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Discrete math - The ceiling of a real number x, denoted by ⌈𝑥⌉, is the unique integer that satisfies the inequality

2

$begingroup$

I have a discrete math question below with a solution written by my teacher. I'm really lost as to what answers I'm trying to find exactly. I don't understand how the teacher got 1 and -1 for the first two values, and I also don't understand how and why the last line turns ⌈(𝟔.𝟑 − ⌈𝟏.𝟕⌉)⌉ into ⌈𝟔.𝟑 − 𝟐⌉ = ⌈𝟒.𝟑⌉ = 𝟓

I would appreciate it if someone could walk me through the solution.

discrete-mathematics inequality

share|cite|improve this question

asked 1 hour ago

GilmoreGirlingGilmoreGirling

515

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Here's what I see.
  $endgroup$
  – Shaun
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Here's a MathJax tutorial :)
  $endgroup$
  – Shaun
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Let there be an integer valued function $f$ such that $$f(x) -1 < x leq f(x), forall x in Bbb R.$$ Then we have $$x leq f(x) < x+1.$$ This shows that $$f(x)=lfloor f(x) rfloor = lfloor x rfloor.$$
  $endgroup$
  – Dbchatto67
  1 hour ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The above argument proves the uniqueness of $lceil x rceil.$
  $endgroup$
  – Dbchatto67
  1 hour ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry in the above argument $lfloor f(x) rfloor$ should be replaced by $lceil f(x) rceil$
  $endgroup$
  – Dbchatto67
  1 hour ago
  

  
  
  
  

add a comment | 

2

$begingroup$

I have a discrete math question below with a solution written by my teacher. I'm really lost as to what answers I'm trying to find exactly. I don't understand how the teacher got 1 and -1 for the first two values, and I also don't understand how and why the last line turns ⌈(𝟔.𝟑 − ⌈𝟏.𝟕⌉)⌉ into ⌈𝟔.𝟑 − 𝟐⌉ = ⌈𝟒.𝟑⌉ = 𝟓

I would appreciate it if someone could walk me through the solution.

discrete-mathematics inequality

share|cite|improve this question

asked 1 hour ago

GilmoreGirlingGilmoreGirling

515

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Here's what I see.
  $endgroup$
  – Shaun
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Here's a MathJax tutorial :)
  $endgroup$
  – Shaun
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Let there be an integer valued function $f$ such that $$f(x) -1 < x leq f(x), forall x in Bbb R.$$ Then we have $$x leq f(x) < x+1.$$ This shows that $$f(x)=lfloor f(x) rfloor = lfloor x rfloor.$$
  $endgroup$
  – Dbchatto67
  1 hour ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The above argument proves the uniqueness of $lceil x rceil.$
  $endgroup$
  – Dbchatto67
  1 hour ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry in the above argument $lfloor f(x) rfloor$ should be replaced by $lceil f(x) rceil$
  $endgroup$
  – Dbchatto67
  1 hour ago
  

  
  
  
  

add a comment | 

$begingroup$

I have a discrete math question below with a solution written by my teacher. I'm really lost as to what answers I'm trying to find exactly. I don't understand how the teacher got 1 and -1 for the first two values, and I also don't understand how and why the last line turns ⌈(𝟔.𝟑 − ⌈𝟏.𝟕⌉)⌉ into ⌈𝟔.𝟑 − 𝟐⌉ = ⌈𝟒.𝟑⌉ = 𝟓

I would appreciate it if someone could walk me through the solution.

discrete-mathematics inequality

share|cite|improve this question

asked 1 hour ago

GilmoreGirlingGilmoreGirling

515

$endgroup$

share|cite|improve this question

asked 1 hour ago

GilmoreGirlingGilmoreGirling

515

share|cite|improve this question

asked 1 hour ago

GilmoreGirlingGilmoreGirling

515

asked 1 hour ago

GilmoreGirlingGilmoreGirling

515

asked 1 hour ago

GilmoreGirlingGilmoreGirling

515

3 Answers
3

active

oldest

votes

1

$begingroup$

A drawing of the function $f(x)=lceil xrceil$ can help:

Taking for granted what the drawing says it's very easy to check any of the substitutions (e.g. clearly $lceil 2.4rceil=lceil 2.15rceil=lceil 2.836rceil=3$ simply reading it)

You can too check the truth of the substitutions right into the inequalities. E. g. $lceil1.7rceil=2$ satisfies $2-1<1.7leq2$

Finally, the very name helps: for a number some integer is its ceiling.

share|cite|improve this answer

answered 1 hour ago

Rafa BudríaRafa Budría

6,0521825

$endgroup$

add a comment | 

1

$begingroup$

You just need to know that $lceil{x}rceil$ is the smallest integer that is greater than or equal to $x$.

Let $x = 0.1$.

Is $0 ≥ x$? No.

