# Thread: Writing down propositions symbolically & state truth values.

1. ## 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 $x$ such that $x^2 - 3x + 2 = 0$.

My Answer: $\exists x \epsilon \mathbb{R}$ $(x^2 - 3x + 2 = 0)$ and this is TRUE for when $x=1$. The negation to this is $\forall x \epsilon \mathbb{R}$ $(x^2 - 3x + 2 \neq 0)$

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

My Answer: $\forall x \epsilon \mathbb{R}$ $\exists y \epsilon \mathbb{R}$ $(x=y^2)$ and this is FALSE because not a single $y$ real value for $y^2$ can equal all real values of $x$. The negation to this is $\exists x \epsilon \mathbb{R}$ $\forall y \epsilon \mathbb{R}$ $(x \neq y^2)$

2. 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 $x$ such that $x^2 - 3x + 2 = 0$.

My Answer: $\exists x \epsilon \mathbb{R}$ $(x^2 - 3x + 2 = 0)$ and this is TRUE for when $x=1$. The negation to this is $\forall x \epsilon \mathbb{R}$ $(x^2 - 3x + 2 \neq 0)$
Right, on both counts.

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

My Answer: $\forall x \epsilon \mathbb{R}$ $\exists y \epsilon \mathbb{R}$ $(x=y^2)$ and this is FALSE
Right.
because not a single $y$ real value for $y^2$ can equal all real values of $x$.
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, $x\in\mathbb{R}$, and you, the defender of that statement, would have to come up with a $y\in\mathbb{R}$ such that $x=y^2$ 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 $x<0$, otherwise it is quite easily possible to come up with such a $y$

The negation to this is $\exists x \epsilon \mathbb{R}$ $\forall y \epsilon \mathbb{R}$ $(x \neq y^2)$
Correct, and this statement is, of course, true.

3. 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, $x\in\mathbb{R}$, and you, the defender of that statement, would have to come up with a $y\in\mathbb{R}$ such that $x=y^2$ 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 $x<0$, otherwise it is quite easily possible to come up with such a $y$

Correct, and this statement is, of course, true.
thank you so much! Understood