# cardinality of a set of natural lattice points versus natural numbers

Given a set D = $\{(a,b)∣a,b ∈ \mathbb{N}\}$ where L is the set of all points in the first quadrant whose coordinates are natural numbers.

Which has more elements, D or $\mathbb{N}$?

I know it has something to do with finding a surjection or injection, which can help decide which has more elements, but it's also possibly a bijection? Not sure which has more or if they have the same cardinality/size.

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What is $L$? It seems you mean $D$ –  Thomas Andrews Dec 1 '13 at 15:24
Depends on what you mean by "formula." –  Thomas Andrews Dec 1 '13 at 15:26
a formula that you can solve. For example, not related to this problem, a formula such as f(x)=2x where the domain is N and the codomain is 2N. something like that where there's an operation applied to the function –  gticecream8 Dec 1 '13 at 17:56

Using the cantor Scherer theorem: if we can find an injection $f$ from $\mathbb N$ to $D$ and a surjection $g$ from $\mathbb N$ to $D$ then we are done.

for the surjection let $f((a,b))= a$.

For the injection let $g((a,b))=a+b+a$.

see this video to see why the injection is indeed an injection.

Infinity is bigger than you think

and see they are the same thing.

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How about $$s(n)=(i+1,j+1)\text{ if }n=2^i3^ju\text{ with }\gcd(6,u)=1$$ or $$s(n)=\left(\left\lfloor \frac{n-\lfloor\sqrt n\rfloor^2}2\right\rfloor,\lfloor\sqrt n\rfloor-\left\lfloor \frac{n-\lfloor\sqrt n\rfloor^2}2\right\rfloor\right)$$ (Adjustments according to your local deifnition of $\mathbb N$ may be necessary).

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