Question

Need help! :D [Question attached below]

Answer 1

\[\lceil \frac{ 2x+1 }{ 4 } \rceil-\lfloor \frac{ 2x+1 }{ 4 } \rfloor=1\]

Answer 2

How?

Answer 3

I mean, how do I explain that?

Answer 4

But that does not work when x = 3.5 yeah?

Answer 5

you are given
\[\lceil x \rceil=least~integer \ge x~and~\lfloor x \rfloor=greatest~integer le~x\]
it always fails when\[\left[ x \right]~ is~ an~ integer.\]

Answer 6

You mean it fails when 'x' is an integer?

Answer 7

But what about when x = 3.5
\[\lceil 2 \rceil - \lfloor 2 \rfloor = 0\]

Answer 8

i mean to say there is an integer in the \[\left[ an~ integer \right]-\left[ same~integer \right]=0\]

Answer 9

[an integer]−[same integer]=0
Shouldn't that be quite obvious?
That is like saying,
x-x = 0
?
How do I solve the question @surjithayer ?

Answer 10

@LastDayWork
@adrynicoleb

Answer 11

I am one of the last people that you should be asking for help in math, js. ._.

Answer 12

Oh okay, judging by your SS, it didn't seem like that.
But then again, SS is not a way to judge people's aptitude!
Thanks for trying though.

Answer 13

Lol. I get most of my SS points from English. It's my best subject. And you're welcome. :3

Answer 14

What surjithayer meant to say is -
for any y ∈ R
ceil(y) - floor(y) = 0 ; if y ∈ Z
ceil(y) - floor(y) = 1 ; if y ∉ Z

Answer 15

Yes, agreed. Then?

Answer 16

I need some time to solve the Q..we simply need to make cases for the terms to be integers and non-integers..

Answer 17

Yeah?
I was just thinking, why couldn't we just graph it an prove it?
The thing is, i am not quite certain if I am allowed to use a graphing calculator.

Answer 18

Graph will 'validate' it for (only) some values of x.
We need to 'prove' it for all x ∈ R

Answer 19

You get a usual greatest integer function when using a graphical calculator but yes, it would actually be better to prove it without it.

Answer 20

The most pathetic solution would be to verify for each of the three cases, namely
1) 2x+1 is a multiple of 4
2) 2x+1 is a multiple of 2 but not 4
3) 2x+1 is not a multiple of 2

I can't think of a better solution..

Answer 21

Okay, so do I use random numbers [following the 3 cases] and try to prove it?

Answer 22

No, we have to use variables
I am solving case (1) "2x+1 is a multiple of 4" as an example

Let 2x + 1 = 4n ; where n ∈ Z
Then the given expression
\[\lceil \frac{ 2x+1 }{ 2 } \rceil-\lceil \frac{ 2x+1 }{ 4 } \rceil+\lfloor \frac{ 2x+1 }{ 4 } \rfloor\]
\[=\lceil 2n \rceil-\lceil n \rceil+\lfloor n \rfloor\]
\[=2n-n+n\]
\[=2n\]

Now, from 2x + 1 = 4n ; we get
\[x = \frac{ 4n-1 }{ 2 }=2n-\frac{ 1 }{ 2 }\]
implies
\[\lceil x \rceil=\lceil 2n-\frac{ 1 }{ 2 } \rceil=2n\]

Hence,
\[\lceil \frac{ 2x+1 }{ 2 } \rceil-\lceil \frac{ 2x+1 }{ 4 } \rceil+\lfloor \frac{ 2x+1 }{ 4 } \rfloor = \lceil x \rceil\]
for case (1)

Answer 23

I understood that! :D
But how do we say it is not divisible by something?
And why aren't we considering decimals now?

Answer 24

"why aren't we considering decimals now?"
Ceil and floor always return an integer; hence there aren't any decimals in my solution.

"how do we say it is not divisible by something?"
Which case are you working with - (2) or (3) ??

Answer 25

2

Answer 26

Use, what I posted before -

What surjithayer meant to say is -
for any y ∈ R
ceil(y) - floor(y) = 0 ; if y ∈ Z
ceil(y) - floor(y) = 1 ; if y ∉ Z

Answer 27

2) 2x+1 is a multiple of 2 but not 4

How shall I write this?

2x + 1 = 2\(\lambda\)
2x + 1 \(\neq~4\mu\)

?

Answer 28

If 2x+1 is not a multiple of 4 , then
(2x+1)/4 is not an integer
implies, the last two terms of the given expression can be replaced with a fixed number..

Answer 29

Did you get my point ??

Answer 30

yes..

-1 !

Answer 31

Good, now tell me what does the expression in case(2) reduces to ??

Answer 32

So wait, are we dividing case 2 to 2 more cases
one where 2x+1 is divisible by 2
and another where it is not divisible by 4 ??

Answer 33

I distributed the value of x into three cases, namely -
1) 2x+1 is a multiple of 4
2) 2x+1 is a multiple of 2 but not 4
3) 2x+1 is not a multiple of 2

Comment on whether (or not) they are
-overlapping
-exhaustive

Answer 34

Have you studied set theory ??

Answer 35

Yes, I know exhaustive, not quite sure about overlapping?
They have a common intersection?

