# Elementary Logic Negation

Thanks for taking a looking at my questions. My class just started a unit on elementary logic, and I want to make sure I'm on the right track with negating these statements.

1) If $S$ and $T$ are vector spaces of $\mathbb{R}^3$, then $S \cap T$ and $S \cup T$ are vector subspaces of $\mathbb{R}^3$.

Negation: There are vector spaces $S$ and $T$ of $\mathbb{R}^3$ such that $S \cap T$ and $S \cup T$ are not vector spaces of $\mathbb{R}^3$.

2) There is a real number $\epsilon > 0$ such that, for any real number $\delta > 0$, we have $\left| \dfrac{2}{3+\delta}-\dfrac{2}{3}\right| < \epsilon$.

Negation: "For any real number $\epsilon > 0$ such that, for any real number $\delta > 0$, we have $\left| \dfrac{2}{3+\delta}-\dfrac{2}{3}\right| \geq \epsilon$."

3) There is a unique positive integer $x$ such that $x^2-2x-8=0$.

Negation: Either there is no positive integer $x$ such that $x^2-2x-8=0$, or there are more than one positive integer $x$ such that $x^2-2x-8=0$.

Thanks again, this is a great community!

The first and third questions are perfectly answered. Congrats!

In the second you made a slight mistake. When there are several quantifiers in a sentence (for any, there is) a negation flips them all, not only the first one.

The negation of the sentence thus becomes "For any real number $\epsilon > 0$ there is a real number $\delta > 0$ such that we have $\left| \dfrac{2}{3+\delta}-\dfrac{2}{3}\right| \geq \epsilon$."

• Thanks, Jsevillamol, I appreciate your help! – jbrow35 Aug 30 '16 at 2:27

I agree with Jsevillamol on every point beside (1). Note the in the negation it is enough that at least on of the sets is not a subspace... It is due to the de morgan laws: $\neg (A\wedge B)$ is equivalent to $\neg A\vee \neg B$.

• Thanks, Gur, I appreciate your thoughts. I think you're right, I went back to my notes on DeMorgan's Laws and it looks like I should OR between the sets. – jbrow35 Aug 30 '16 at 2:29
• Oops, you are totally right. Good catch. – Jsevillamol Aug 30 '16 at 15:15

There are some errors. 1) You began all right: negating the implicite universal quantifier to get the the existential statement, then negating the implication $A \rightarrow B$ to get a statement of the form $A \wedge \neg B$. BUT: You your negation of the statement B (i.e. "intersection and union are ... ") is not correct. Its negation should be "intersection OR union are not...".

2)This is a statement of the form $\exists \epsilon \forall \delta \varphi(\epsilon,\delta)$. Its negation should have form $\forall \epsilon \exists \delta \neg \varphi(\epsilon,\delta)$.

3)Seems OK.

• Thanks, l_j, I appreciate your time. I think you're correct about #1, I went back and made the correction. – jbrow35 Aug 30 '16 at 2:30