-1
$\begingroup$

I am trying to prove that this is false: $-m + n \geq |m + n|$ where m is a positive number and n is any number.

Can I subtract n from both sides? If not, how do I show that is false?

  • 0
    $|m+n| \ge |n| - |m| = |n| - m$, and so $n \ge |n|$, but this means $n = |n|$, so $n \ge 0$. Now this means $n+m \ge 0$, so $-m+ n\ge m+n$, thus $2m \le 0$, i.e. $m \le 0$. A contradiction. – 2017-01-15

3 Answers 3

2

Since $m$ is a positive number, we know that either $m+n\geq 0$ or $m+n<0$. Consider these two cases:

  • If $m+n\geq 0$, then $|m+n|=m+n$, so your inequality is equivalent to $-m+n\geq m+n$, subtracting $n$ from both sides gives $-m\geq m$, which is not true for a positive number.

  • If $m+n<0$, then $|m+n|=-m-n$, so your inequality simplifies to $-m+n\geq -m-n$. Adding $m$ to both sides gives $n\geq -n$, which is not true since $n$ must be a negative number. (Why must $n$ be negative).

In either case, if you assume the given inequality is true, you reach a contradiction, so the inequality cannot be true.

  • 0
    What do you mean by "m+n = -m-n"? – 2017-01-16
  • 1
    @Goldname The absolute value bars were missing. Thanks for spotting the typo. – 2017-01-16
0

Hint: Choose $m=n> 0$ and conclude

  • 0
    It depends on how the statement is quantified. The inequality is *always* false. – 2017-01-15
0

Assuming $m>0$, the inequality can be rewritten $$ n\ge|m+n|+m $$ which is clearly false if $n<0$.

If $n\ge0$, then $|m+n|=m+n$ and the inequality becomes $$ n\ge 2m+n $$ that is, $0\ge2m$, which is false.