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Show that if $x,y\in [0,1], a,b\geq 0$ and $|x-y|\leq a + b$ then there exists $z\in [0,1]$ such that $|x-z|\leq a$ and $|z-y|\leq b.$

I tried to come up with a general $z$ that works but I am unable to do so. Any hints would be much appreciated.

Student
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  • As a hint about how to proceed, try to find the unique $z$ when $|x - y| = a + b$. It will lie somewhere in the interval $[x, y]$ (or $[y, x]$), so there must be some $\lambda \in [0, 1]$ such that $z = \lambda x + (1 - \lambda) y$. Try to find that $\lambda$ as a function of $a$ and $b$. – Theo Bendit Jan 07 '19 at 23:38
  • Also, I assume $a, b \ge 0$, otherwise the result is clearly false. :-) – Theo Bendit Jan 07 '19 at 23:40
  • @TheoBendit OP stated that $a,b\geq 0$. He just missed it in the title. – Kavi Rama Murthy Jan 07 '19 at 23:44

2 Answers2

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If $a,b\in\mathbb{R}$ have no restriction is false because you can choose one of them negative, so you can't find $z$ which satisfy the request.

However, if $a,b\ge 0$ you can reason like this: suppose $x<y$ and take $z=\frac{a}{a+b}y+\frac{b}{a+b}x$.

By choice we have $z\in[x,y]\subseteq[0,1]$, and moreover \begin{align} |x-z|=z-x &=\frac{a}{a+b}y+\frac{b}{a+b}x-x=\\ &=\frac{a}{a+b}y+\frac{b-a-b}{a+b}x=\\ &=\frac{a}{a+b}y-\frac{a}{a+b}x=\\ &=\frac{a}{a+b}(y-x) \end{align}

So $|z-x|=\frac{a}{a+b}|y-x|<\frac{a}{a+b}(a+b)=a$. In the same way you can prove $|z-y|<b$.

Theo Bendit
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ecrin
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Consider the intervals $[x-a,x+a]$ and $[y-b,y+b]$. If they have a point in common then a common point $z$ will do the trick. Prove that neither interval lies completely to the left of the other , i.e. we cannot have $y+b <x-a$ or $x+a<y-b$. This proves that the intervals intersect. Consider the case $x-a \leq y-b\leq x+a \leq y+b$. If the interval $[y-b,x+a]$ does not intersect $[0,1]$ then either $y-b >1$ or $x+a<0$ and both these are clearly false. Hence we can choose $z$ to be between $0$ and $1$. Similarly argument holds in the other case.

Kavi Rama Murthy
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