Deciphering a number system

Question

If I were to say that:

2    +   2   =   22
2    +   q   =   q2
q    +   2   =   q2
22   +   q2  =   q22
2q2  +   q   =   q2q
qqq  +   22  =   qq22

2    x   2   =   22
2    x   q   =   2q
q    x   2   =   2q
2q   x   q   =   2q22
q2q  x   q2  =   q22222
qq   x   qq  =   q222q

could you then tell me what 2qq2 x qq2 is?

Some notes:

• the 'x' in the above is ordinary mathematical multiplication, the '+' is addition, and the '=' is equality.
• each string of 2's and q's is a natural number. Every string is a number, and every number except 0 is a string (including the empty string).
• you do not need every piece of information in the box to determine an answer, though the answer will be consistent with every piece of information in the box.

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Since someone got a correct answer, using different reasoning than I did, I'll add my own reasoning below for those who are interested:

First off, the empty string is 1. If some string x corresponds to some integer $n$, then 2x corresponds to $2n$ and qx corresponds to $2n + 1$. So 2 = $2 \times 1$ = $2$, q = $2 \times 1 + 1$ = $3$ and so on. qq2 = $2 \times (2 \times 2 + 1) + 1$ = 11, and 2qq2 thus = $2 \times 11$ = $22$, the product of which is $242$ = $2 \times (2 \times (2 \times (2 \times (2 \times (2 \times (2 + 1) + 1) + 1))) + 1)$ = 2q22qqq.

Answer 1

My answer is

2q22qqq

The system:

Replace 2 with $0$ and q with $1$.  Add another $1$ at the end, then reverse the order of the string of bits and interpret it as a binary integer.  For example,
2 = $\phantom{00}10 = \phantom{0}2$
22 = $\phantom{0}100 = \phantom{0}4$
222 = $1000 = \phantom{0}8$, etc.
q = $\phantom{00}11 = \phantom{0}3$
qq = $\phantom{0}111 = \phantom{0}7$
qqq = $1111 = 15$, etc.
q2 = $\phantom{0}101 = \phantom{0}5$
2q = $\phantom{0}110 = \phantom{0}6$, and so on.

How I determined this answer:

Given that we are talking about ordinary addition and multiplication, and only the numbers are enciphered, I can say that, for any a,  a+a =$2~\times$a, a+a+a= $3~\times$a, a+a+a+a =$4~\times$a, etc.  So 2+2 equals $2~\times$2.  But 2+2 = 2×2, so 2 equals $2$.

2+2 = 2×2 = 22, so 22 equals $4$.

q ≠$0$, because if q =$0$, then 2+q (which equals q+2 by the commutative property of addition) would equal 2.  (Also, the OP suggested that $0$ was not representable.)  q ≠$1$, because if q =$1$, then 2×q (which equals q×2 by the commutative property of multiplication) would equal 2.  I interpreted the OP’s comment that “every natural number is representable” to mean “every natural number is representable uniquely”, so q ≠ 2($2$) and q ≠$4$.  I guessed that maybe q was $3$, which led to the conclusion that 2+q = $2+3=5$, so q2 must be $5$, and 2×q = $2\times 3=6$, so 2q must be $6$.  So, I had
2 = $2$
q = $3$
22 = $4$
q2 = $5$
2q = $6$
(some of which were guesses).  Between that, and the OP’s hint that some number might be representable as an empty string, I guessed the rest.  All the given equations are (now) consistent with this scheme.

The specific question:

2qq2 = $10110 = 22$
qq2 = $\phantom{0}1011 = 11$

Working in binary:\begin{align}10110\\\times\qquad 1011\\\hline10110\\10110\phantom{0}\\+~~10110\phantom{000}\\\hline11110010\end{align} Working in decimal: $22 \times 11 = 242 = 128 + 64 + 32 + 16 + 2 = 11110010_2$

Either way, strip off the leading $1$ and reverse the remaining bits to get $0100111 \leftrightarrow$ 2q22qqq.