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I need to solve one equation, but I dont know how to solve equation with floor functions

0

$begingroup$

Im a student with not such a knowledge to solve equations with floor functions. I want to ask, if it is even possible and if it so, how is possible to prove this equation to be true.

- ⌊(n+m)/G⌋ = ⌊(2g-n-m)/G⌋-1

and where :

 G= b^r 
 g= b^r - 1 

when needed , r and b can be replaced by any natural number

Edit1 :
I only need to prove it when G= 2^r and g = 2^r-1 where r is variable

Edit2 : One of the variables, n or m can be fixed.

proof-verification binary floor-function binary-operations

share|cite|improve this question

edited Mar 31 at 13:24

Patrik Bašo

asked Mar 31 at 13:18

Patrik BašoPatrik Bašo

206

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is n,m,a,b,g,G,r? All different variables? There are way too many variables for this to be remotely solvable.
  $endgroup$
  – Don Thousand
  Mar 31 at 13:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I already changed it , there are just variables n,m,G and g . But i only need to prove it when G= 2^r and g = 2^r-1 where r is variable
  $endgroup$
  – Patrik Bašo
  Mar 31 at 13:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  also it is possible to fix one of the variable so for example n=0
  $endgroup$
  – Patrik Bašo
  Mar 31 at 13:22
  

  
  
  
  

add a comment | 

0

$begingroup$

Im a student with not such a knowledge to solve equations with floor functions. I want to ask, if it is even possible and if it so, how is possible to prove this equation to be true.

- ⌊(n+m)/G⌋ = ⌊(2g-n-m)/G⌋-1

and where :

 G= b^r 
 g= b^r - 1 

when needed , r and b can be replaced by any natural number

Edit1 :
I only need to prove it when G= 2^r and g = 2^r-1 where r is variable

Edit2 : One of the variables, n or m can be fixed.

proof-verification binary floor-function binary-operations

share|cite|improve this question

edited Mar 31 at 13:24

Patrik Bašo

asked Mar 31 at 13:18

Patrik BašoPatrik Bašo

206

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is n,m,a,b,g,G,r? All different variables? There are way too many variables for this to be remotely solvable.
  $endgroup$
  – Don Thousand
  Mar 31 at 13:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I already changed it , there are just variables n,m,G and g . But i only need to prove it when G= 2^r and g = 2^r-1 where r is variable
  $endgroup$
  – Patrik Bašo
  Mar 31 at 13:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  also it is possible to fix one of the variable so for example n=0
  $endgroup$
  – Patrik Bašo
  Mar 31 at 13:22
  

  
  
  
  

add a comment | 

$begingroup$

Im a student with not such a knowledge to solve equations with floor functions. I want to ask, if it is even possible and if it so, how is possible to prove this equation to be true.

- ⌊(n+m)/G⌋ = ⌊(2g-n-m)/G⌋-1

and where :

 G= b^r 
 g= b^r - 1 

when needed , r and b can be replaced by any natural number

Edit1 :
I only need to prove it when G= 2^r and g = 2^r-1 where r is variable

Edit2 : One of the variables, n or m can be fixed.

proof-verification binary floor-function binary-operations

share|cite|improve this question

edited Mar 31 at 13:24

Patrik Bašo

asked Mar 31 at 13:18

Patrik BašoPatrik Bašo

206

$endgroup$

- ⌊(n+m)/G⌋ = ⌊(2g-n-m)/G⌋-1

 G= b^r 
 g= b^r - 1 

share|cite|improve this question

edited Mar 31 at 13:24

Patrik Bašo

asked Mar 31 at 13:18

Patrik BašoPatrik Bašo

206

share|cite|improve this question

edited Mar 31 at 13:24

Patrik Bašo

asked Mar 31 at 13:18

Patrik BašoPatrik Bašo

206

asked Mar 31 at 13:18

Patrik BašoPatrik Bašo

206

asked Mar 31 at 13:18

Patrik BašoPatrik Bašo

206

1 Answer
1

active

oldest

votes

0

$begingroup$

If $G=2^r$; $g=2^r-1$ and $n+m=k$ then we have
$$- Biglfloorfrack2^rBigrfloor = Biglfloorfrac2^r+1-2-k2^rBigrfloor-1$$
$$Biglfloorfrack2^rBigrfloor+Biglfloorfrac2^r+1-2-k2^rBigrfloor=1$$
$$Biglfloorfrack2^rBigrfloor+Biglfloor2-frack+22^rBigrfloor=1$$
As $kgt0$ we need one of the floor expressions to evaluate to $0$ and the other evaluate to $1$ in order for this to be true. This occurs when
$$2^rle kle2^r+1-2$$
So a valid solution is any values of $n,m$ such that
$$Gle n+mle2g$$
If you fix one of the variables, for example $n=0$ then we have a range of solutions
$$Gle mle2g$$

share|cite|improve this answer

edited Mar 31 at 13:38

answered Mar 31 at 13:29

Peter ForemanPeter Foreman

7,2371318

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thank you so much, now I know I did some error before this conclusion, because it is not fitting to my case, I think I will make another thread completely describing my case, but anyways, thank you man
  $endgroup$
  – Patrik Bašo
  Mar 31 at 13:39
  

