Similarity and Dominance.

Clash Royale CLAN TAG#URR8PPP

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I have problem card($mathbb Ntimes mathbb N$) $=$ card($mathbb N$).

So we have to show that $mathbb Ntimes mathbb N$ $approx$ $mathbb N$.

($Aapprox$ B that mean there exists a bijection between A and B , where A and B be sets.)

We define a bijection that is $f(m,n)=2^m-1(2n-1)$ for all $m,ninmathbb Ntimesmathbb N$

So i want to show that card ($mathbb Ntimes mathbb Ntimesmathbb N$) $=$ card($mathbb N$), i can't define a bijection between $mathbb Ntimes mathbb Ntimesmathbb N$ and $mathbb N$.

Please, hlep me.

Moreover , please hint me to show card ($mathbb N^n$) $=$ card($mathbb N$) for all $nin mathbb N$.

elementary-set-theory

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edited Aug 3 at 16:55

Andrés E. Caicedo

63.2k7151236

asked Jul 17 at 2:21

ruler maroon

15

add a comment | 

up vote
0
down vote

favorite

I have problem card($mathbb Ntimes mathbb N$) $=$ card($mathbb N$).

So we have to show that $mathbb Ntimes mathbb N$ $approx$ $mathbb N$.

($Aapprox$ B that mean there exists a bijection between A and B , where A and B be sets.)

We define a bijection that is $f(m,n)=2^m-1(2n-1)$ for all $m,ninmathbb Ntimesmathbb N$

So i want to show that card ($mathbb Ntimes mathbb Ntimesmathbb N$) $=$ card($mathbb N$), i can't define a bijection between $mathbb Ntimes mathbb Ntimesmathbb N$ and $mathbb N$.

Please, hlep me.

Moreover , please hint me to show card ($mathbb N^n$) $=$ card($mathbb N$) for all $nin mathbb N$.

elementary-set-theory

share|cite|improve this question

edited Aug 3 at 16:55

Andrés E. Caicedo

63.2k7151236

asked Jul 17 at 2:21

ruler maroon

15

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

I have problem card($mathbb Ntimes mathbb N$) $=$ card($mathbb N$).

So we have to show that $mathbb Ntimes mathbb N$ $approx$ $mathbb N$.

($Aapprox$ B that mean there exists a bijection between A and B , where A and B be sets.)

We define a bijection that is $f(m,n)=2^m-1(2n-1)$ for all $m,ninmathbb Ntimesmathbb N$

So i want to show that card ($mathbb Ntimes mathbb Ntimesmathbb N$) $=$ card($mathbb N$), i can't define a bijection between $mathbb Ntimes mathbb Ntimesmathbb N$ and $mathbb N$.

Please, hlep me.

Moreover , please hint me to show card ($mathbb N^n$) $=$ card($mathbb N$) for all $nin mathbb N$.

elementary-set-theory

share|cite|improve this question

edited Aug 3 at 16:55

Andrés E. Caicedo

63.2k7151236

asked Jul 17 at 2:21

ruler maroon

15

I have problem card($mathbb Ntimes mathbb N$) $=$ card($mathbb N$).

So we have to show that $mathbb Ntimes mathbb N$ $approx$ $mathbb N$.

($Aapprox$ B that mean there exists a bijection between A and B , where A and B be sets.)

We define a bijection that is $f(m,n)=2^m-1(2n-1)$ for all $m,ninmathbb Ntimesmathbb N$

So i want to show that card ($mathbb Ntimes mathbb Ntimesmathbb N$) $=$ card($mathbb N$), i can't define a bijection between $mathbb Ntimes mathbb Ntimesmathbb N$ and $mathbb N$.

Please, hlep me.

