The renown Cantor's Diagonal Argument shows that although we can easily map the set of natural numbers to the set of real numbers in many ingenious ways, we can never find a surjection $f:\mathbb N\rightarrow\mathbb R$. A surjection here means $f(\mathbb N)=\mathbb R$. Instead, as Cantor proved, we can do no better than $f(\mathbb N)\subset\mathbb R$.
This means that any counting function $f:\mathbb N\rightarrow\mathbb R$ assigning enumerations to the real numbers such that the real numbers can be written as a list $\mathbb R=\{f(1),f(2),...\}$ is impossible. The set $\{f(1),f(2),...\}$ can contain some of the real numbers, but not all.
In this sense, which is called cardinality (see link in comments), the set of real numbers is bigger.
There are various proofs of Cantor's Diagonal Argument. One is to prove the more general statement that given a set $A$, the so-called Power Set $\mathcal P(A)$ containing all subsets of $A$ always has bigger cardinality than $A$.
So assume we have found an enumeration $f:\mathbb N\rightarrow\mathcal P(\mathbb N)$. Then $f(n)$ is some subset of $\mathbb N$ as for instance $\{2,4,6,8,...\}$. If $f$ were surjective it should be possible for any element $B\in\mathcal P(\mathbb N)$ to find some $b\in \mathbb N$ with $f(b)=B$.
Consider $B\in\mathcal P(\mathbb N)$ defined as
$$
B=\{n\in\mathbb N\mid n\notin f(n)\}
$$
Assume we have found $b$ such that $f(b)=B$. Suppose $b\in B=f(b)$, but then the above definition implies that $b$ should not be an element of $B$. So we must have $b\notin B=f(b)$, but then the above definition tells us that $b$ should be in $B$. Contradiction! For any function $f$ the set $B$ above is well defined, but as just shown no element of $\mathbb N$ can map to it. We must have $f(\mathbb N)\subset\mathcal P(\mathbb N)\setminus B$, so in particular $f(\mathbb N)\neq\mathcal P(\mathbb N)$ for any choice of $f$.
Let us connect the above argument to the real numbers. We can make a simple mapping $g:\mathcal P(\mathbb N)\rightarrow\mathbb R$ that takes a given subset of $B=\{n_1,n_2,...\}\in\mathcal P(\mathbb N)$ of natural numbers and maps it to
$$
g(B)=10^{-n_1}+10^{-n_2}+...=\sum_{n\in B}10^{-n}
$$
For instance $B=\{2,4,6,8,...\}$ would map to $g(B)=0.010101\overline{01}$. This mapping is injective meaning that different sets $B$ map to different real numbers $g(b)$. So $\mathcal P(\mathbb N)$ has the same cardinality as the subset $g(\mathcal P(\mathbb N))$ of the real numbers meaning that we have the cardinality inequalities
$$
|\mathbb N|<|\mathcal P(\mathbb N)|=|g(\mathcal P(\mathbb N))|\leq|\mathbb R|
$$
In fact a different choice of $g$ shows that $|\mathcal P(\mathbb N)|=|\mathbb R|$, but that is a different story.
I'm going to do something I rarely do, namely defining $\Bbb N$ to include $0$ as an element. It makes the treatment herein a little easier or at least a little more Peano-conventional; feel free to adjust it, as an exercise, to start $\Bbb N$ at $1$.
From $$0\in\Bbb N,\,\forall n\in\Bbb N (Sn\in\Bbb N),\,a+0=a,\,a+Sb=S(a+b),\,a\times 0=a,\,a\times Sb=a\times b+a$$and $$\varphi(0)\land\forall n(\varphi(n)\to\varphi(n+1))\to\forall n\in\Bbb N(\varphi(n))$$we'll prove $\forall a,\,b\in\Bbb N(a+b\in\Bbb N)$, and as a separate theorem $\forall a,\,b\in\Bbb N(a\times b\in\Bbb N)$. For both proofs we'll induct on $b$.
First, addition. The case $b=0$ is trivial because $a+0=a\in\Bbb N$. Now we just need the inductive step; if it works when $b=k$ then $a+k\in\Bbb N\implies a+Sk=S(a+k)\in\Bbb N$.
Next, multiplication; the addition result is actually needed in the inductive step. Again, $b=0$ is trivial because $a\times 0=0\in\Bbb N$. Now the inductive step, assuming $a\times k$ exists; then $a\times Sk=(a\times k)+a$ is the sum of two elements of $\Bbb N$, completing the proof by the above result.
You may also wish as an exercise to use this result about multiplication to in turn prove $a^b\in\Bbb N$ using $a^0=1,\,a^{Sb}=a^b\times a$.
Best Answer
The set of natural numbers is denoted $\mathbb N$ and the set of pairs of natural numbers is denoted $\mathbb{N \times N}$. These two sets have the same cardinality between them. What you suggest with your $15$ example is a bijection $\mathbb{N \to N \times N}$.
Given a number $a \in \mathbb N$ write it as a decimal number. Map this to $(b, c)$ where $b$ are the even digits of $a$ and $c$ are the odd (counting from the $1$'s place). So $15 \mapsto (1, 5)$, $112 \mapsto (1, 12)$, and $1000001 \mapsto (0, 1001)$.
Given a pair of numbers $(b, c)$ we map $(b, c) \mapsto a$ where $a$ is the number created by interleaving the digits of $b$ and $c$, starting with the first digit of $c$. If the numbers are not the same length we add zeros to make them so before interleaving. So $(10, 1240) = (0010, 1240) \mapsto 1021400$ and $(1, 12) \mapsto 112$.
These two operations are inverse to each other.