# Math Help - Floor and Ceiling stuff.

1. ## Floor and Ceiling stuff.

Okey, im fairly sure this is simple but i've not done maths in ages and we have horrendous maths lecturers and tutorial people who's reply when our whole table couldn't do it was 'Guys, it's easy, why dont you get it?'

Now im not asking for the answer, just some help if that's ok .

The question is:

$
f(x) = \lceil\frac{x}{4}\rceil - \lfloor \frac{x}{4} + \frac{1}{4} \rfloor
$

Now the part in brackets is meant to be floored. And the part not in brackets is mean to be ceiling.

Question:
Prove:
a.) If n is an integer then f( x + 4n) = f(x)
b.) If 0 < x < 3 then f(x) = 1 and if 3 < x < 4 then f (x) = 0

Ermmm...we've been taught...very briefly proof by induction so i'm assuming this needs to be used? But i have no idea how to apply proof by induction to this...if you add 1 to the f(x + 4n) to get f ( x + 4n + 1) you're just going to get f(x + 1) as the final result and that just seems pointless to me an unproved :S.

Any help would be appriciated!!! Even if you could just link me to something handy =). Thankyou so much x

2. $$f(x) = \frac{x}{4} - \left\lfloor {\frac{x}{4} + \frac{1}{4}} \right\rfloor$$
gives $f(x) = \frac{x}{4} - \left\lfloor {\frac{x}{4} + \frac{1}{4}} \right\rfloor$

You may want to edit the question.

3. Originally Posted by AshleyT
Okey, im fairly sure this is simple but i've not done maths in ages and we have horrendous maths lecturers and tutorial people who's reply when our whole table couldn't do it was 'Guys, it's easy, why dont you get it?'

Now im not asking for the answer, just some help if that's ok .

The question is:
$f(x) = \frac{x}{4} - \left\lfloor {\frac{x}{4} + \frac{1}{4}} \right\rfloor$
Question:
Prove:
a.) If n is an integer then f( x + 4n) = f(x)
b.) If 0 < x < 3 then f(x) = 1 and if 3 < x < 4 then f (x) = 0
Here is some hints.
$\left( {\forall n \in \mathbb{Z}} \right)\left[ {\left\lfloor {a + n} \right\rfloor = \left\lfloor a \right\rfloor + n} \right]$

$\left\lfloor {\frac{{x + 4n}}{4} + \frac{1}{4}} \right\rfloor = \left\lfloor {\frac{x}{4} + \frac{1}{4} + n} \right\rfloor$

4. Originally Posted by Plato
Here is some hints.
$\left( {\forall n \in \mathbb{Z}} \right)\left[ {\left\lfloor {a + n} \right\rfloor = \left\lfloor a \right\rfloor + n} \right]$

$\left\lfloor {\frac{{x + 4n}}{4} + \frac{1}{4}} \right\rfloor = \left\lfloor {\frac{x}{4} + \frac{1}{4} + n} \right\rfloor$
Thankyou i edited my post .
Okey, im probally going to ask a stupid question now(sorry) but the hint you gave me is for use with induction? I got what was in the bracket but only 'showed' they were equal so far.

5. Originally Posted by AshleyT
the hint you gave me is for use with induction? I got what was in the bracket but only 'showed' they were equal so far.
Frankly I do not understand the above question.
If it is about the properties of the floor function, I think you should have proved them before being asked this question.

Now, I did not understand that the question included the ceiling function.
The is a property you need.
$\left( {\forall n \in \mathbb{Z}} \right)\left[ {\left\lceil {a + n} \right\rceil = \left\lceil a \right\rceil + n + 1} \right]$

Again you should have proved this before being asked this question.

6. Originally Posted by Plato
Frankly I do not understand the above question.
If it is about the properties of the floor function, I think you should have proved them before being asked this question.

Now, I did not understand that the question included the ceiling function.
The is a property you need.
$\left( {\forall n \in \mathbb{Z}} \right)\left[ {\left\lceil {a + n} \right\rceil = \left\lceil a \right\rceil + n + 1} \right]$

Again you should have proved this before being asked this question.
The question was worded wrongly in the end which is why i struggled so much :S...didn't have to prove anything, just 'show that'....

Thanks for your help though plato =)