Write the contrapositive and negation of the statement

[docnet]

Homework Statement

Write the contrapositive and negation of the statement.

Relevant Equations

$\forall x\in\mathbb{R}, (x\geq 3)\Rightarrow (x^2>5)$

The contrapositive statement is
$$\forall x\in\mathbb{R}, (x^2\leq 5)\Rightarrow (x<3)$$
The negation statement is
$$\forall x\in\mathbb{R}, (x<3)\Rightarrow (x^2\leq 5)$$

 

[fishturtle1]

For the negation: Let $D$ be a set and $P(x)$ be a statement about the elements of $D$. Then $$\neg \forall x \in D : P(x) \equiv \exists x \in D : \neg P(x)$$
and
$$\neg \exists x \in D : P(x) \equiv \forall x \in D : \neg P(x)$$

These equivalences follow by DeMorgan's laws. Note also that $\neg (p \Longrightarrow q) \equiv p \land \neg q$. (You can write the truth tables to prove they are equivalent).

 

[docnet]

Okay, I understand. Question: Would the meaning of the statement $\neg (p \Longrightarrow q) \equiv p \land \neg q$ stay the same if we replace $\Longrightarrow$ by $\land$? Also, what is a truth table (seriously)?

Original statement: $\forall x\in\mathbb{R}, (x\geq 3)\Rightarrow (x^2>5)$
The negated statement
$$\exists x\in\mathbb{R}, (x\geq 3)\Rightarrow (x^2\leq 5)$$
The contraposited statement
$$\forall x\in\mathbb{R}, (x<3)\Rightarrow (x^2\leq 5)$$
The negated then contraposited statement
$$\exists x\in\mathbb{R}, (x<3)\Rightarrow (x^2> 5)$$
The contraposited then negated statement
$$\exists x\in\mathbb{R}, (x<3)\Rightarrow (x^2> 5)$$
The Negated then contraposited then negated statement
$$\forall x\in\mathbb{R}, (x<3)\Rightarrow (x^2\leq 5)$$
And so on...

So, contraposition and and negation are involutory operations on logical statements.

Also, contrapositation and and negation are commutative operations on logical statements.

So, if the original statement is true,

1. the contrapositive is true.
2. the negated is false.
3. the negated contrapositive is false.
4. the contrapositive negated is false.
5. And so on...

I think this is consistent with DeMorgan's laws.

 

[fishturtle1]

Question: Would the meaning of the statement ¬(p⟹q)≡p∧¬q stay the same if we replace ⟹ by ∧?

The statement $\neg (p \land q) \equiv p \land \neg q$ is not true. Can you write the truth table for these two statements (see below)?

Also, what is a truth table (seriously)?

A truth table for a logical statement is a table that says when a statement is true and when it is false. Here's an example for $p \Longrightarrow q$ and $p \land \neg q$. There are two variables $p$ and $q$. So there's four possibilities to consider:

$\begin{array}{|c|c|c|c|c|} \hline p & q & p \Longrightarrow q & \neg (p \Longrightarrow q) & p \land \neg q \\ \hline T & T & T & F & F \\ \hline T & F & F & T & T \\ \hline F & T & T & F & F \\ \hline F & F & T & F & F \\ \end{array}$

Note that the columns for $\neg (p \Longrightarrow q)$ and $p \land \neg q$ are the same , i.e., they take on the same truth values given any assignment of $p$ and $q$. We say they are logically equivalent and denote this by $\neg (p \Longrightarrow q) \equiv p \land \neg q$.

 

[fishturtle1]

Original statement: ∀x∈R,(x≥3)⇒(x2>5)
The negated statement
∃x∈R,(x≥3)⇒(x2≤5)

If $P(x)$ is the statement $(x \ge 3) \Rightarrow x^2 > 5$, what is $\neg P(x)$? Hint: Let $p(x)$ be the statement $x \ge 3$ and $q(x)$ be the statement $x^2 > 5$. Then we want to negative $p(x) \Rightarrow q(x)$.

 

[docnet]

Question: I do not understand how you got the last two entries of the column $p\Longrightarrow q$. Does $p\Longrightarrow q$ mean if $p$ is true, that implies $q$ is true?