Is $1 ≥ x$? Yes.

Is $2 ≥ x$? Yes. However, it is not the smallest integer as $1$ also satisfies this condition.

Try this procedure with $x = -1.7$ and note that $-1.7 color{red}{≤} -1$.

share|cite|improve this answer

answered 1 hour ago

Toby MakToby Mak

3,70011128

$endgroup$

add a comment | 

1

$begingroup$

Let there be an integer valued function $f$ such that $$f(x)-1 < x leq f(x), forall x in Bbb R.$$ Then we have $$x leq f(x) < x+1, forall x in Bbb R.$$ Now two cases may arise which are

$(1)$ $x in Bbb Z.$

$(2)$ $x notin Bbb Z.$

If $x in Bbb Z$ then since $f$ is an integer valued function with $f(x) in [x,x+1)$ it follows that $f(x) = x.$ But then we have $f(x) = lceil f(x) rceil = lceil x rceil.$

Now if $x notin Bbb Z$ then $exists$ a unique integer $n$ with $x<n<x+1.$ So $n$ is the least integer just exceeding $x.$ Therefore $$lceil x rceil = n. (1)$$ On the other hand since $f$ is integer valued with $f(x) in [x,x+1)$ and the only integer in the interval $[x,x+1)$ is $n$ so it also follows that $$f(x)=n. (2)$$ Combining $(1)$ and $(2)$ it follows that $$f(x) = lceil x rceil$$ as required.

This proves the uniqueness of the ceiling function $lceil x rceil.$

share|cite|improve this answer

answered 32 mins ago

Dbchatto67Dbchatto67

3,571626

$endgroup$

add a comment | 

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1

$begingroup$

A drawing of the function $f(x)=lceil xrceil$ can help:

Taking for granted what the drawing says it's very easy to check any of the substitutions (e.g. clearly $lceil 2.4rceil=lceil 2.15rceil=lceil 2.836rceil=3$ simply reading it)

You can too check the truth of the substitutions right into the inequalities. E. g. $lceil1.7rceil=2$ satisfies $2-1<1.7leq2$

Finally, the very name helps: for a number some integer is its ceiling.

share|cite|improve this answer

answered 1 hour ago

Rafa BudríaRafa Budría

6,0521825

$endgroup$

add a comment | 

1

$begingroup$

A drawing of the function $f(x)=lceil xrceil$ can help:

Taking for granted what the drawing says it's very easy to check any of the substitutions (e.g. clearly $lceil 2.4rceil=lceil 2.15rceil=lceil 2.836rceil=3$ simply reading it)

You can too check the truth of the substitutions right into the inequalities. E. g. $lceil1.7rceil=2$ satisfies $2-1<1.7leq2$

Finally, the very name helps: for a number some integer is its ceiling.

share|cite|improve this answer

answered 1 hour ago

Rafa BudríaRafa Budría

6,0521825

$endgroup$

add a comment | 

$begingroup$

A drawing of the function $f(x)=lceil xrceil$ can help:

Taking for granted what the drawing says it's very easy to check any of the substitutions (e.g. clearly $lceil 2.4rceil=lceil 2.15rceil=lceil 2.836rceil=3$ simply reading it)

You can too check the truth of the substitutions right into the inequalities. E. g. $lceil1.7rceil=2$ satisfies $2-1<1.7leq2$

Finally, the very name helps: for a number some integer is its ceiling.

share|cite|improve this answer

answered 1 hour ago

Rafa BudríaRafa Budría

6,0521825

$endgroup$

share|cite|improve this answer

answered 1 hour ago

Rafa BudríaRafa Budría

6,0521825

answered 1 hour ago

Rafa BudríaRafa Budría

6,0521825

answered 1 hour ago

Rafa BudríaRafa Budría

6,0521825

1

$begingroup$

You just need to know that $lceil{x}rceil$ is the smallest integer that is greater than or equal to $x$.

Let $x = 0.1$.

Is $0 ≥ x$? No.

Is $1 ≥ x$? Yes.

Is $2 ≥ x$? Yes. However, it is not the smallest integer as $1$ also satisfies this condition.

Try this procedure with $x = -1.7$ and note that $-1.7 color{red}{≤} -1$.

share|cite|improve this answer

answered 1 hour ago

Toby MakToby Mak

3,70011128

$endgroup$

add a comment | 

1

$begingroup$

You just need to know that $lceil{x}rceil$ is the smallest integer that is greater than or equal to $x$.

Let $x = 0.1$.

Is $0 ≥ x$? No.

Is $1 ≥ x$? Yes.

Is $2 ≥ x$? Yes. However, it is not the smallest integer as $1$ also satisfies this condition.

Try this procedure with $x = -1.7$ and note that $-1.7 color{red}{≤} -1$.

share|cite|improve this answer

answered 1 hour ago

Toby MakToby Mak

3,70011128

$endgroup$

add a comment | 

$begingroup$

You just need to know that $lceil{x}rceil$ is the smallest integer that is greater than or equal to $x$.

