Are there any $x,y,z$ in $\mathbb{R}$ for which the following equations hold?

Question

Are there any $x,y,z \in \mathbb{R}$ for which the following equations hold? $$|x+1| \leq 2\\ |y+1| \leq 3\\ |y-z| \leq 1$$

With the given we know that $x$ is between $[-3,1]$, $y$ is between $[-4,2]$, and $(y-z)$ is between $[-1,1]$.

Using a triangle inequality:

$$|(x+1)+(-y-1)|= |x-y| \leq |x+1| + |y+1|$$

there i can see that |x-y| can be maximum 7; and on the right side |x+1| can me maximum 2; and |y+1| maximum 3; which is a contradiction since 7 is never less than 5.

maybe this is proven but am not sure The anwser to this question is no.

any feedback is welcome Thanks.

Yes From class. and the answer was given, but no explaination — Abel, Jan 12 '14 at 13:56
It seems to me that you can forget about the first condition. Namely, $|x+1|\leq 2.$ This is trivially satisfiable, and since $x$ does not feature in any of the other equations, you can simply forget about it. — goblin GONE, Jan 12 '14 at 15:02
That is all that was given. i was told it could be shown with the triangle innequality used above... — Abel, Jan 12 '14 at 16:22

score 1 · Accepted Answer · edited Jan 12 '14 at 15:06

1

For every $$x \in [-3,1]$$ and $$y \in [-4,2]$$ it is possible to select some $z$: $$z \in [y-1,y+1]$$

edited Jan 12 '14 at 15:06

hardmath

37,015

answered Jan 12 '14 at 13:49

V-X

665

1

This points out that the choice of $x \in [-3,1]$ does not restrict the further choices of $y,z$, and that indeed $z$ can easily be chosen based only on a prior choice for $y \in [-4,2]$. – hardmath Jan 12 '14 at 15:07

Are there any $x,y,z$ in $\mathbb{R}$ for which the following equations hold?

1 Answers1