The explanation to be given to the student depends upon whether you are teaching him Pure Mathematics or Applied Mathematics. I'd like to write my answer from the Pure Mathematical point of view.
In Mathematics a Number is a Mathematical object, in a less strict language, a symbol. E.g. the natural numbers $2$ and $3$, are symbols which represent the concept of being two and being three. The most common use of the natural numbers is to count things, e.g. $5$ eggs -- $7$ books. Mathematics doesn't bother where these objects(natural numbers) are used in real life or not. Mathematics studies the properties of these objects. Natural numbers $a,b$ and $c$ always follow the law:
$$\begin{cases}
\text{If}\ \ a+c=b+c \\
\ \ \ \ \ \ \ \ \ \ \ \ a=b \tag{Law - I}
\end{cases}$$
The above mentioned law can be understood intuitively by recalling the concept of numerical equality. Let $a,b$ and $c$ represent the number of apples in three different groups. Now, if it is given that $a+c=b+c$, then the only way this can be true is that a and b represent the same number of apples. Law - I can never be proved, so we take it as an axiom.
Another law that these numbers follow is, $$\begin{cases}
\text{If}\ \ a+c> or <b+c \\
\ \ \ \ \ \ \ \ \ \ \ \ a> or <b \tag{Law - II}
\end{cases}$$
Then, I'd explain the concept of subtraction. The symbol $a-b$ is defined by the equation,
$$(a-b)+b=a$$
In case $a=b$ or $a<b$, we don't have any natural number for $a-b$. Again Mathematics does not care where the concept of subtraction is used in real life. It does not care if the $a-a$ or $a-b$( when $a<b$) have any use in real life or not.
We had the law $1$, if $a+c=b+c$ then $a=b$. By the principle of permanence we define the law to be true for, if $(r-r)+c=(p-q)+c$, then $(r-r)=(p-q)$--whatever the meaning of this symbolic equation is. This definition eventualy implies that any symbol $x-x$ doesn't depend upon the value of $x$. So we choose the symbol $0$ for that expression. Also $a-a=b-b=x-x=0.$ By the definition of subtraction we have, $(b-b)+b=b$. This is equivalent to $(a-a)+b=b$
Having defined $0$, I'd explain the negative numbers. A negative number $-b$ is a brief symbol for $0-b$. By the principle of permanence we define all the laws of natural numbers to be true for the symbol $-b$, e.g. commutative, associative etc; whatever the meaning in real life may be assigned to this symbol, it will obey all those laws, even the law-1 and law-2.
$a+(-b)=a+(0-b)=a+0-b=(a+0)-b=a-b$. So $a-a=a+(-a)$.
Consider two numbers $r$ and $s$, such that $b>a$. We have,
$$\ \ \ \ \ \ \ \ \ \ \ b>a$$
$$ \implies \ \ 0+b>0+a$$
$$ \implies (a-a)+b>(b-b)+a$$
$$ \implies (-a)+(a+b)>(-b)+(a+b)$$
$$ \implies -a>-b$$
Briefly, if $b>a$ then $-a>-b$. In particular for the objects $3,4,-3$ and $-4$ we have $4>3$ so $-3>-4$.