# inequality problem

• Mar 17th 2010, 01:13 AM
donnagirl
inequality problem
The real number c is positive and the real numbers a and b are such that $\displaystyle ab > 0$ and $\displaystyle a > b$. How can I disprove the case $\displaystyle a - c < b + c$ for all such a, b, and c? (The case $\displaystyle \frac{c}{a} < \frac{c}{b}$ works). When I tried it, I can't seem to get numbers that disprove it??
• Mar 17th 2010, 01:44 AM
HallsofIvy
Quote:

Originally Posted by donnagirl
The real number c is positive and the real numbers a and b are such that $\displaystyle ab > 0$ and $\displaystyle a > b$. How can I disprove the case $\displaystyle a - c < b + c$ for all such a, b, and c? (The case $\displaystyle \frac{c}{a} < \frac{c}{b}$ works). When I tried it, I can't seem to get numbers that disprove it??

How hard did you try? How about a= 100, b= 1, c= 1?
• Mar 17th 2010, 01:46 AM
Hello donnagirl
Quote:

Originally Posted by donnagirl
The real number c is positive and the real numbers a and b are such that $\displaystyle ab > 0$ and $\displaystyle a > b$. How can I disprove the case $\displaystyle a - c < b + c$ for all such a, b, and c? (The case $\displaystyle \frac{c}{a} < \frac{c}{b}$ works). When I tried it, I can't seem to get numbers that disprove it??

I'm not quite sure I understand the problem here, because it seems very easy to find a counter-example. In order that $\displaystyle a-c>b+c$, we must ensure that $\displaystyle a-b > 2c$. So, for instance:
$\displaystyle a=10, b = 1, c = 3$
gives $\displaystyle a-c=7$ and $\displaystyle b+c = 4$. So $\displaystyle a-c> b+c$.