The $2$-adic absolute value gives an ultrametric on $\mathbb{Q}$, in which every triangle is isosceles and every point of a ball is a centre

For every $j \in \mathbb{N}$ one has $e_j = j + j$ and $o_j = \sigma(j+j)$, by induction on $j$: at $j = 0$, $e_0 = 0 = 0 + 0$ and $o_0 = \sigma(0) = \sigma(0+0)$; and if $e_j = j+j$ and $o_j = \sigma(j+j)$, then $e_{\sigma(j)} = \sigma(\sigma(e_j)) = \sigma(\sigma(j+j)) = \sigma(j) + \sigma(j)$ and $o_{\sigma(j)} = \sigma(\sigma(o_j)) = \sigma(\sigma(\sigma(j+j))) = \sigma(\sigma(j)+\sigma(j))$, using $\sigma(p) + q = \sigma(p+q)$ and $p + \sigma(q) = \sigma(p+q)$.

No integer $k$ satisfies $0 < k < 1$: such a $k$ would be positive, hence the image of a natural $n \ne 0$, so $n \ge 1$ and, the embedding being order preserving, $k \ge 1$, contradicting $k < 1$.

A product of two odd integers is odd: $(2a+1)(2c+1) = 2(2ac + a + c) + 1$ by ring arithmetic.

Every natural number is $j + j$ for exactly one $j$, or $\sigma(j+j)$ for exactly one $j$, and never both: this is the disjoint-union statement for the ranges of $e$ and $o$, rewritten through step 1.1.

No integer is both even and odd: if $2s = 2t + 1$ then the integer $k := s - t$ satisfies $2k = 1$, and $k \le 0$ would give $2k \le 0 < 1$ while $k \ge 1$ would give $2k \ge 2 > 1$, so $0 < k < 1$, which step 1.2 forbids.

Every integer is even or odd: an integer $m \ge 0$ is the image of a natural $n$, which by step 2.1 is $j+j$ or $\sigma(j+j)$, so $m$ is $2t$ or $2t+1$ with $t$ the image of $j$, the embedding preserving addition and successors; and if $m < 0$ then $-m > 0$ is $2t$ or $2t+1$, whence $m = 2(-t)$ or $m = 2(-t-1)+1$.

The representation is unique: if $2^{\,j}u = 2^{\,j'}u'$ with $u, u'$ odd and, without loss of generality, $j \le j'$, then $2^{\,j'} = 2^{\,j}2^{\,j'-j}$ and cancelling the nonzero factor $2^{\,j}$ gives $u = 2^{\,j'-j}u'$; if $j' > j$ then $j' - j \ge 1$ and $u = 2\big(2^{\,j'-j-1}u'\big)$ is even as well as odd, which step 2.2 forbids; so $j = j'$ and then $u = u'$.

Every nonzero integer $m$ is $2^{\,j}u$ with $j \in \mathbb{N}$ and $u$ odd: apply strong induction to the property $P(n)$ that every nonzero integer whose absolute value is the image of $n$, that is every nonzero $m$ equal to the image of $n$ or to its negative, has such a representation. Given $n$ and $P(k)$ for all $k < n$, note $n \ne 0$ since $m \ne 0$; by step 2.1 either $n = \sigma(j+j)$, in which case $m$ is $\pm(2t+1)$ and hence odd by the computation of step 3.1, so $m = 2^0 m$ works, or $n = j + j$ with $j \ne 0$, in which case $m = 2m'$ for the nonzero integer $m'$ that is the image of $j$ or its negative, and $j < n$ because $j \ge 1$ gives $n = j + j \ge j + 1 > j$, so $P(j)$ supplies $m' = 2^{\,i}u$ and $m = 2^{\,i+1}u$. Every nonzero integer is the image of some natural or its negative, so the conclusion holds for all of them.

For nonzero integers $a, c$ with $a + c \ne 0$: $v_2(a+c) \ge \min\{v_2(a), v_2(c)\}$. Indeed write $a = 2^{\,s}u$ and $c = 2^{\,s'}u'$ with $u, u'$ odd and, without loss of generality, $s \le s'$; then $a + c = 2^{\,s}\big(u + 2^{\,s'-s}u'\big)$, the bracket is a nonzero integer, so by step 4.1 it equals $2^{\,i}q$ with $q$ odd, whence $a + c = 2^{\,s+i}q$ and, by the uniqueness of step 3.2, $v_2(a+c) = s + i \ge s = \min\{s,s'\}$.

