is this range proof i made sound? so to prouve a value $v$ is in range $[0,2n−1]$ we convince the verifier that $v$ is represented by a binary vector $a∈\{0,1\}^n$ so that $&lt;a,2n&gt;=v$
$//$ $2n∈Zn$
is the vector of powers of $2$ satch that $2n={2^0,2^1,2^2,,,,,2^n}$
we have $r∈Zn$ a random vector for blinding
we also have $G$
an eleptice curve generator and $A,R1,R2,R3,R4,V$ are eliptice curve points such that :
$A=&lt;a,1&gt;G,R1=&lt;r,1&gt;G,R2=&lt;2n,r&gt;G,R3=&lt;a,r&gt;G,R4=&lt;r,r&gt;G
$ and $V=vG$
//
for an example $R3$ equals : $R3=(a_1*r_1+a_2∗r_2....+a_n∗r_n)G$
the prouve goes like this:
$-$ the prouver send $A,R1,R2,R3,R4$ to the verifier
$-$ the verifier send back a challenge : $x$
$-$the prouver compute and send:
$fx=xa+r$
$-$ the verifier verify:
$&lt;fx,1&gt;G=?=xA+R1$  $//$ check that $fx$ was constructed corectly
$&lt;fx,fx&gt;G=?=x^2A+xR3+R4$        $//$ check that $a$  is a binary vector because a binary vector is the only vector where $&lt;a,a&gt;=&lt;a,1&gt;$
$&lt;fx,2n&gt;G=?=xV+R2$  $//$ check that $&lt;a,2n&gt;=v$
is this prouve sound? i'm a beginner so probably not
thanks and let me know if anything is unclear
If you're using the syntax $&lt;a, b&gt;$ to mean dot-product, then the assumption that you make:

a binary vector is the only vector where $&lt;a,a&gt;=&lt;a,1&gt;$

is incorrect.  It is correct in the ring $\mathbb{Z}$, however we're not in the integers, we're in a finite field.
Counterexamples in finite fields are easy to find; for example, in $GF(11)$ (picked solely because it's large enough to be nontrivial, but small enough for the computation to be easy), we find that for $a = \{ 2, 5 \}$, we have:
$$&lt;a, a&gt; = 2 \cdot 2 + 5 \cdot 5 = 4 + 3 = 7$$
$$&lt;a, 1&gt; = 2 \cdot 1 + 5 \cdot 1 = 2 + 5 = 7$$