Mathematical logic and contrapositives.

Question

I have the following statement

If $x^2=4$, then $x=2$ or $x=-2$ I have to write its corresponding contrapositive.

I know that this should be stated as follows:

If $x$ is not equal to $2$ and $-2$, then $x^2$ is not equal to $4$'.

However, I learned something about conditional statements and it has confused me. I learned that I can write the conditional statement above also as: $$ x^2=4 \quad \Rightarrow \quad x=2 \text{ or } x=-2, $$ this is supposedly logically equivalent to $x^2$ is not equal to $4$ or $x=2$ or $x=-2$. Then to write the contrapositive of this would lead me to believe I should write $x$ is not equal to $2$ and $-2$ and $x^2=4$, however I know that this isn't correct. Can someone explain how to manipulate if-then statements to lead to contrapositives and negations?

Answer 1

I guess you are confusing the terminology.

• Implication: $P \implies Q$.
• Contrapositive: $\neg Q \implies \neg P$.
• Inverse: $\neg P \implies \neg Q$.
• Converse: $Q \implies P$.
• Negation: $\neg (P \implies Q)$.
• Logical equivalence: $\neg P \lor Q$ and $\neg(P \land \neg Q)$.

In your case $(x^2 = 4) \implies (x = 2 \quad\text{or}\quad x = -2)$ you have

• contrapositive $(x \ne 2 \quad\text{and}\quad x \ne -2) \implies (x^2 \ne 4)$,
• logically equivalent to $(x^2 \ne 4) \quad\text{or}\quad (x = 2 \quad\text{or}\quad x = -2)$,
• also, logically equivalent to $\text{not}\;[(x^2 = 4) \quad\text{and}\quad (x \ne 2 \quad\text{and}\quad x \ne -2)]$.

The equivalence $\neg(P \land \neg Q)$ is usual in a proof by contradiction: supposing $P$ and no $Q$ we obtain a contradiction. So you can conclude $P \implies Q$, beacuse is logically equivalent to $\neg(P \land \neg Q)$, and $P \land \neg Q$ leads a contradiction.

The contrapositive is usefull when it is easier to prove than $P \implies Q$.

The converse is used when one need to show $P \iff Q$, becuase we have to prove $P \implies Q$ and $Q \implies P$.

Edit. Another way to use the logical equivalence $\neg P \lor Q$ is in a proof by contradiction. For instance, if you want to prove

For every rational number $\epsilon > 0$, there exists a non-negative rational number $x$ such that $x^2 < 2 < (x + \epsilon)^2$.

Then suppose for sake a contradiction that there is no non-negative rational number $x$ such that $x^2 < 2 < (x + \epsilon)^2$, i.e., for all $x$ you have $x^2 \ge 2 \text{ or } 2 \ge (x + \epsilon)^2$. Thus you can use the logical equivalences $$x^2 < 2 \implies 2 \ge (x + \epsilon)^2$$ $$2 < (x + \epsilon)^2 \implies x^2 \ge 2$$ to lead a contradiction (maybe using induction).

Answer 2

Christian Gz has a very instructive answer.

To add to his reply, note that your original statement is incorrect. It is not the case that $$x^2=4 \implies (x=-2 \;\; \text{or} \;\;x=2).$$ Because $x$ cannot be both $-2$ and $2$ simultaneously. While it is the case we know this is so, it is not reflected in the statement. The proper, but more subtle, statement is $$x^2=4 \implies (x=-2 \;\; \text{xor} \;\;x=2).$$ This new statement will be both valid and sound under any truth values of the predicates. To construct your contra-positive I remind you the negation of a exclusive or(xor) is the bi-conditional(if and only if).

Answer 3

It's just a confusion about the contraposition law. It states the following:

$\vDash (\phi \rightarrow \psi) \leftrightarrow (\neg \psi \rightarrow \neg \phi)$

Hence contraposition states an equivalence between two conditional statements, saying that the latter is logically equivalent to the inverse and converse of the former.

Now you learned another equivalence, you learned that:

$\vDash (\phi \rightarrow \psi) \leftrightarrow (\neg \phi \vee \psi)$

But from this you can only infer that:

$\vDash (\neg \phi \vee \psi) \leftrightarrow (\neg \psi \rightarrow \phi)$

And not try to move by such "improper contraposition" from the equivalent disjunction. In case of any doubts, it's always good to see how this law is defined.

Answer 4

Let $p$ be the statement "$x^2 = 4$" and $q$ be the statement "$x = 2$ or $x = -2$". Then we can rewrite the given statement as $p \implies q$. The contrapositive is $\neg q \implies \neg p$. The negation of $q$ is $x \not= 2$ and $x \not = -2$ by DeMorgan's laws. The negation of $p$ is $x^2 \not=4$. So we can rewrite the contrapositive as "if $x \not= 2$ and $x \not = -2$, then $x^2 \not=4$," which is obviously true.

Answer 5

So, I use Polish/Lukasiewicz notation.

For the statement represented by the formula Cpq, CNqNp is the contrapositive.

Now, if ANpq is an abbrevation for Cpq, then it follows that the contrapositive of ANpq is ANNqNp. We just replace "C" with "AN" in "Cpq" and "CNqNp" to see this. Along with Genomeme's analysis this allows us to infer that the contrapositive of

"Either it is not the case that x$^2$=4 or (x=2 xor x=-2)"

is

"Either it is not, not the case that (x=2 xor x=-2) or it is not the case that x$^2$=4."

Eliminating the double negations this becomes:

"Either it the case that (x=2 xor x=-2) or it is not the case that x$^2$=4."