find the total Q hello i have a problem with exercise
the problem is follow:
Consider the set $Q$ of integers defined as follows:
$1 ∈ Q$
If $b ∈ Q$, then $2b-1 ∈ Q$
If $b ∈ Q$, then $2b +1 ∈ Q$
What is the total $Q$? Prove your answer carefully.
I have found a solution but I have not shown with any particular methodology. Is there concrete evidence; thank you very much
 A: Let's add some rigor to David's post.
Claim: The set of odd positive integers is contained in $Q$.
Proof: (Induction) We are given $1 \in Q$. Now assume all odd positive integers less than $2n+1$ are in $Q$ (inductive hypothesis). Consider $n$. Either $n$ is odd so that $n<2n+1$ and thus by hypothesis $n \in Q$ and therefore, by the third condition $2n+1\in Q$. Or $n$ is even. In this case $n+1$ is odd and $n+1<2n+1$ thus by hypothesis $n+1\in Q$. Thus by the second condition, $2(n+1)-1=2n+1 \in Q$. Therefore, by induction all positive odd integers are included in $Q$.
Claim 2: These are the only elements of $Q$ (i.e. $Q={ n \in \mathbb{Z} \;|\; n>0 \;\mathrm{and}\; n \;\mathrm{odd}}$).
Proof: Since $1>0$ and given $n\geq 1$ then $2n\pm 1 \geq 1$, we have that all elements in $Q$ must be positive. Next, since $1$ is odd and $2n\pm 1$ is odd for an integer, it must be that all the elements of $Q$ are odd. Therefore, $Q$ is precisely the set we claimed it is.
A: Start with $1\in Q$
Applying the other conditions to $\{ 1\}$, we see that $3\in Q$.
So $\{1,3\}\subseteq Q$.
Applying the conditions to each new element of the above, we see
$
5
$ and $7$ are in $Q$.
So
$$
 \{1,3,5,7\}\subseteq Q.
$$
Applying the conditions to each new element of the above, we see
$
9
$, $11$, $13$, and $15$ are in $Q$.
So
$$
 \{1,3,5,7, 9, 11, 13, 15\}\subseteq Q.
$$
The next step will give you
$$
 \{1,3,5,7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31\}\subseteq Q.
$$
And you can continue in this manner to show that $Q$ contains all odd integers.
Since the conditions always give odd integers, $Q$ is in fact the set of odd integers. (That is, if $Q$ is determined by the conditions. As stated, there is the possibility that other even numbers are in it.  But the conditions won't give new even numbers).
