average of an inequality what is the result of $<1 + <1$? ;
$<1 + <1 = <2$? or <1 + <1 = <1?
Now what is that number divided by 2?
Either we have:
<2 / 2 = <1 or 
<1 / 2 = <0.5
 A: I have to infer that by "$<1$" you mean the set of all real numbers less than $1$.  That's more commonly written as the open interval $(-\infty,1)$.  If $A$ and $B$ are sets, then there is a well-defined set addition operation, defined by
$$
A + B = \left\{a+b \mid a \in A,\ b\in B\right\}
$$
If $x < 1$ and $y < 1$, then $x+y < 2$.  This tells us that $(-\infty,1) + (-\infty,1) \subseteq (-\infty,2)$.  Moreover, if $z < 2$, then $z/2 < 1$, and since $z = z/2 + z/2$, we know $(-\infty,2) \subseteq (-\infty,1) + (-\infty,1)$.  Taking the two together, we have
$$
    (-\infty,1) + (\infty,1) = (-\infty,2)
$$
or "$\mathop{<}1 + \mathop{<}1 = \mathop{<}2$" as you wrote it.
Similarly, we can define for a set $A$ and number $\alpha$ the set
$$
    \alpha A = \left\{\alpha a \mid a \in A\right\}
$$
It's not hard to show that
$$
    \frac{1}{2}(-\infty,2) = (-\infty,1)
$$
But to be less formal, and reading your whole question, it seems you want to know what assumptions you can make about the average of two numbers if you know their bounds.  If they are both less than one, then yes, their average is less than one.
A: I will take the question to mean this: If all we know about two number is that they are both less then $1$, then what can we conclude about their sum?
In the first place, it is less than $2$.
The harder question, is whether we can conclude any more than that.  And the answer to that is: no.  The reason is this: Suppose $y<2$.  Then $y/2 < 1$.  So add $y/2$ to itself and get $y$.  In other words, any number at all that is less than $2$ can be the sum of two numbers less than $1$.
A: Basic property of inequalities: if $a<b$ and $c<d$ then $a+b<c+d$. Why? Cause when adding $c$ to the left hand side (LHS) and $d$ to the right hand side (RHS), the increment on the LHS is smaller than the increment on the RHS, so the inequality is preserved.
