prove by cases that $|r + s| \leq |r| + |s|$ for all real numbers.

Question

Prove by cases that $|r + s| \leq |r| + |s| $.

I'm new to proofs and picked up a discrete math book and I can't seem to figure out how to go about solving this.

Any help would be appreciated.

Answer 1

Case 1) $r,s>0$

Case 2) $r,s<0$

Case 3) $r>0, s<0$

The other cases hold trivially, and so there is no need to mention them.

In each of the cases, use the definition of the absolute value.

Answer 2

It says do it by cases so consider the cases

1) $r \ge 0; s\ge 0$

$|r|=r$ and $|s|=s$ and $r+s \ge 0$ so $|r+s| = r+s$.

So $|r+s| = r+s = |r| + |s|$ and that's it.

2) $r \ge 0$ and $s < 0$.

$|r| = r$ and $|s| =-s$ and $r+s = r-|s|$ may be negative, postive, or zero depending on how $r$ and $|s|$ compare

2a) $r\ge 0$ and $s < 0$ and $|s|\le r$.

$|r| =r$ and $|s|=-s$ and $r+s \ge 0$ so $|r+s| = r+s = r-|s|$.

So $|r+s| = r-|s| < r+|s| = |r| + |s|$.

2b) $r \ge 0$ and $s< 0$ and $|s|> r$

$|r|=r$ and $|s|=-s$ and $r+s < 0$ so $|r+s| = -(r+s) = -r -s =|s| -r$.

And $|r+s| = |s| -r \le |s| + r = |r| + |s|$.

3) $r < 0$ and $s \ge 0$.

That's done exactly like 2a) and 2b) we don't have to repeat them.

4) $r < 0$ and $s < 0$.

$|r| =-r$ and $|s| =-s$ and $r+s < 0$ so $|r+s| = -r - s$.

So $|r+s| = -r-s = |r| + |s|$.

All done.

....

But it'd be slicker and faster, if you are comfotable, to do it more generally.

1) $r$ and $s$ have the same sign and $r+s$ is actually adding two values

Then $|r+s| = \pm (r+s) = \pm r \pm s = |r| +|s|$.

2) $r$ and $s$ have different signs and $r+s$ is actually subtracting two values.

The $r = \pm |r|$ and $s=\mp |s|$ and $r+s = \pm |r| \mp |s| = \pm(|r|-|s|)$.

$|r+s| = |r| -|s|$ if $|r| \ge |s|$ or $|r+s| = |s|-|r|$ if $|s| \ge |r|$.

In either event both of those are $\le |r| + |s|$.