  
  
  
  

add a comment | 

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0

$begingroup$

If $G=2^r$; $g=2^r-1$ and $n+m=k$ then we have
$$- Biglfloorfrack2^rBigrfloor = Biglfloorfrac2^r+1-2-k2^rBigrfloor-1$$
$$Biglfloorfrack2^rBigrfloor+Biglfloorfrac2^r+1-2-k2^rBigrfloor=1$$
$$Biglfloorfrack2^rBigrfloor+Biglfloor2-frack+22^rBigrfloor=1$$
As $kgt0$ we need one of the floor expressions to evaluate to $0$ and the other evaluate to $1$ in order for this to be true. This occurs when
$$2^rle kle2^r+1-2$$
So a valid solution is any values of $n,m$ such that
$$Gle n+mle2g$$
If you fix one of the variables, for example $n=0$ then we have a range of solutions
$$Gle mle2g$$

share|cite|improve this answer

edited Mar 31 at 13:38

answered Mar 31 at 13:29

Peter ForemanPeter Foreman

7,2371318

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thank you so much, now I know I did some error before this conclusion, because it is not fitting to my case, I think I will make another thread completely describing my case, but anyways, thank you man
  $endgroup$
  – Patrik Bašo
  Mar 31 at 13:39
  

  
  
  
  

add a comment | 

0

$begingroup$

If $G=2^r$; $g=2^r-1$ and $n+m=k$ then we have
$$- Biglfloorfrack2^rBigrfloor = Biglfloorfrac2^r+1-2-k2^rBigrfloor-1$$
$$Biglfloorfrack2^rBigrfloor+Biglfloorfrac2^r+1-2-k2^rBigrfloor=1$$
$$Biglfloorfrack2^rBigrfloor+Biglfloor2-frack+22^rBigrfloor=1$$
As $kgt0$ we need one of the floor expressions to evaluate to $0$ and the other evaluate to $1$ in order for this to be true. This occurs when
$$2^rle kle2^r+1-2$$
So a valid solution is any values of $n,m$ such that
$$Gle n+mle2g$$
If you fix one of the variables, for example $n=0$ then we have a range of solutions
$$Gle mle2g$$

share|cite|improve this answer

edited Mar 31 at 13:38

answered Mar 31 at 13:29

Peter ForemanPeter Foreman

7,2371318

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thank you so much, now I know I did some error before this conclusion, because it is not fitting to my case, I think I will make another thread completely describing my case, but anyways, thank you man
  $endgroup$
  – Patrik Bašo
  Mar 31 at 13:39
  

  
  
  
  

add a comment | 

$begingroup$

If $G=2^r$; $g=2^r-1$ and $n+m=k$ then we have
$$- Biglfloorfrack2^rBigrfloor = Biglfloorfrac2^r+1-2-k2^rBigrfloor-1$$
$$Biglfloorfrack2^rBigrfloor+Biglfloorfrac2^r+1-2-k2^rBigrfloor=1$$
$$Biglfloorfrack2^rBigrfloor+Biglfloor2-frack+22^rBigrfloor=1$$
As $kgt0$ we need one of the floor expressions to evaluate to $0$ and the other evaluate to $1$ in order for this to be true. This occurs when
$$2^rle kle2^r+1-2$$
So a valid solution is any values of $n,m$ such that
$$Gle n+mle2g$$
If you fix one of the variables, for example $n=0$ then we have a range of solutions
$$Gle mle2g$$

share|cite|improve this answer

edited Mar 31 at 13:38

answered Mar 31 at 13:29

Peter ForemanPeter Foreman

7,2371318

$endgroup$

share|cite|improve this answer

edited Mar 31 at 13:38

answered Mar 31 at 13:29

Peter ForemanPeter Foreman

7,2371318

answered Mar 31 at 13:29

Peter ForemanPeter Foreman

7,2371318

answered Mar 31 at 13:29

Peter ForemanPeter Foreman

7,2371318