Mathematics · Glossary

What is Statement, connectives?

Also known as: statement

Definition 1.1 University Mathematics — Year 1 · Chapter 1 — Logic, Sets and Maps

A statement (or proposition) is a sentence which is either true (T) or false (F) — exactly one of the two. From statements PP and QQ one forms:

  • the negation ¬P\lnot P (“not PP”), true exactly when PP is false;
  • the conjunction PQP \land Q (“PP and QQ”), true exactly when both are true;
  • the disjunction PQP \lor Q (“PP or QQ”), true exactly when at least one is true (this “or” is inclusive);
  • the implication P    QP \implies Q, false exactly when PP is true and QQ is false;
  • the equivalence P    QP \iff Q, true exactly when PP and QQ have the same truth value.

Examples

Example 1.6 (Negating everyday mathematical sentences)

Let f ⁣:RRf \colon \R \to \R. The sentence “ff is increasing” reads

xR, yR,xy    f(x)f(y),\forall x \in \R,\ \forall y \in \R,\quad x \leq y \implies f(x) \leq f(y) ,

and its negation, by Proposition 1.5 plus the rule ¬(P    Q)    P¬Q\lnot(P \implies Q) \iff P \land \lnot Q:

xR, yR,xy  and  f(x)>f(y):\exists x \in \R,\ \exists y \in \R,\quad x \leq y \ \text{ and }\ f(x) > f(y) :

a single witnessing pair suffices. Likewise “ff is bounded” is MR, xR, f(x)M\exists M \in \R,\ \forall x \in \R,\ \abs{f(x)} \leq M, with negation

MR, xR,f(x)>M:\forall M \in \R,\ \exists x \in \R,\quad \abs{f(x)} > M :

whatever bound is proposed, some point beats it. The insight: a correct negation never contains “not” applied to a quantified block — it is a new positive statement, in which the roles are exchanged: one now produces the witnesses one previously received.

Example 1.7 (Order of quantifiers)

The order of different quantifiers matters:

xR, yR, y>xis true (take y=x+1),\forall x \in \R,\ \exists y \in \R,\ y > x \quad\text{is true (take } y = x+1\text{),}
yR, xR, y>xis false (no real number exceeds all reals).\exists y \in \R,\ \forall x \in \R,\ y > x \quad\text{is false (no real number exceeds all reals).}

In the first statement yy may depend on xx; in the second, one single yy must work for all xx. Two identical quantifiers, on the other hand, always commute.

Example 1.13 (Proving unique existence)

A statement !x, P(x)\exists!\,x,\ P(x) is two statements, proved separately: existence (exhibit or construct some x0x_0 with P(x0)P(x_0)) and uniqueness (assume P(x)P(x) and P(x)P(x'), deduce x=xx = x'). Sample: there is a unique real xx with x3+x=2x^3 + x = 2. Existence: x0=1x_0 = 1 works, since 1+1=21 + 1 = 2. Uniqueness: if x3+x=x3+xx^3 + x = x'^3 + x', then

0=(x3x3)+(xx)=(xx)(x2+xx+x2+1),0 = (x^3 - x'^3) + (x - x') = (x - x')\,\bigl(x^2 + xx' + x'^2 + 1\bigr),

and the second factor is positive (it equals (x+x2)2+34x2+11\bigl(x + \tfrac{x'}2\bigr)^2 + \tfrac34 x'^2 + 1 \geq 1), so x=xx = x'. Note the division of labor: existence used a lucky guess, uniqueness used algebra valid for arbitrary solutions — neither argument does the other’s job, and forgetting the second half is a standing temptation once a solution has been found.

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