The valuation of a nonzero rational is well defined: if $a/b = c/d$ with $a, b, c, d$ nonzero integers then $ad = cb$, and writing $a = 2^{\,s}u$, $b = 2^{\,t}w$, $c = 2^{\,s'}u'$, $d = 2^{\,t'}w'$ with all four of $u,w,u',w'$ odd gives $ad = 2^{\,s+t'}(uw')$ and $cb = 2^{\,s'+t}(u'w)$ with $uw'$ and $u'w$ odd by step 1.3; the uniqueness of step 3.2 applied to the nonzero integer $ad$ forces $s + t' = s' + t$, that is $s - t = s' - t'$.

Basic properties of $|\cdot|_2$: for $x \ne 0$ the value $2^{-v_2(x)}$ is a positive real, since $2 > 0$ and powers and inverses of positives are positive; so $|x|_2 > 0$ for $x \ne 0$ and $|x|_2 = 0$ exactly when $x = 0$. Moreover $|-x|_2 = |x|_2$, because $-x = (-a)/b$ and $-\big(2^{\,s}u\big) = 2^{\,s}(-u)$ with $-u = 2(-t-1)+1$ odd when $u = 2t+1$, so $v_2(-x) = v_2(x)$.

Strong triangle inequality for $|\cdot|_2$: $|x+y|_2 \le \max\{|x|_2, |y|_2\}$. If $x = 0$, or $y = 0$, or $x + y = 0$, this is immediate from step 6.1. Otherwise write $x = a/b$ and $y = c/b$ over a common nonzero denominator $b$, with $a, c$ nonzero integers, so that $x + y = (a+c)/b$ with $a + c \ne 0$; then $v_2(x) = v_2(a) - v_2(b)$, $v_2(y) = v_2(c) - v_2(b)$ and $v_2(x+y) = v_2(a+c) - v_2(b)$, so step 5.1 gives $v_2(x+y) \ge \min\{v_2(x), v_2(y)\}$; finally $t \mapsto 2^{\,t}$ is strictly increasing on $\mathbb{Z}$, since for $t < t'$ one has $2^{\,t'} = 2^{\,t}2^{\,t'-t}$ with $2^{\,t} > 0$ and $2^{\,t'-t} > 1$, so $2^{-v_2(x+y)} \le 2^{-\min\{v_2(x),v_2(y)\}} = \max\{2^{-v_2(x)}, 2^{-v_2(y)}\}$.

Claim 1: $d_2(x,y) = |x-y|_2$ vanishes exactly when $x = y$ by step 6.1, is symmetric because $|y-x|_2 = |-(x-y)|_2 = |x-y|_2$, and satisfies $d_2(x,z) = |(x-y)+(y-z)|_2 \le \max\{|x-y|_2, |y-z|_2\} = \max\{d_2(x,y), d_2(y,z)\}$ by step 7.1; being nonnegative, it also satisfies the ordinary triangle inequality, the maximum of two nonnegative reals being at most their sum. So $d_2$ is an ultrametric on $\mathbb{Q}$.

Claim 2: suppose $d_2(x,y) \ne d_2(y,z)$ and, without loss of generality, $d_2(x,y) < d_2(y,z)$. Then $d_2(x,z) \le \max\{d_2(x,y), d_2(y,z)\} = d_2(y,z)$; and $d_2(y,z) \le \max\{d_2(y,x), d_2(x,z)\} = \max\{d_2(x,y), d_2(x,z)\}$, where the maximum cannot be $d_2(x,y)$, since that would give $d_2(y,z) \le d_2(x,y) < d_2(y,z)$; so $d_2(y,z) \le d_2(x,z)$ and the two are equal, that is $d_2(x,z) = \max\{d_2(x,y), d_2(y,z)\}$.

Claim 3: let $y \in B(x,r)$, so $d_2(x,y) < r$. For $z \in B(y,r)$ the strong triangle inequality gives $d_2(x,z) \le \max\{d_2(x,y), d_2(y,z)\} < r$, so $z \in B(x,r)$; and for $z \in B(x,r)$ it gives $d_2(y,z) \le \max\{d_2(y,x), d_2(x,z)\} < r$, so $z \in B(y,r)$. Hence $B(y,r) = B(x,r)$.

Claims 1, 2 and 3 are established by steps 8.1, 9.1 and 9.2, so the $2$-adic distance is an ultrametric on $\mathbb{Q}$ in which every triangle is isosceles and every point of a ball is a centre.