Logical symbolism. Necessary and sufficient conditions.

Let $α, β, . . . -$ be certain statements or assertions, that is, narrative sentences, about each of which it can be said whether it is true or false.

The notation $\overset{―}{α}$ means "not $α$ ", that is, the negation of the statement $α$ .

The notation $α \Rightarrow β$ means "statement $α$ implies statement $β$ ". ( $\Rightarrow -$ implication symbol).

The notation $α \Leftrightarrow β$ means "Statement $α$ is equivalent to statement $β$ ", that is, $α$ implies $β$ , and $β$ implies $α$ (" $\Leftrightarrow -$ " equivalence symbol).

The notation $α \land β$ means " $α$ and $β$ " ( $\land -$ conjunction symbol).

The notation $α \lor β$ means " $α$ or $β$ " ( $\lor -$ disjunction symbol).

The notation $\forall x \in X,,, α$ means "for every element $x \in X$ , statement $α$ holds" ( $\forall -$ universal quantifier).

The notation $\exists x \in X,,, α$ means "there exists an element $x \in X$ for which statement $α$ holds" ( $\exists -$ existential quantifier).

The notation $\exists! x \in X,,, α$ means "there exists a unique element $x \in X$ for which statement $α$ holds."

Examples.

Read the statements below, clarify their meaning, and determine whether they are true or false (where symbols $x,, y,, z,, a,, b,, c$ represent real numbers).

1. $\forall x, \exists y, (x + y = 3) .$

Solution.

The statement $\forall x, \exists y, (x + y = 3)$ means that for every element $x$ , there exists an element $y$ such that the equation $x + y = 3$ holds.

This statement is true. Indeed, for any element $x$ , there exists such an element $y$ and it equals $3 - x .$

Answer: the statement is true.

2. $\forall x (x^{2} > x \Leftrightarrow x > 1 \lor x < 0) .$

Solution.

The statement $\forall x (x^{2} > x \Leftrightarrow x > 1 \lor x < 0)$ means that for every element $x$ , the statement $x^{2} > x$ is true if and only if either $x > 1$ or $x < 0$ . Let's check the truth of this statement.

If in the inequality $x^{2} > x$ we let $x = 0$ , then the inequality $0 > 0$ holds, which is false. Therefore, $x \neq 0.$

Dividing both sides of the inequality $x^{2} > x$ by $x$ , we get $x > 1$ if $x > 0$ and $x < 1$ if $x < 0$ . Let's take the intersection of the obtained sets $(x > 1) \cap (x > 0)$ and $(x < 1) \cap (x < 0) .$

We obtain that either $x > 1$ or $x < 0$ . This completes the proof.

Answer: the statement is true.

3. $\forall x$ ( $x > 2$ $\land$ $x > 3$ $\Leftrightarrow$ $2 < x \leq 3) .$

Solution.

The statement $\forall x$ ( $x > 2 \land$ $x > 3 \Leftrightarrow$ $2 < x \leq 3)$ means that for every element $x$ , the statement " $x > 2$ and $x > 3$ " is true if and only if the condition $2

The set of elements $x$ for which the conditions $x > 2$ and $x > 3$ are simultaneously satisfied is the intersection of sets $\infty) \cap (3, \infty) = (3, \infty) .$ And the condition $2 < x \leq 3$ corresponds to the expression $x \in (2, 3] .$ Obviously, the obtained sets do not coincide. Therefore, the expression $\forall x (x > 2 \land x > 3 \Leftrightarrow 2 < x \leq 3)$ is false.

Answer: the statement is false.

To determine the exact meaning of the following statements and write them using logical symbols, formulate and write their negations:

4. The number $x_{0}$ is a solution of the equation $f (x) = 0.$

Solution.

The statement "The number $x_{0}$ is a solution of the equation $f (x) = 0$ " means that at the point $x_{0}$ , the function $f (x)$ takes the value $0.$ Using logical symbolism, this can be written as $f (x_{0}) = 0.$

Negation: at the point $x_{0}$ , the function $f (x)$ does not take the value $0$ , or $f (x_{0}) \neq 0.$

Answer: $f (x_{0}) = 0,$ $f (x_{0}) \neq 0.$

5. The number $m$ is the smallest element of the set $X .$

Solution.

This statement means that the number $m$ belongs to the set $X$ , and all other elements of the set $X$ are greater than or equal to $m .$ Let's write this using logical symbols: $(m \in X) \land (\forall x \in X (m \leq x)) .$

Negation: the number $m$ does not belong to the set $X$ , or there exists an element of the set $X$ that is less than $m .$ Let's write this using logical symbols:

$(m\notin X)\vee (\exists x\in X (x

Answer: $(m \in X) \land (\forall x \in X (m \leq x)),$ $(m\notin X)\vee (\exists x\in X (x

6. Число $m \in Z$ является делителем числа $n \in Z$ или в краткой записи $m | n .$

Solution.

This statement means that there exists an integer $k$ such that $k m = n .$ Let's write this using logical symbols:

$\exists k \in Z (k m = n) .$

Negation: for any integer $k$ , $k m \neq n .$ Or $\forall k \in Z (k m \neq n) .$

Answer: $\exists k \in Z (k m = n),$ $\forall k \in Z (k m \neq n) .$

Homework:

Read the statements below, clarify their meaning, and determine whether they are true or false (where symbols $x,, y,, z,, a,, b,, c$ represent real numbers).

7 $\exists y, \forall x,, (x + y = 3) .$