# Discrete/Logic

• Mar 14th 2008, 03:35 AM
g4me
Discrete/Logic
Heres the problem

http://img219.imageshack.us/img219/9035/60272145mo7.jpg

Heres my attempt. I'm not sure if I have done it correctly, if not, am I on the right track?

http://img219.imageshack.us/img219/4056/53371463lk4.jpg

Edit: My answer is meant to be true, not false
• Mar 14th 2008, 03:49 AM
Plato
$\left[ {0 \in \mathbb{R}\backslash \mathbb{R}^ - } \right]\left[ {\forall y \in \mathbb{R}\backslash \mathbb{R}^ - } \right]\left( {0^2 < 1 + y} \right)$
• Mar 14th 2008, 04:15 AM
g4me
What are these R symbols?

Was I totally wrong? Is what you wrote the solution?
• Mar 14th 2008, 07:17 AM
Plato
Quote:

Originally Posted by g4me
What are these R symbols?
Was I totally wrong? Is what you wrote the solution?

Although that is no absolute set of symbols in mathematics some notations are widely accepted. The symbol $\mathbb{R}$ is used to denote the real numbers. $\mathbb{R}^+$ denotes the positive reals and $\mathbb{R}^-$ denotes the negative reals.
Thus $\mathbb{R}= \mathbb{R}^+ \cup \mathbb{R}^- \cup {0}$ so the nonnegative reals are $\mathbb{R}\backslash \mathbb{R}^ - = \mathbb{R}^ + \cup \{ 0\}$.

As to the “Was I totally wrong?”, I really have no idea what you did. I cannot follow it. The symbolic statement reads “There is some nonnegative real number, x, having the property that for every real number y then $x^2.
You see that zero has that property; so does $\frac {1} {2}$.