# Writing down propositions symbolically & state truth values.

• Apr 13th 2010, 12:51 AM
jvignacio
Writing down propositions symbolically & state truth values.
Hey guys, I need help with a study question. "Write down the following propositions symbolically and state their values. Then write down the negation of each propositions."

a) There is a real number $\displaystyle x$ such that $\displaystyle x^2 - 3x + 2 = 0$.

My Answer: $\displaystyle \exists x \epsilon \mathbb{R}$ $\displaystyle (x^2 - 3x + 2 = 0)$ and this is TRUE for when $\displaystyle x=1$. The negation to this is $\displaystyle \forall x \epsilon \mathbb{R}$$\displaystyle (x^2 - 3x + 2 \neq 0) b) For every real number \displaystyle x there is a real number \displaystyle y such that \displaystyle x=y^2. My Answer: \displaystyle \forall x \epsilon \mathbb{R} \displaystyle \exists y \epsilon \mathbb{R} \displaystyle (x=y^2) and this is FALSE because not a single \displaystyle y real value for \displaystyle y^2 can equal all real values of \displaystyle x. The negation to this is \displaystyle \exists x \epsilon \mathbb{R} \displaystyle \forall y \epsilon \mathbb{R} \displaystyle (x \neq y^2) Are my answers OK? thanks • Apr 13th 2010, 03:26 AM Failure Quote: Originally Posted by jvignacio Hey guys, I need help with a study question. "Write down the following propositions symbolically and state their values. Then write down the negation of each propositions." a) There is a real number \displaystyle x such that \displaystyle x^2 - 3x + 2 = 0. My Answer: \displaystyle \exists x \epsilon \mathbb{R} \displaystyle (x^2 - 3x + 2 = 0) and this is TRUE for when \displaystyle x=1. The negation to this is \displaystyle \forall x \epsilon \mathbb{R}$$\displaystyle (x^2 - 3x + 2 \neq 0)$

Right, on both counts.

Quote:

b) For every real number $\displaystyle x$ there is a real number $\displaystyle y$ such that $\displaystyle x=y^2$.

My Answer: $\displaystyle \forall x \epsilon \mathbb{R}$ $\displaystyle \exists y \epsilon \mathbb{R}$ $\displaystyle (x=y^2)$ and this is FALSE
Right.
Quote:

because not a single $\displaystyle y$ real value for $\displaystyle y^2$ can equal all real values of $\displaystyle x$.
Wrong! The point is not that there must exist a single y that does it for all x. The idea is that if the statement were true, then one could provide an arbitrary, but fixed, $\displaystyle x\in\mathbb{R}$, and you, the defender of that statement, would have to come up with a $\displaystyle y\in\mathbb{R}$ such that $\displaystyle x=y^2$ holds. If someone comes up with a different x, you are allowed to choose a different y. This is because the existential quantifier for y occurs in the scope of the quantifier of x. So at the point where the y must be show to exist, the value of x has already been decided upon and cannot be changed anymore, so to speak.

This is impossible only if $\displaystyle x<0$, otherwise it is quite easily possible to come up with such a $\displaystyle y$

Quote:

The negation to this is $\displaystyle \exists x \epsilon \mathbb{R}$ $\displaystyle \forall y \epsilon \mathbb{R}$ $\displaystyle (x \neq y^2)$
Correct, and this statement is, of course, true.
• Apr 13th 2010, 03:41 AM
jvignacio
Quote:

Originally Posted by Failure
Right, on both counts.

Right.

Wrong! The point is not that there must exist a single y that does it for all x. The idea is that if the statement were true, then one could provide an arbitrary, but fixed, $\displaystyle x\in\mathbb{R}$, and you, the defender of that statement, would have to come up with a $\displaystyle y\in\mathbb{R}$ such that $\displaystyle x=y^2$ holds. If someone comes up with a different x, you are allowed to choose a different y. This is because the existential quantifier for y occurs in the scope of the quantifier of x. So at the point where the y must be show to exist, the value of x has already been decided upon and cannot be changed anymore, so to speak.

This is impossible only if $\displaystyle x<0$, otherwise it is quite easily possible to come up with such a $\displaystyle y$

Correct, and this statement is, of course, true.

thank you so much! Understood :)