Logical symbolism. Necessary and sufficient conditions.

Literature: Collection of problems in mathematics. Part 1. Edited by A. V. Efimov, B. P. Demidovich.

Let $\alpha, \beta, ... -$ 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 $\overline{\alpha}$ means "not $\alpha$", that is, the negation of the statement $\alpha$.

The notation ${\alpha}\Rightarrow\beta$ means "statement $\alpha$ implies statement $\beta$". ($\Rightarrow -$ implication symbol).

The notation ${\alpha}\Leftrightarrow\beta$ means "Statement $\alpha$ is equivalent to statement $\beta$", that is, $\alpha$ implies $\beta$, and $\beta$ implies $\alpha$ ("$\Leftrightarrow-$" equivalence symbol).

The notation ${\alpha}\wedge\beta$ means "$\alpha$ and $\beta$" ($\wedge -$ conjunction symbol).

The notation ${\alpha}\vee\beta$ means "$\alpha$ or $\beta$" ($\vee -$ disjunction symbol).

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

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

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

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

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.

1.87. $\forall x(x^2>x\Leftrightarrow x>1\vee x<0).$

Solution.

The statement $\forall x(x^2>x\Leftrightarrow x>1\vee x<0)$ 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.

1.88. $\forall x (x>2\wedge x>3\Leftrightarrow 2

Solution.

The statement $\forall x (x>2\wedge x>3\Leftrightarrow 22$ 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 $(2,, \infty)\cap(3,, \infty)=(3, ,\infty).$ And the condition $22\wedge x>3\Leftrightarrow 2

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:

1.92.(a). 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: $f(x_0)=0,$ $f(x_0)\neq 0.$

1.93.(b). 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: $(m\in X)\wedge(\forall x\in X (m\leq x)).$

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

1.94.(а). 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 $km=n.$ Let's write this using logical symbols:

$\exists k\in \mathbb{Z} (km=n).$

Negation: for any integer $k$, $km\neq n.$ Or $\forall k\in \mathbb{Z} (km\neq n).$

Answer: $\exists k\in \mathbb{Z} (km=n),$ $\forall k\in \mathbb{Z} (km\neq n).$

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).

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