$\begin{array}{|c|c|c|c|c|} \hline p & q & p \Longrightarrow q & \neg (p \land q) & p \land \neg q \\ \hline T & T & T & F & F \\ \hline T & F & F & T & T \\ \hline F & T & T & F & F \\ \hline F & F & T & F & F \\ \end{array}$

The statement $¬(p∧q)≡p∧¬q$ is not true. Can you write the truth table for these two statements (see below)?

Yes, I will try.

$\begin{array}{|c|c|c|c|c|} \hline p & q & p \land q & \neg (p \land q) & p \land \neg q \\ \hline T & T & T & F & F \\ \hline T & F & F & T & T \\ \hline F & T & F & T & F \\ \hline F & F & F & T & F \\ \end{array}$

 

[fishturtle1]

Question: I do not understand how you got the last two entries of the column p⟹q. Does p⟹q mean if p is true, that implies q is true?

To your first question, it is by definition. To your second question, yes: If $p \Longrightarrow q$ is true and $p$ is true, then we can conclude $q$ is true.

Your truth tables are correct.

 

[docnet]

If $P(x)$ is the statement $(x \ge 3) \Rightarrow x^2 > 5$, what is $\neg P(x)$? Hint: Let $p(x)$ be the statement $x \ge 3$ and $q(x)$ be the statement $x^2 > 5$. Then we want to negative $p(x) \Rightarrow q(x)$.

If $P(x)$ is the statement $p\Longrightarrow q$,

\begin{align}\neg P(x)=&\neg(p\Longrightarrow q)\\
=&p\land \neg q\\
=&(x\geq 3)\land (x^2\leq 5)\end{align}

 

[docnet]

Re-attempt the original problem using this new tool:

Original statement: $\forall x\in\mathbb{R}, (x\geq 3)\Rightarrow (x^2>5)$
The negated statement
$$\exists x\in\mathbb{R}, (x\geq 3)\land (x^2\leq 5)$$
The contraposited statement
$$\forall x\in\mathbb{R}, (x<3)\Rightarrow (x^2\leq 5)$$

 

[fishturtle1]

Re-attempt the original problem using this new tool:

Original statement: $\forall x\in\mathbb{R}, (x\geq 3)\Rightarrow (x^2>5)$
The negated statement
$$\exists x\in\mathbb{R}, (x\geq 3)\land (x^2\leq 5)$$
The contraposited statement
$$\forall x\in\mathbb{R}, (x<3)\Rightarrow (x^2\leq 5)$$

The negation is correct, but the contrapositive is not. In fact, your contrapositive is correct in the OP!

 

[PeroK]

Homework Statement:: Write the contrapositive and negation of the statement.
Relevant Equations:: $\forall x\in\mathbb{R}, (x\geq 3)\Rightarrow (x^2>5)$

The contrapositive statement is
$$\forall x\in\mathbb{R}, (x^2\leq 5)\Rightarrow (x<3)$$
The negation statement is
$$\forall x\in\mathbb{R}, (x<3)\Rightarrow (x^2\leq 5)$$

I've never studied the formal logic and notation in post #2. The way I would understand this is:
$$\forall x\in\mathbb{R}, (x\geq 3)\Rightarrow (x^2>5)$$Says that whenever $x \ge 3$ we must have $x^2 > 5$. This means we can't find $x$ where $x \ge 3$ and $x^2 \le 5$. And that means that if $x^2 \le 5$ then we must have $x < 3$. That's gives the contraposition (as you have):
$$\forall x\in\mathbb{R}, (x^2\leq 5)\Rightarrow (x<3)$$In practical mathematical terms, this means that to prove the original statement we could either prove the statement itself or its contraposition, as the two are equivalent.

For the negation, we have to find the statement that is true iff the original statement is false. If the statement is false we can find $x \ge 3$ where $x^2 < 5$. And, if we find such an $x$ then the statement is false. The negation, therefore, is:
$$\exists x \in \mathbb R: x \ge 3 \ \text{and} \ x^2 < 5$$In practical mathematical terms, if we want to disprove the statement, then we have to find such an $x$. Or, if we assume such an $x$ exists and reach a contradiction, then we have proved the original statement by contradiction.

