Expanding an irrational number in form of a simple continued fraction each of the terms $a_i$ in $[a_0;a_1, a_2, a_3, \dots ]$ are 'uniquely' determined since the lowest integer less than or equal to some positive number is unique. But does the uniqueness hold in the reverse direction? That is, is it true that if both $x=[a_0;a_1, a_2, a_3, \dots ]$ and $x=[b_0;b_1, b_2, b_3, \dots ]$ then $a_i=b_i$? If so, how to prove it?
PS The same problem can happen for $x$ as a rational but if I can know the method for an irrational case so may I work on the rational case.
It is unique in both directions, you can find the proof here. In short, in a mathematical induction you can show that if $k$ first partial quotients agree then the $(k+1)$-th must as well, as both $a_{k+1}$ and $b_{k+1}$ are the integer parts of the same number. In other words, as $a_{k+2}$ and $b_{k+2}$ are at least 1 each, $[a_{k+1}, a_{k+2}, \ldots]$ and $[b_{k+1}, b_{k+2}, \ldots]$, for which it holds that $a_{k+1} \le [a_{k+1}, a_{k+2}, \ldots] < a_{k+1}+1$ and $b_{k+1} \le [b_{k+1}, b_{k+2}, \ldots] < b_{k+1}+1$, could never agree if $b_{k+1} \ne a_{k+1}$. The same argument, assuming $a_1, b_1 \ne 0$ (in your notation, as in the link that would be $a_2$, $b_2$), also secures the initial step of the induction.