Is NxN Truly Equivalent to N When Using Gaussian Integers?

[cragar]

I thought of a way to use Gaussian integers to show that NxN~N
We look at (1+i)(1-i) and this corresponds to the coordinate (1,1)
then (1+2i)(1-2i)-->(1,2) then (1+3i)(1-3i)-->(1,3)... and you keep doing this, so we have injected NxN into N.

 

[cragar]

actually there is a problem with this (x,y) and (y,x) get mapped to the same integer

 

[Mark44]

It looks to me like your mapping goes from N to N x N. Is that what you intended? (1 + i)(1 - i) = 1 - i² = 1 + 1 = 2. So here the integer 2 is mapped to (1, 1). Did you mean for it to go the other way?

 

[HallsofIvy]

The fundamental problem is that N x N is NOT equivalent to N, it has the same cardinality as the set of rational numbers. It appears that your assignment is "one-to-one" but not "onto".

 

[micromass]

The fundamental problem is that N x N is NOT equivalent to N, it has the same cardinality as the set of rational numbers. It appears that your assignment is "one-to-one" but not "onto".

But $\mathbb{N}$ is equivalent to $\mathbb{N}\times\mathbb{N}$...

 

[Jorriss]

The fundamental problem is that N x N is NOT equivalent to N, it has the same cardinality as the set of rational numbers. It appears that your assignment is "one-to-one" but not "onto".

The rationals and the naturals do have the same cardinality.