<s>[INST] What's the correct definition of "conjugate", and do we identify them? 
What does a "conjugate" in math mean?

I know the definition says if we have $x+y$, then $x-y$ is its conjugate.
So, $5+ \sqrt{5}$ and $5-\sqrt{5}$ are conjugate of one another. 
Are $8$ and $2$ conjugate of one another as they can be $5+3$ and $5-3$?
Now, $5$ and $\sqrt{2}$ are not conjugate. How? We can write them like $\frac{5 + \sqrt{2}+(5-\sqrt{2})}{2}$ and $\frac{5 + \sqrt{2}-(5-\sqrt{2})}{2}$.
Now, they are.
So, what's the concept?
 [/INST] Conjugates are usually of two different domains. For example, in your case of $5 \pm \sqrt{2}$, the first number is rational, the second is not. The reason why this is useful is when the product and sum of conjugates is easier to deal with than the conjugates themselves. Again, in your example, the sum of the two, and their product, are rational (and actually are integers).
Another such example are complex conjugates. E.g. $a \pm bi$ where $a,b$ are reals and $i = \sqrt{-1}$. Then, sum of the conjugates is real, their difference is imaginary (still easier to deal with than general complex) and their product is real:
$$
(a+bi)(a-bi) = a^2 + b^2.
$$
Of course you can write any two reals as conjugates, but generally that does not buy you much advantage unless you are trying to get rid of some property which either conjugates have but their sum or product will not.
Another example. Consider simplifying
$$
\frac{1}{a + b\sqrt{c}}
 = \frac{1}{a + b\sqrt{c}} \times \frac{a - b\sqrt{c}}{a - b\sqrt{c}}
 = \frac{a - b\sqrt{c}}{a^2-b^2c}
 = \frac{a}{a^2-b^2c} - \frac{b}{a^2-b^2c} \sqrt{c},
$$
which for real $a,b,c>0$ ends up with a square root in the numerator instead of a square root in the denominator, typically a much simpler expression to manage. To illustrate on specific numbers,
$$
\frac{1}{3 + 2\sqrt{2}} = 3 - 2\sqrt{2}.
$$
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