Center of interval belongs to interval

Question

How can I prove that

$$ \frac{a+b} {2}\in [a , b], \quad a<b, \quad(a, b) \in \mathbb{R}^2 $$

Thank you,

Tom

Answer 1

$$a<b\Rightarrow a+a<a+b\Rightarrow 2a<a+b\Rightarrow a<\frac{a+b}{2}$$ Similarly $$a<b\Rightarrow a+b<b+b\Rightarrow a+b<2b\Rightarrow\frac{a+b}{2}<b$$ So, $$\frac{a+b}{2}\in\left[a,b\right]$$

Answer 2

To say that $(a + b)/2 \in [a,b]$ is the same as saying that $(a + b)/2 \geq a$ and $(a + b)/2 \leq b$. Note that we know $a \lt b$, and so we know that $(b - a) > 0$, $(a - b) < 0$.

$(a + b)/2 = (2a + (b - a))/2 = a + (b-a)/2$; since we know that $(b - a)>0$, we have $(a + b)/2 > a$, and a similar argument shows that $(a+b)/2 < b$.

This turns out to be an important result as we develop the topology of the reals: that the midpoint of an interval is always contained within the interval (and moreover, the midpoint of an open interval is strictly contained within the interval)