Which of the following statements are true and which are false? In each case give a valid reason for saying so.
$s:$ If $x$ and $y$ are integers such that $x > y,$ then $-x < -y.$
$s:$ If $x$ and $y$ are integers such that $x > y,$ then $-x < -y.$
(A) Given the inequality $x > y$.
According to the rules of inequality,if we multiply both sides of an inequality by a negative number,the direction of the inequality sign reverses.
Multiplying both sides by $-1$,we get:
$(-1) \times x < (-1) \times y$
$-x < -y$
Thus,the given statement $s$ is true.
According to the rules of inequality,if we multiply both sides of an inequality by a negative number,the direction of the inequality sign reverses.
Multiplying both sides by $-1$,we get:
$(-1) \times x < (-1) \times y$
$-x < -y$
Thus,the given statement $s$ is true.
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Similar Questions
Let $p, q, r$ be three statements such that the truth value of $(p \wedge q) \rightarrow (\sim q \vee r)$ is $F$. Then the truth values of $p, q, r$ are respectively:
For the statement: "If a quadrilateral $ABCD$ is a rhombus,then its opposite sides are parallel",its contrapositive and converse are respectively given by:
Among the two statements:
$(S1): (p \Rightarrow q) \wedge (q \wedge (\sim q))$ is a contradiction and
$(S2): (p \wedge q) \vee ((\sim p) \wedge q) \vee (p \wedge (\sim q)) \vee ((\sim p) \wedge (\sim q))$ is a tautology.
$(S1): (p \Rightarrow q) \wedge (q \wedge (\sim q))$ is a contradiction and
$(S2): (p \wedge q) \vee ((\sim p) \wedge q) \vee (p \wedge (\sim q)) \vee ((\sim p) \wedge (\sim q))$ is a tautology.
Which of the following statements is logically equivalent to $\sim (p \leftrightarrow q)$?
Negation of the inverse of the following statement pattern $(p \wedge q) \rightarrow (p \vee \sim q)$ is
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