Velleman exercise 6 in section 4.3

Question

I am stuck on exercise 6 section 4.3, in Daniel J. Velleman's book "How To Prove It".

I just need to prove the following, but cannot do it. The free variables $r$ and $s$ are arbitrary positive real numbers.

$\forall x \in \mathbb{R} \forall z \in \mathbb{R} ( |x-z|<r+s \rightarrow \exists y \in \mathbb{R} (|x-y|<r \wedge |y-z|<s) )$

Can anyone help me find a way to prove it?

Thanks!

edited: changed a $y$ to $z$. thanks coffeemath!

Answer 1

If the distance from $x$ to $z$ is less than $r+s$, the open intervals $(x-r,x+r)$ and $(z-s,z+s)$ have a nonvoid intersection. Pick up any $y$ in this intersection.

Edit: to be specific, let us assume $z>x$ for example. Take $y=\dfrac{x+z}{2}+\dfrac{r-s}{2}$ (midpoint between the right endpoint of $(x-r,x+r)$ and the left endpoint of $(z-s,z+s)$).

Then, it is easy to see that conditions:

$$|x-y|<r \ \ \wedge \ \ |y-z|<s$$

are fulfilled . Explicitly for the first one:

$$x-y=x-\dfrac{x+z}{2}+\dfrac{r-s}{2}=\dfrac{z-x}{2}+\dfrac{r-s}{2}<\dfrac{r+s}{2}+\dfrac{r-s}{2}=r$$

@coffemath I see that, once more, we have the same way of thinking... (I had not seen your post).

Answer 2

With the $z$ where the $y$ was on the left: Let $I=(x-r,x+r),\ J=(z-s,z+s).$ The assumption $|x-z|<r+s$ implies $I \cap J$ is nonempty. Let $y$ be any point in the overlap.

Under the assumption $|x-z|<r+s,$ a specific point $y$ in $I \cap J$ is given by $$y=\frac{sx+rz}{s+r}.$$ To check this, note that $$|x-y|=r \cdot \frac{|x-z|}{r+s},\\ |y-z|=s \cdot \frac{|x-z|}{r+s}.$$ Since the fractions on the right of each of these are strictly less than $1$ we indeed have $|x-y|<r,\ |y-z|<s.$