# Help me prove or disprove these claims

#### Selena

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
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].

#### parasio

(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 $$\displaystyle \left\lfloor x\right\rfloor+\left\lfloor y\right\rfloor\leq\left\lfloor x+y\right\rfloor$$.

#### Selena

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!

#### parasio

(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 $$\displaystyle \left\lfloor x\right\rfloor+\left\lfloor y\right\rfloor\leq\left\lfloor x+y\right\rfloor$$.

Similar threads