# prove inequality

• Sep 2nd 2008, 09:06 PM
Dubulus
prove inequality
ok heres the prob....the statement below is not always true for x, y (symbol for real numbers). give an example where it is false, and add a hypothesis on y that makes it true. "if x and y are nonzero real numbers and x>y then (-1/x)>(-1/y)"
• Sep 2nd 2008, 09:18 PM
Dubulus
(-1/-1) > (-1/-2) makes
1 > 1/2 which is true
• Sep 2nd 2008, 09:26 PM
particlejohn
Choose $\displaystyle x = 2, y = -1$. Add hypothesis: $\displaystyle y$ is same sign as $\displaystyle x$.
• Sep 3rd 2008, 08:33 AM
kwah
Consider the four cases {{ note: for some reason, > is showing as an upside down ? }}:

Quote:

$\displaystyle \begin{tabular}{c c l} x & y & comment/result\\ + & + & True\\ + & - & LHS -ve and RHS +ve therefore false\\ - & + & in breach of x > y which is given\\ - & - & True (see particlejohn's post below)\\ \end{tabular}$

From this, you can see that both $\displaystyle x$ and $\displaystyle y$ must have the same sign.

EDIT: apologies.. particlejohn is correct and my post has been edited accordingly =]
• Sep 3rd 2008, 12:56 PM
particlejohn
Quote:

Originally Posted by kwah
Consider the four cases {{ note: for some reason, > is showing as an upside down ? }}:

From this, you can see that both $\displaystyle x$ and $\displaystyle y$ must be positive.

Choose $\displaystyle x = -3$ and $\displaystyle y = -4$ (e.g. $\displaystyle x > y$). Then $\displaystyle \frac{1}{3} > \frac{1}{4}$.