Moreover , please hint me to show card ($mathbb N^n$) $=$ card($mathbb N$) for all $nin mathbb N$.

elementary-set-theory

share|cite|improve this question

edited Aug 3 at 16:55

Andrés E. Caicedo

63.2k7151236

asked Jul 17 at 2:21

ruler maroon

15

share|cite|improve this question

share|cite|improve this question

edited Aug 3 at 16:55

Andrés E. Caicedo

63.2k7151236

edited Aug 3 at 16:55

Andrés E. Caicedo

63.2k7151236

edited Aug 3 at 16:55

Andrés E. Caicedo

63.2k7151236

63.2k7151236

asked Jul 17 at 2:21

ruler maroon

15

asked Jul 17 at 2:21

ruler maroon

15

asked Jul 17 at 2:21

ruler maroon

15

15

add a comment | 

add a comment | 

1 Answer
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Hint: We can define a bijection $g:Bbb N^3 to Bbb N^2$ by
$$
g(p,q,r) = (f(p,q),r)
$$
Note that $f circ g: Bbb N^3 to Bbb N$ is a bijection, since it is a composition of bijections.

In general, we can use this trick to get bijections between $Bbb N^n$ and $Bbb N^n-1$.

share|cite|improve this answer

edited Jul 17 at 2:36

answered Jul 17 at 2:34

Omnomnomnom

121k784170

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank, i try it.
  – ruler maroon
  Jul 17 at 2:36
  

  
  
  
  

add a comment | 

Your Answer

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote

Hint: We can define a bijection $g:Bbb N^3 to Bbb N^2$ by
$$
g(p,q,r) = (f(p,q),r)
$$
Note that $f circ g: Bbb N^3 to Bbb N$ is a bijection, since it is a composition of bijections.

In general, we can use this trick to get bijections between $Bbb N^n$ and $Bbb N^n-1$.

share|cite|improve this answer

edited Jul 17 at 2:36

answered Jul 17 at 2:34

Omnomnomnom

121k784170

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank, i try it.
  – ruler maroon
  Jul 17 at 2:36
  

  
  
  
  

add a comment | 

up vote
1
down vote

Hint: We can define a bijection $g:Bbb N^3 to Bbb N^2$ by
$$
g(p,q,r) = (f(p,q),r)
$$
Note that $f circ g: Bbb N^3 to Bbb N$ is a bijection, since it is a composition of bijections.

In general, we can use this trick to get bijections between $Bbb N^n$ and $Bbb N^n-1$.

share|cite|improve this answer

edited Jul 17 at 2:36

answered Jul 17 at 2:34

Omnomnomnom

121k784170

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank, i try it.
  – ruler maroon
  Jul 17 at 2:36
  

  
  
  
  

add a comment | 

up vote
1
down vote

up vote
1
down vote

Hint: We can define a bijection $g:Bbb N^3 to Bbb N^2$ by
$$
g(p,q,r) = (f(p,q),r)
$$
Note that $f circ g: Bbb N^3 to Bbb N$ is a bijection, since it is a composition of bijections.

In general, we can use this trick to get bijections between $Bbb N^n$ and $Bbb N^n-1$.

share|cite|improve this answer

edited Jul 17 at 2:36

answered Jul 17 at 2:34

Omnomnomnom

121k784170

Hint: We can define a bijection $g:Bbb N^3 to Bbb N^2$ by
$$
g(p,q,r) = (f(p,q),r)
$$
Note that $f circ g: Bbb N^3 to Bbb N$ is a bijection, since it is a composition of bijections.

In general, we can use this trick to get bijections between $Bbb N^n$ and $Bbb N^n-1$.

share|cite|improve this answer

edited Jul 17 at 2:36

answered Jul 17 at 2:34

Omnomnomnom

121k784170

share|cite|improve this answer

share|cite|improve this answer

edited Jul 17 at 2:36

edited Jul 17 at 2:36

edited Jul 17 at 2:36

answered Jul 17 at 2:34

Omnomnomnom

121k784170

answered Jul 17 at 2:34

Omnomnomnom

121k784170

answered Jul 17 at 2:34

Omnomnomnom

121k784170

121k784170

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank, i try it.
  – ruler maroon
  Jul 17 at 2:36
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank, i try it.
  – ruler maroon
  Jul 17 at 2:36
  

  
  
  
  

Thank, i try it.
– ruler maroon
Jul 17 at 2:36

Thank, i try it.
– ruler maroon
Jul 17 at 2:36

add a comment | 

 

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