Answer 36

Then all of them are exhaustive!

Answer 37

Yea, two overlapping implies they have a common intersection..or in other words, some values of x are common b/w the cases..

Answer 38

*omit two

Answer 39

So here you are considering all possible integer cases of 2x+1 so the cases ARE exhaustive yeah? I understood the 1st case clearly.

So in the second case the solution would be

\[\lceil \frac{2x+1}{2} \rceil - 1\]

And 2x+1 is a multiple of 2 so that should result in an integer value.

Answer 40

So far so good..

Answer 41

:D

Answer 42

So what do you finally get ??

Answer 43

So..

That was \[\lceil \frac{2x+1}{2} \rceil - 1\]\[\lceil x + \frac{1}{2} \rceil - 1\]
And as it is a ceiling function it will equal : x

Answer 44

Why don't you consider 2x+1 = 2k ; & then solve it like I did in case (1)

Answer 45

* where k ∈ Z

Answer 46

Then it'll be just k-1

Answer 47

Find it in terms of ceil(x) or floor(x)..

Answer 48

But isn't that what I did there before?

First Case:

In the first case you found what the value of the whole expression was in terms of 'n'.
And then you found the value of ceil(x) or floor(x) and proved that it was also equal to the expression in terms of 'n'.

Second Case:

Here I took it in terms of k. And found that it equals = k-1
Now, we know that

2x+1 = 2k
x = (2k - 1)/2
x = (k - 1/2)

So floor(x) = floor(k -1/2) = k - 1 which is equal to the value we found earlier.
So case 2 is proved?

Answer 49

Yea..that's ^^ what I wanted you to do..

Now, go for case (3)

Answer 50

Third case:
2x+1\(\neq 2\lambda\)
That means if it is not a multiple of 2 it is not a multiple of 4

And thus the expression now is
\[\lceil \frac{2x+1}{2} \rceil - 1\]

Now how should I put it in terms of the multiple?

Answer 51

Use
\[\frac{ 2x+1 }{ 2 }=I+f\]
Where
\[I=\lfloor \frac{ 2x+1 }{ 2 } \rfloor\]
\[f=\frac{ 2x+1 }{ 2 }-\lfloor \frac{ 2x+1 }{ 2 } \rfloor\]

Answer 52

Fractional part integer function?

Answer 53

Yep

Answer 54

But then how do I proceed, I still am sorta confused for this one. :/

Answer 55

\[\frac{ 2x+1 }{ 2 }= I + f\]
implies
\[x=I+f-\frac{ 1 }{ 2 }\]

Can you foresee the rest of the solution ?

Answer 56

No, not exactly, I don't know how to proceed. :/

Answer 57

The given expression will reduce to I

Now for f ≥ (1/2) ; floor(x) = I
for f ≤ (1/2) ; ceil(x) = I

Answer 58

I do understand that. I just sort of lost you with I and f.
What are we actually doing with that?

Answer 59

It was just to ease the solution..to deal separately with decimal and whole number..

Answer 60

What I did here, was it right?

Answer 61

It wasn't wrong, but it appeared a dead-end to me..
I mean - What would have been your next step ??

Answer 62

I would have written

\[\lceil \frac{2x + 1}{2} \rceil - 1 ~~as~~ \lceil x + 1/2 \rceil - 1~~=~~x+1 -1 = x\]

Answer 63

In general,
\[\lceil x+\frac{ 1 }{ 2 } \rceil \neq x+1\]

Answer 64

Is it because we are not sure, whether x is an integer or not?

Answer 65

Yep

Answer 66

So you told me to use this because it is not divisible 2x+! is not divisible by 2.
And thus it may give decimal values?

Answer 67

Yep

Answer 68

btw, can I ask you one more question?
How and why did you come up with those cases?

Answer 69

The question involved floor and ceil functions. So I came up with cases where I can be certain whether the arguments are integers or not.

It's necessary to ensure that the set of cases are exhaustive (for this Q, covers all x ∈ R)
and sometimes, it's also necessary to ensure that the cases AREN'T mutually overlapping..

Answer 70

Are we considering here, that

2x+1 \(\in\) Z ?

Answer 71

Only for case (1) and (2)

Answer 72

And for case 3 it belongs to all real numbers indivisible by 2?

Answer 73

its better to say that - for case (3) it belongs to all real numbers not covered in case (1) and (2)

Answer 74

Okay, let me just try it out now.

Answer 75

Aren't cases a and b overlapping?

In b it asks for multiples of 2
In a it asks for multiples of 4. [Which is also a multiple of 2]

Answer 76

In case (1) we are working with integers of the form (4n + 0)
In case (2) we are working with integers of the form (4n + 2)

Answer 77

Do you still think these two cases are overlapping ??

Answer 78

No I mean in case 1 : 2x + 1 can be 8
Even in case 2 : 2x + 1 can be 8

There is an intersection isn't there?

Answer 79

In case (2) ; 2x+1 is a multiple of 2 but not a multiple of 4
Hence 2x+1 can't be 8

Answer 80

OHH OKAY! :D
Lol, i am so dumb. *poker face*

Alright, I'll try it out now. :)

Answer 81

Every1's a dumb..in a way.. ;)
Take your time..

Answer 82

So you are basically just dividing it like this: