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What I mean to say is that " Is the Cardinal number of the set containing natural numbers same as the Cardinal number of the set containing the natural numbers divisible by 17 (or any other number ) "

I feel like the answer should be true in a case where the set of natural numbers would contain an infinite number of elements and the set containing the multiples of 17 would have an infinite number of multiples as well. Thus the Cardinal number of both sets should be same, right ?

Kevin Peter
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  • The fact that both sets are infinite is not enough. You need come up with some bijective map between the two sets. – math Jan 24 '17 at 02:47
  • You have the right idea but there are more than one infinity. Example the $\mathbb N$ has a different cardinality than $\mathbb R$. To show that they have the same cardinality is to show there is a 1-1 correspondence. If every $n \in \matnbb N$ there is a $17*n \in ${multiples of 17}. and for every $M \in ${multiples of 17}$ there is a $M/17 \in \mathbb N$ so that is a 1-1 corespondence. So the do have the same cardinality. (Oh, it's a bit more complicated as we need to deal with negative multiples. There's a trick but the answer is the same.) – fleablood Jan 24 '17 at 02:49
  • Do you know what a bijective map is? The answer to your question, if I understand it, is yes, but to prove it, we need to go a little deeper. – The Count Jan 24 '17 at 02:49
  • Despite all the comments, if you have an infinite subset of $\Bbb N$ it has cardinality $\aleph_0$, the same as $\Bbb N$. This relies on the fact that $\aleph_0$ is the smallest infinite cardinality, so if you remove elements but don't drop to finite you haven't changed. So multiples of $n$, squares, $879$th powers, etc. all have the same cardinality. – Ross Millikan Jan 24 '17 at 03:56

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It's not as easy as saying that both sets are of infinite size, as there are plenty of examples of two infinite sized sets that are do not have the same cardinality, e.g. The set of real numbers is infinite, but its cardinality is greater than that of the natural numbers.

To show that two sets do have the same cardinility, you have to show that there exists a bijection between the two sets that covers all elements. In your case that is actualy quite easy:

Pair up 0 with 0, 1 with 17, 2 with 34, etc.

Bram28
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  • Technically that is a bijection between the natural numbers and the non-negative multiples... (just to nitpick...) – fleablood Jan 24 '17 at 02:51
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When taking about infinity, one should be cautious. Not all infinities are equal.

Here, for cardinality, we say that that infinite sets have the same cardinality if and only if there exists a one-to-one mapping between each set.

In your example, we can create such a mapping.

$$\begin{array}[ccccc] .0 &1 & 2 & 3 & 4 & \dots \\ 17 \times 0 & 17\times 1&17\times 2&17\times 3&17\times 4 & \dots \end{array}$$

In this mapping, one nonnegative integer corresponds to one unique multiple of $17$, and vice-versa.

We say that there is a bijection between these sets (the bijection being the mapping we found), so, by definition of infinite sets of same cardinality, they have the same cardinality.

However, note that not all infinite sets have the same cardinality. The classical example is the set of natural numbers and the set of real numbers (check out Cantor's diagonal argument on Wikipedia)

Mariuslp
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