Velleman exercise 6 in section 4.3 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!
 A: 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).
A: 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 $|x-y|<r,\ |y-z|<s.$
A: I would see this as a simple calculation driven by the desire to simplify.  (In this answer, all variables range over $\;\mathbb R\;$.)$
\newcommand{\calc}{\begin{align} \quad &}
\newcommand{\op}[1]{\\ #1 \quad & \quad \unicode{x201c}}
\newcommand{\hints}[1]{\mbox{#1} \\ \quad & \quad \phantom{\unicode{x201c}} }
\newcommand{\hint}[1]{\mbox{#1} \unicode{x201d} \\ \quad & }
\newcommand{\endcalc}{\end{align}}
\newcommand{\Ref}[1]{\text{(#1)}}
\newcommand{\then}{\rightarrow}
\newcommand{\when}{\leftarrow}
\newcommand{\equiv}{\leftrightarrow}
\newcommand{\true}{\text{true}}
\newcommand{\false}{\text{false}}
\newcommand{\abs}[1]{\left|#1\right|}
\newcommand{\mx}{\mathbin\max}
\newcommand{\mn}{\mathbin\min}
$
Starting at the most complex side, i.e., the right hand side of $\;\then\;$, we try to eliminate $\;y\;$, and we calculate as follows:
$$\calc
    \exists y (\abs{x−y} < r \land \abs{y−z} < s)
\op\equiv\hint{basic property of $\;\abs\cdot\;$, twice}
    \exists y (-r < x−y < r \;\land\; -s < y−z < s)
\op\equiv\hint{isolate $\;y\;$ in each inequality -- to make it easier to remove}
    \exists y (y < x+r \;\land\; x−r < y \;\land\; z-s < y \;\land\; y < z+s)
\op\equiv\hint{combine inner and outer inequalities}
    \exists y ((x-r) \mx (z-s) < y \;\land\; y < (x+r) \mn (z+s))
\op{\tag{*}\equiv}\hint{between two reals there is always another real}
    (x-r) \mx (z-s) < (x+r) \mn (z+s)
\op\equiv\hint{split into four separate inequalities}
    x-r < x+r \;\land\; x−r < z+s \;\land\; z-s < x+r \;\land\; z-s < z+s
\op\equiv\hints{inner inequalities: bring $\;z,x\;$ and $\;r,s\;$ together;}\hint{outer inequalities: use $\;r,s>0\;$}
    −r-s < z-x \;\land\; z-x < r+s
\op\equiv\hint{basic property of $\;\abs\cdot\;$}
    \abs{z-x} < r+s
\endcalc$$
Note that key step $\;\Ref{*}\;$ also would hold if you were working within $\;\mathbb Q\;$.
