Every positive integer can be expressed uniquely as $N = a_0 + a_1 2 + a_n 2^n $ and all $ a_i $ belongs to $\{0, 1\}$ I am trying assignment problems of an Institute in which I don't study. I am unable to think about this problem. 


  Problem ->Prove that every positive integer can be expressed uniquely as $N = a_0 + a_1 2 + a_n 2^n $ and all $ a_i $ belongs to $\{0, 1\}$. 


Can someone please help.
The statement is false. For example, the number $N=12$ cannot be written as $12=a_0 + 2a_1 + 2^n a_n$ where $a_i\in\{0,1\}$, and it is easy to show this since:


$a_n$ must be $1$, since otherwise, $a_0 + 2a_1 + 2^n a_n=a_0+2a_1 &lt; 3 &lt; 12$.
$n$ must be equal to $3$, since $n\leq 2$ means that $a_0 + 2a_1 + 2^n a_n$ is at most $7$, and $n\geq 4$, then $a_0 + 2a_1 + 2^n a_n$ is at least $16$.
$12=a_0 + 2a_1 + 2^n a_n$ means that $a_0=0$, since the left side is even, and if $a_0=1$, then the right side is odd.
Therefore, $12=2a_1 + 2^3 a_3 = 2a_1 + 8a_3$, meaning $6=a_1+4a_3$, which is impossible (either $a_1=0$, in which case $6=4a_3$, which is impossible since $a_3$ must be $0$ or $1$, or $a_1=1$, in which case $a_1+4a_3$ is odd, while $6$ is even).
The statement is false. For example, the number $N=12$ cannot be written as $12=a_0 + 2a_1 + 2^n a_n$ where $a_i\in\{0,1\}$, and it is easy to show this since:


$a_n$ must be $1$, since otherwise, $a_0 + 2a_1 + 2^n a_n=a_0+2a_1 &lt; 3 &lt; 12$.
$n$ must be equal to $3$, since $n\leq 2$ means that $a_0 + 2a_1 + 2^n a_n$ is at most $7$, and $n\geq 4$, then $a_0 + 2a_1 + 2^n a_n$ is at least $16$.
$12=a_0 + 2a_1 + 2^n a_n$ means that $a_0=0$, since the left side is even, and if $a_0=1$, then the right side is odd.
Therefore, $12=2a_1 + 2^3 a_3 = 2a_1 + 8a_3$, meaning $6=a_1+4a_3$, which is impossible (either $a_1=0$, in which case $6=4a_3$, which is impossible since $a_3$ must be $0$ or $1$, or $a_1=1$, in which case $a_1+4a_3$ is odd, while $6$ is even).