Let $x = 0.1$.

Is $0 ≥ x$? No.

Is $1 ≥ x$? Yes.

Is $2 ≥ x$? Yes. However, it is not the smallest integer as $1$ also satisfies this condition.

Try this procedure with $x = -1.7$ and note that $-1.7 color{red}{≤} -1$.

share|cite|improve this answer

answered 1 hour ago

Toby MakToby Mak

3,70011128

$endgroup$

share|cite|improve this answer

answered 1 hour ago

Toby MakToby Mak

3,70011128

answered 1 hour ago

Toby MakToby Mak

3,70011128

answered 1 hour ago

Toby MakToby Mak

3,70011128

1

$begingroup$

Let there be an integer valued function $f$ such that $$f(x)-1 < x leq f(x), forall x in Bbb R.$$ Then we have $$x leq f(x) < x+1, forall x in Bbb R.$$ Now two cases may arise which are

$(1)$ $x in Bbb Z.$

$(2)$ $x notin Bbb Z.$

If $x in Bbb Z$ then since $f$ is an integer valued function with $f(x) in [x,x+1)$ it follows that $f(x) = x.$ But then we have $f(x) = lceil f(x) rceil = lceil x rceil.$

Now if $x notin Bbb Z$ then $exists$ a unique integer $n$ with $x<n<x+1.$ So $n$ is the least integer just exceeding $x.$ Therefore $$lceil x rceil = n. (1)$$ On the other hand since $f$ is integer valued with $f(x) in [x,x+1)$ and the only integer in the interval $[x,x+1)$ is $n$ so it also follows that $$f(x)=n. (2)$$ Combining $(1)$ and $(2)$ it follows that $$f(x) = lceil x rceil$$ as required.

This proves the uniqueness of the ceiling function $lceil x rceil.$

share|cite|improve this answer

answered 32 mins ago

Dbchatto67Dbchatto67

3,571626

$endgroup$

add a comment | 

1

$begingroup$

Let there be an integer valued function $f$ such that $$f(x)-1 < x leq f(x), forall x in Bbb R.$$ Then we have $$x leq f(x) < x+1, forall x in Bbb R.$$ Now two cases may arise which are

$(1)$ $x in Bbb Z.$

$(2)$ $x notin Bbb Z.$

If $x in Bbb Z$ then since $f$ is an integer valued function with $f(x) in [x,x+1)$ it follows that $f(x) = x.$ But then we have $f(x) = lceil f(x) rceil = lceil x rceil.$

Now if $x notin Bbb Z$ then $exists$ a unique integer $n$ with $x<n<x+1.$ So $n$ is the least integer just exceeding $x.$ Therefore $$lceil x rceil = n. (1)$$ On the other hand since $f$ is integer valued with $f(x) in [x,x+1)$ and the only integer in the interval $[x,x+1)$ is $n$ so it also follows that $$f(x)=n. (2)$$ Combining $(1)$ and $(2)$ it follows that $$f(x) = lceil x rceil$$ as required.

This proves the uniqueness of the ceiling function $lceil x rceil.$

share|cite|improve this answer

answered 32 mins ago

Dbchatto67Dbchatto67

3,571626

$endgroup$

add a comment | 

$begingroup$

Let there be an integer valued function $f$ such that $$f(x)-1 < x leq f(x), forall x in Bbb R.$$ Then we have $$x leq f(x) < x+1, forall x in Bbb R.$$ Now two cases may arise which are

$(1)$ $x in Bbb Z.$

$(2)$ $x notin Bbb Z.$

If $x in Bbb Z$ then since $f$ is an integer valued function with $f(x) in [x,x+1)$ it follows that $f(x) = x.$ But then we have $f(x) = lceil f(x) rceil = lceil x rceil.$

Now if $x notin Bbb Z$ then $exists$ a unique integer $n$ with $x<n<x+1.$ So $n$ is the least integer just exceeding $x.$ Therefore $$lceil x rceil = n. (1)$$ On the other hand since $f$ is integer valued with $f(x) in [x,x+1)$ and the only integer in the interval $[x,x+1)$ is $n$ so it also follows that $$f(x)=n. (2)$$ Combining $(1)$ and $(2)$ it follows that $$f(x) = lceil x rceil$$ as required.

This proves the uniqueness of the ceiling function $lceil x rceil.$

share|cite|improve this answer

answered 32 mins ago

Dbchatto67Dbchatto67

3,571626

$endgroup$

share|cite|improve this answer

answered 32 mins ago

Dbchatto67Dbchatto67

3,571626

answered 32 mins ago

Dbchatto67Dbchatto67

3,571626

answered 32 mins ago

Dbchatto67Dbchatto67

3,571626