We just learned about Aleph-naught today and I read about it on wikipedia but I do not know how to go about solving this problem in my homework:
Prove that N(natural numbers) has the same cardinality as 3N(3times the natural numbers)
How would I go about this? I understand that bijections are involved but I do not know how to prove this proof.
On
$\mathbb{N} = \{0, 1, 2, 3, 4, \ldots , x, \ldots \}$
$\mathbb{3N} = \{0, 3, 6, 9, 12, \ldots , 3x, \ldots \}$
If it is true that there exists a bijection: $ \exists R : \mathbb{N}\mathop{\leftrightarrow}^R \mathbb{3N}$, then for every element in the natural numbers, $\mathbb{N}$, there is a one-to-one correspondant element in $3\mathbb{N}$ and hence they have the same cardinallity.
I will take $0 \in \Bbb N$, though that doesn't change much. $\Bbb N = \{0,1,2,3,4,\dots \}$ and $\Bbb {3N} = \{0,3,6,9,12\}$ and you want to find a bijection between these two sets. There seems an easy way to that. Given an $n \in \Bbb N$, what element of $\Bbb {3N}$ would would associate with it?