 

[docnet]

I've never studied the formal logic and notation in post #2. The way I would understand this is:
$$\forall x\in\mathbb{R}, (x\geq 3)\Rightarrow (x^2>5)$$Says that whenever $x \ge 3$ we must have $x^2 > 5$. This means we can't find $x$ where $x \ge 3$ and $x^2 \le 5$. And that means that if $x^2 \le 5$ then we must have $x < 3$. That's gives the contraposition (as you have):
$$\forall x\in\mathbb{R}, (x^2\leq 5)\Rightarrow (x<3)$$In practical mathematical terms, this means that to prove the original statement we could either prove the statement itself or its contraposition, as the two are equivalent.

For the negation, we have to find the statement that is true iff the original statement is false. If the statement is false we can find $x \ge 3$ where $x^2 < 5$. And, if we find such an $x$ then the statement is false. The negation, therefore, is:
$$\exists x \in \mathbb R: x \ge 3 \ \text{and} \ x^2 < 5$$In practical mathematical terms, if we want to disprove the statement, then we have to find such an $x$. Or, if we assume such an $x$ exists and reach a contradiction, then we have proved the original statement by contradiction.

That is the clearest explanation of the contraposition and negation statement I read so far.

Question: Is the contrapositive statement a method that is used in high level research mathematics?

 

[PeroK]

Question: Is the contrapositive statement a method that is used in high level research mathematics?

All the time. Using the contraposition is a fundamental idea. And extremely useful.

 

[sysprog]

Re-attempt the original problem using this new tool:

Original statement: $\forall x\in\mathbb{R}, (x\geq 3)\Rightarrow (x^2>5)$
The negated statement
$$\exists x\in\mathbb{R}, (x\geq 3)\land (x^2\leq 5)$$
The contraposited statement
$$\forall x\in\mathbb{R}, (x<3)\Rightarrow (x^2\leq 5)$$

The statement that you call the 'negated statement' is not the negated statement, and the statement that you called the 'contrapositive' is not the contrapositive.

Using an existential quantifier statement to negate a universal quantifier statement takes the form:

$\forall x[P_x]$,​

with the negation being,​

(for an atomic statement)​

$\exists x[\neg P_x]$,​

​

(or for either an atomic statement or a composite statement)​

$\exists x\neg[P_x]$.​

The contrapositive has the following form:

if the implication statement that is to be contraposited is​

$(p \Rightarrow q)$,​

then the contrapositive is​

$(\neg q \Rightarrow \neg p)$.​

The rule for deriving the contrapositive of an implication statement is called modus tollens.

In English:

negation $-$

all $x$ is $P$,​

there exists an $x$ such that $x$ is not $P$.​

​

contrapositive $-$

if $p$ then $q$,​

if not $q$ then not $p$.​

 

[docnet]

(for an atomic statement)​

$\exists x[\neg P_x]$,​

​

(or for either an atomic statement or a composite statement)​

$\exists x\neg[P_x]$.​

Thank you. I do not understand how the notation $[\neg P_x]$ and $\neg[P_x]$ are different.

 

[sysprog]

Thank you. I do not understand how the notation $[\neg P_x]$ and $\neg[P_x]$ are different.

They aren't different for an atomic statement, such as $P_x$; however, for a composite statement, such as $P_x \wedge R_x$, they are different:

$\forall x[\neg P_x \wedge R_x]$, means 'for all $x$, $x$ is not $P$, and $x$ is $R$',

while

$\forall x\neg[P_x \wedge R_x]$, means 'for all $x$, it is not the case that: $x$ is $P$ and $x$ is $R.$

 

[docnet]

They aren't different for an atomic statement, such as $P_x$; however, for a composite statement, such as $P_x \wedge R_x$, they are different:

$\forall x[\neg P_x \wedge R_x]$, means 'for all $x$, $x$ is not $P$, and $x$ is $R$',

while

$\forall x\neg[P_x \wedge R_x]$, means 'for all $x$, it is not the case that: $x$ is $P$ and $x$ is $R.$

Thank you. I understand.