Last element in the world

We can write the operation as $ab+a+b=(a+1)(b+1)-1$ In general for $A_{1776}=\left\{a_1,a_2,a_3,\ldots,a_{1776}\right\}$ we have $A_{1775}=\left\{(a_1+1)(a_2+1)-1,a_3,\ldots,a_{1776}\right\}$ $A_{1774}=\left\{(a_1+1)(a_2+1)(a_3+1)-1,a_4,\ldots,a_{1776}\right\}$ And so on until $A_{1}=\left\{(a_1+1)(a_2+1)\ldots(a_{1776}+1)-1\right\}$

When $a_n=\frac1n$ for $n=1,2,\ldots,1776$ we have the last element to be $(\frac11+1)(\frac12+1)\ldots(\frac{1}{1776}+1)-1$ $=\frac21\times\frac32\times\ldots\times\frac{1777}{1776}-1$ $=1777-1$ $=\boxed{1776}$

Start the operation with the first two elements, therefore the new first term $a'(1)$ is given by: $a'_1 = ab+a+b = (1)((\frac{1}{2})+1+\frac{1}{2}=2$ .

Do the operation with $a'(1)$ and $a_3$ , the second new first term $a'(2)$ is: $a'(2) = (2)(\frac{1}{3}) + 2 + \frac{1}{3}=3$ .

From above it can be seen that the new first term after $n$ operations is $a'(n)=n+1$ . Therefore, after $1775$ operations the only term left $a'(1775)=1775+1=\boxed{1776}$ .

Let $a,b$ be two numbers chosen from the initial set $A_{1776}$ . Then note that the operation yields a number $ab+a+b=(a+1)(b+1)-1$ . Now, we make the following observations in the second step, which shows that only the following ways are possible for the operation to take place in the second step:

1) The two numbers taken are $(a+1)(b+1)-1$ and another number $c$ , from the other $1774$ elements, resulting in the new number, $[(a+1)(b+1)-1+1](c+1)-1=(a+1)(b+1)(c+1)-1$

2) The two numbers taken are $c,d$ from the rest of the $1774$ elements, resulting in a new number $(c+1)(d+1)-1$ .

This observation allows us to claim the following:

Claim: After any step $k,\ 1775\ge k\ge 1$ , an element of the set $A_{1776-k}$ will have the following form: $\prod_{j=1}^r (a_{j}+1)-1$ where $1\le r\le 1776,\ a_j\in A_{1776}$

Proof: We prove the claim via induction on step number, $k$ . The claim is already verified to be true for $k=1,2$ . Let it be true for $k=n$ . Now, take any two elements $c,d$ of $A_{1776-n}$ . The resulting number, after the operation, will be $(c+1)(d+1)-1$ . By induction hypotheses, $\exists 1\le r_1,r_2\le 1776,\ \{a_i\}_{i=1}^{r_1},\ \{b_j\}_{j=1}^{r_2},\ a_i,b_j\in A_{1776}$ such that $c=\prod_{i=1}^{r_1} (a_i+1)-1,\ d=\prod_{j=1}^{r_2} (b_j+1)-1$ . Thus, the new element is of the form $\prod_{i=1}^{r_1+r_2}(c_i+1)-1$ , where $c_i=a_i,\ 1\le i\le r_1,\ c_i=b_i,\ r_1+1\le i\le r_1+r_2$ . Thus the claim is true for $k=n+1$ , and hence is proved. $\blacksquare$

Thus, we see that the only element of $A_1$ is going to be $\prod_{i=1}^{1776}(1+1/i)-1=1777-1=\boxed{1776}$ .