Help me prove or disprove these claims

Feb 2010
21
0
Heya everyone, I need help proving or disproving these claims:



This definition of the floor totally confused me, I don't know how to start this problem as I don't recognize anything such as axioms or formulas from the claim. Please give me some pointers in the right direction, thanks in advance!

(Yeah I know this is my third question, there are just so many problems and I'm running into a lot of dead ends) ^^
 

CRGreathouse

Forum Staff
Nov 2006
16,046
936
UTC -5
I suggest rewriting each real number as an integer plus a real number between 0 and 1:
x = X + e, 0 <= e < 1.

So ii) would be
[x * z] = [(X + e)z] = [Xz + ez].

Now if the first part turns out to be true :)wink:) then you can simplify this with
[Xz + ez] = [ez] + Xz
since X and z are integers, so their product is an integer too. Now you just have to ask: is there some number in [0, 1) and some integer that, when multiplied together, give a number 1 or bigger? If so, ii) is false; if not, ii) is true.

Similarly, rewrite iii) as [(X + e) + (Y + f)] = [X + Y + e + f].
 
Feb 2009
172
5
(i) is true
Take \(\displaystyle x\in\mathbb R\) and \(\displaystyle z\in\mathbb Z\). By the definition \(\displaystyle \left\lfloor x\right\rfloor\leq x\). Hence \(\displaystyle \left\lfloor x\right\rfloor+z\leq x+z\). But \(\displaystyle \left\lfloor x\right\rfloor+z\in\mathbb Z\) hence again by the definition \(\displaystyle \left\lfloor x\right\rfloor+z\leq\left\lfloor x+z\right\rfloor\). Observe now that \(\displaystyle \left\lfloor x+z\right\rfloor\leq x+z\) hence \(\displaystyle \left\lfloor x+z\right\rfloor-z\leq x\). But \(\displaystyle \left\lfloor x+z\right\rfloor-z\in\mathbb Z\) wich gives us that \(\displaystyle \left\lfloor x+z\right\rfloor-z\leq\left\lfloor x\right\rfloor\). Hence \(\displaystyle \left\lfloor x+z\right\rfloor\leq\left\lfloor x\right\rfloor+z\). Then \(\displaystyle \left\lfloor x+z\right\rfloor=\left\lfloor x\right\rfloor+z\).

(ii) is false since \(\displaystyle \left\lfloor \frac{1}{2}\cdot 2\right\rfloor=1\) and \(\displaystyle \left\lfloor \frac{1}{2}\right\rfloor\cdot 2=0\cdot 2=0\).

(iii) is true

Take \(\displaystyle x,y\in\mathbb R\). By definition \(\displaystyle \left\lfloor x\right\rfloor\leq x\) and \(\displaystyle \left\lfloor y\right\rfloor\leq y\) hence, \(\displaystyle \left\lfloor x\right\rfloor+\left\lfloor y\right\rfloor\leq x+y\). But \(\displaystyle \left\lfloor x\right\rfloor+\left\lfloor y\right\rfloor\in\mathbb Z\) then by definition \(\displaystyle \left\lfloor x\right\rfloor+\left\lfloor y\right\rfloor\leq\left\lfloor x+y\right\rfloor\).
 
Feb 2010
21
0
CRGreathouse said:
I suggest rewriting each real number as an integer plus a real number between 0 and 1:
x = X + e, 0 <= e < 1.

So ii) would be
[x * z] = [(X + e)z] = [Xz + ez].

Now if the first part turns out to be true :)wink:) then you can simplify this with
[Xz + ez] = [ez] + Xz
since X and z are integers, so their product is an integer too. Now you just have to ask: is there some number in [0, 1) and some integer that, when multiplied together, give a number 1 or bigger? If so, ii) is false; if not, ii) is true.

Similarly, rewrite iii) as [(X + e) + (Y + f)] = [X + Y + e + f].
So small x is the real number, and big X is the floor right?
And shouldn't ii) be [x * z] = [x * e] * z?
I'm sorry I don't quite follow the second part =x


parasio said:
(i) is true
Take \(\displaystyle x\in\mathbb R\) and \(\displaystyle z\in\mathbb Z\). By the definition \(\displaystyle \left\lfloor x\right\rfloor\leq x\). Hence \(\displaystyle \left\lfloor x\right\rfloor+z\leq x+z\). But \(\displaystyle \left\lfloor x\right\rfloor+z\in\mathbb Z\) hence again by the definition \(\displaystyle \left\lfloor x\right\rfloor+z\leq\left\lfloor x+z\right\rfloor\). Observe now that \(\displaystyle \left\lfloor x+z\right\rfloor\leq x+z\) hence \(\displaystyle \left\lfloor x+z\right\rfloor-z\leq x\). But \(\displaystyle \left\lfloor x+z\right\rfloor-z\in\mathbb Z\) wich gives us that \(\displaystyle \left\lfloor x+z\right\rfloor-z\leq\left\lfloor x\right\rfloor\). Hence \(\displaystyle \left\lfloor x+z\right\rfloor\leq\left\lfloor x\right\rfloor+z\). Then \(\displaystyle \left\lfloor x+z\right\rfloor=\left\lfloor x\right\rfloor+z\).

