that -x ≤ −⌊x⌋ < -x + 1

Clearly, −⌊x⌋ = ⌈-x⌉

I found this proof online, i think it's kind of bogus since the result of the proof isn't what we set out to prove, the signs are misplaced. The steps seem correct up until the conclusion.
 
twoski said:
Proof for ⌊−x⌋ = −⌈x⌉

By the definition of the floor function, x – 1 < ⌊x⌋ ≤ x

It follows that −x + 1 > −⌊x⌋ ≥ −x

Next, let ⌈x⌉ = n where x ≤ n < x + 1

We note that -x ≤ −⌊x⌋ < -x + 1

Clearly, −⌊x⌋ = ⌈-x⌉
The above is correct, but the last step could be made clearer:
By definition, y ≤ ⌈y⌉< y + 1 for all y. Set y = -x:
-x ≤ ⌈-x⌉< -x + 1
So −⌊x⌋ and ⌈-x⌉ are both integers in the range [-x, -x+1). Since only one integer can be in that range, they are equal.
the result of the proof isn't what we set out to prove, the signs are misplaced.
It is the second part of the OP.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Back
Top