# Thread: Help me prove or disprove these claims

1. ## Help me prove or disprove these claims

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) ^^

2. Originally Posted by 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) ^^
All the definition is saying is that the floor of x is the integer which is less than x and greater than or equal to every integer less than x.

However that will probably not help since by the look of the question you are expected to prove or disprove the given claims using mindless and meaning free rule following in the great formallist/Bourbakist tradition.

CB

3. Originally Posted by 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) ^^
This is an excellent example in formalizing definitions.

See, if you can do that with the definition of square root.

Let me first of all rewrite the above definition using different variables for convenience purposes .

$\displaystyle \forall a\forall b[a\in R\Longrightarrow [a]=b\Longleftrightarrow b\in Z\wedge b\leq a\wedge\forall c(c\in Z\wedge c\leq a\Longrightarrow c\leq b)]$.................................................. ..............................................1

Before now proving (i) we must first prove a very valuable theorem :

$\displaystyle \forall x( x\in R\Longrightarrow [x]\in Z\wedge [x]\leq x\wedge\forall c(c\in Z\wedge c\leq x\Longrightarrow c\leq[x])$.................................................. .................................................. ...2

For that proof simply put in (1) a =x and b =[x] ,and we have :

$\displaystyle x\in R\Longrightarrow [x]=[x]\Longleftrightarrow [x]\in Z\wedge [x]\leq x\wedge\forall c(c\in Z\wedge c\leq x\Longrightarrow c\leq [x])$.

And since [x]=[x] we have as a result formula (2).

Now to prove (i).

Let xεR and yεZ ,and

If in(1) put a = x+y and b = [x]+y we have :

$\displaystyle x+y\in R\Longrightarrow [x+y]=[x]+y\Longleftrightarrow [x]+y\in Z\wedge [x]+y\leq x+y$$\displaystyle \wedge\forall c(c\in Z\wedge c\leq x+y\Longrightarrow c\leq [x]+y).................................................. ...............................................3 Since (x+y)εR from (3) we have : \displaystyle [x+y]=[x]+y\Longleftrightarrow [x]+y\in Z\wedge [x]+y\leq x+y$$\displaystyle \wedge\forall c(c\in Z\wedge c\leq x+y\Longrightarrow c\leq [x]+y)$

Hence to prove [x+y]=[x]+y we must prove :

1) ([x]+y)εZ

2) $\displaystyle [x]+y\leq x+y$

3) $\displaystyle \forall c(c\in Z\wedge c\leq x+y\Longrightarrow c\leq [x]+y$

From (2) and since xεR ,[x]εΖ ,hence ([x]+y)εΖ ,since yεΖ.

From (2) again and since xεR we have:

$\displaystyle [x]\leq x$ and since $\displaystyle y\leq y$ ,$\displaystyle [x]+y\leq x+y$.

Now to prove :

if cεZ and $\displaystyle c\leq x+y$ ,then $\displaystyle c\leq [x]+y$.

I let you prove this one.

4. This is the essence of the floor function: Every real number is either an integer or is between to consecutive integers.
$\displaystyle \left( {\forall z \in \Re } \right)\left[ {\left\lfloor z \right\rfloor \leqslant z < \left\lfloor z \right\rfloor + 1} \right]$
It is easy to see that $\displaystyle n \in \mathbb{Z} \Rightarrow \quad \left\lfloor z \right\rfloor + n \leqslant z + n$ which in turn implies $\displaystyle \left\lfloor z \right\rfloor + n \leqslant \left\lfloor {z + n} \right\rfloor$.
If it were the case that $\displaystyle \left\lfloor z \right\rfloor + n < \left\lfloor {z + n} \right\rfloor \leqslant z + n$ then we get $\displaystyle \left\lfloor z \right\rfloor < \left\lfloor {z + n} \right\rfloor - n \leqslant z$.
But there is no integer between $\displaystyle \left\lfloor z \right\rfloor$ and $\displaystyle z$.

Let $\displaystyle z=0.9~\&~n=2$. Explore $\displaystyle \left\lfloor {z \cdot n} \right\rfloor ~\&~\left\lfloor {z} \right\rfloor n$