(ii) is false since \(\displaystyle \left\lfloor \frac{1}{2}\cdot 2\right\rfloor=1\) and \(\displaystyle \left\lfloor \frac{1}{2}\right\rfloor\cdot 2=0\cdot 2=0\).

(iii) is true

Take \(\displaystyle x,y\in\mathbb R\). By definition \(\displaystyle \left\lfloor x\right\rfloor\leq x\) and \(\displaystyle \left\lfloor y\right\rfloor\leq y\) hence, \(\displaystyle \left\lfloor x\right\rfloor+\left\lfloor y\right\rfloor\leq x+y\). But \(\displaystyle \left\lfloor x\right\rfloor+\left\lfloor y\right\rfloor\in\mathbb Z\) then by definition \(\displaystyle \left\lfloor x\right\rfloor+\left\lfloor y\right\rfloor\leq\left\lfloor x+y\right\rfloor\).
Hi can you please explain i) step by step? I don't quite follow, specifically this part:

But \(\displaystyle \left\lfloor x\right\rfloor+z\in\mathbb Z\) hence again by the definition \(\displaystyle \left\lfloor x\right\rfloor+z\leq\left\lfloor x+z\right\rfloor\)
(wasn't it "Hence \(\displaystyle \left\lfloor x\right\rfloor+z\leq x+z\)"?)

and

But \(\displaystyle \left\lfloor x+z\right\rfloor-z\in\mathbb Z\) wich gives us that \(\displaystyle \left\lfloor x+z\right\rfloor-z\leq\left\lfloor x\right\rfloor\)
Thanks!
 
Feb 2009
172
5
(i) is true
Take \(\displaystyle x\in\mathbb R\) and \(\displaystyle z\in\mathbb Z\). By the definition (part 2) \(\displaystyle \left\lfloor x\right\rfloor\leq x\). Hence \(\displaystyle \left\lfloor x\right\rfloor+z\leq x+z\). Since \(\displaystyle \left\lfloor x\right\rfloor+z\in\mathbb Z\) it follows from the definition (part 3) that \(\displaystyle \left\lfloor x\right\rfloor+z\leq\left\lfloor x+z\right\rfloor\). By definition (part 2) \(\displaystyle \left\lfloor x+z\right\rfloor\leq x+z\) hence \(\displaystyle \left\lfloor x+z\right\rfloor-z\leq x\). But \(\displaystyle \left\lfloor x+z\right\rfloor-z\in\mathbb Z\) then by definition (part 2) \(\displaystyle \left\lfloor x+z\right\rfloor-z\leq\left\lfloor x\right\rfloor\). Hence \(\displaystyle \left\lfloor x+z\right\rfloor\leq\left\lfloor x\right\rfloor+z\).

Then we have that \(\displaystyle \left\lfloor x+z\right\rfloor\leq\left\lfloor x\right\rfloor+z\) and \(\displaystyle \left\lfloor x\right\rfloor+z\leq\left\lfloor x+z\right\rfloor\) hence \(\displaystyle \left\lfloor x+z\right\rfloor=\left\lfloor x\right\rfloor+z\).

(ii) is false since \(\displaystyle \left\lfloor \frac{1}{2}\cdot 2\right\rfloor=1\) and \(\displaystyle \left\lfloor \frac{1}{2}\right\rfloor\cdot 2=0\cdot 2=0\).

(iii) is true

Take \(\displaystyle x,y\in\mathbb R\). By definition \(\displaystyle \left\lfloor x\right\rfloor\leq x\) and \(\displaystyle \left\lfloor y\right\rfloor\leq y\) hence, \(\displaystyle \left\lfloor x\right\rfloor+\left\lfloor y\right\rfloor\leq x+y\). But \(\displaystyle \left\lfloor x\right\rfloor+\left\lfloor y\right\rfloor\in\mathbb Z\) then by definition \(\displaystyle \left\lfloor x\right\rfloor+\left\lfloor y\right\rfloor\leq\left\lfloor x+y\right\rfloor\).