3

For all $x$, if $x^2$ is even, then $x$ is even.

The contrapositive to this statement is:

For all $x$, if $x$ is odd, then $x^2$ is odd.

Why do we ignore the "For all $x$" and not say "For some $x$..."?

Eric Wofsey
  • 330,363
ianc1339
  • 489
  • 3
  • 12
  • 6
    The contrapositive of "if $x^2$ is even, then $x$ is even" is "if $x$ is odd, then $x^2$ is odd'. It only makes sense to talk about the contrapositive of an implication (a.k.a. conditional statement). – azif00 Mar 06 '21 at 04:25
  • 3
    Right. The "for all" is not part of the implication. – amsmath Mar 06 '21 at 04:34
  • 1
    Also: don't forget--the contrapositive is equivalent to the original statement. On the other hand, negating a statement does involve changing the quantifier(s).. – ryang Mar 06 '21 at 05:29

2 Answers2

1

Symbolically

$~~~~\forall x \in N: [x^2 \in E \implies x \in E]$$

$~~~\equiv ~~~\forall x \in N: [x \notin E \implies x^2 \notin E]$

$~~~\equiv ~~~\forall x \in N: [x \in O \implies x^2 \in O]$

where $N$ is the set of natural numbers, $E$ is the set of even numbers, and $O$ is the set of odd numbers.

We use the contrapositive of the original implication. The quantifier is unaffected.

Dan Christensen
  • 14,529
1

Because the contrapositive refers to an equivalent form of the implication, it is thus a tautological equivalence. It refers to the "realm" of propositional logic and not first order logic.

For example, I'm sure you know that an equivalent form of $\phi \to \psi$ is $\lnot \phi \lor \psi$. Would you not then substitute in $\forall x (Fx \to Qx)$ for $\forall x (\lnot Fx \lor Qx)$?

Simone
  • 620