Let $$A = \{0, 3, 4, 6, 7, 8, 9, 10\}$$ and $$R$$ be the relation defined on $$A$$ such that $$R\{(x,y) \in A \times A: x-y$$ is odd positive integer or $$x-y = 2\}$$. The minimum number of elements that must be added to the relation $$R$$, so that it is a symmetric relation, is equal to ______.

To solve this as written, we must find every pair $$(x, y)$$ that satisfies the given rules and then count how many "reverse" pairs $$(y, x)$$ are missing.

The set is $$A = \{0, 3, 4, 6, 7, 8, 9, 10\}$$.

Condition A: $$x - y = 2$$

Comparing all elements where the difference is exactly 2:

1. $$(6, 4)$$
2. $$(8, 6)$$
3. $$(9, 7)$$
4. $$(10, 8)$$
5. From 3: $$(3, 0)$$
6. From 4: $$(4, 3)$$
7. From 6: $$(6, 3)$$
8. From 7: $$(7, 0), (7, 4), (7, 6)$$
9. From 8: $$(8, 3), (8, 7)$$
10. From 9: $$(9, 0), (9, 4), (9, 6), (9, 8)$$
11. From 10: $$(10, 3), (10, 7), (10, 9)$$
12. A negative number can never be a "positive odd integer."
13. A negative number can never equal $$2$$.

(Total: 4 pairs)

Condition B: $$x - y$$ is an odd positive integer

We look for $$x > y$$ where the result is $$1, 3, 5, 7, 9...$$:

(Total: 15 pairs)

For $$R$$ to be symmetric, if $$(x, y) \in R$$, then $$(y, x)$$ must be in $$R$$.

In the current relation, $$x$$ is always greater than $$y$$ ($$x - y = 2$$ or $$x - y = \text{positive odd}$$). Therefore, in every existing pair, $$x > y$$.

For any reverse pair $$(y, x)$$, the difference $$y - x$$ would be negative.

This means none of the reverse pairs are currently in $$R$$.

Total elements in $$R = 4 \text{ (from Cond A)} + 15 \text{ (from Cond B)} = 19$$.

To make it symmetric, you must add the reverse of every element.

Mathematically, the number to add is 19.