Statement that $a \geqslant a$ Is it legitimate to make a statement that $a \geqslant a$? This sign means greater or equal, and  surely the second part (equal) will always hold. But maybe someone will disagree and say that I must have used $a = a$ instead.
 A: As other people have pointed out, the notation $a \leq b$ means "$a < b$ or $a = b$." 
For example, the statement $x \leq 5$ means "$x < 5$ or $x = 5$."
Now if a person knew for sure that, say, $x < 5$ or that, say, $x = 5$, they would probably say this, rather than saying, less informatively, that $x \leq 5$. The usefulness of a statement such as "$x \leq 5$ comes from the fact that the person may not know which of these statements is true.
Since the statement 
$$x \leq 5$$ 
means "$x < 5$ or $x = 5$", the statement is certainly true when $x = 5$. Therefore the statement above must also be true when we replace $x$ with $5$:
$$5 \leq 5.$$
The reason that this statement, though indisputably true, now appears so odd is that, now that we know what $x$ is, there is no longer any point in hedging our bets. We might as well just say that $5 = 5$. 
The difficulty is not related to the fact that the statement "$5 < 5$ or $5 = 5$" is hard to interpret logically. It is that on a pragmatic level, we can no longer understand why a person would not just come out and say which of the two possibilities is the correct one.
But if mathematics can allow a statement such as "$x \leq 5$," it must also allow the statement "$5 \leq 5$," in which $x$ has been replaced by a number.
A: Assuming that the "context" of your question is mathematical logic, we must start with a language.
First-order number theory can be the appropriate one. This language has five primitive symbols : $0$, $S$ (the successor function), $+$ and $.$ (the sum and product functions) and the $=$ (equal) predicate, plus countable many individual variables ($v_i$).
We can assume also first-order Peano's Axioms.
In this language, the following are well-formed formulae :

$v_1 = v_2$, $\lnot v_1 = v_2$, $0=0 \lor \lnot 0=1$, $v_1 + v_2 = v_3$, and so on.

In this language, $>$ is not a primitive symbol; we must introduce it as an abbreviation :

$x > y$ stand for : $\exists z (x = y+S(z))$.

Now, we can introduce the expression $x \ge y$ as an abbreviation for : $x > y \lor x = y$.
The interesting thing to do is to prove (starting from Peano's Axioms) the following theorem of f-o number theory :

$\vdash \forall x (x \ge x)$.

A: $a \ge a$ means a is greater than OR equal to a. 
Logically speaking, if one of the conditions "a is greater than a" or "equal to a" holds, then the statement "a is greater than OR equal to a" will hold by the generalization principle of logic.
Since we already have "$a=a$", then "a is greater than OR equal to a" is true.
