Write the converse of the following statement:
If a number $n$ is even,then $n^{2}$ is even.
If a number $n$ is even,then $n^{2}$ is even.
(N/A) The converse of the statement "If $p$,then $q$" is "If $q$,then $p$.
Given statement: If $n$ is even,then $n^{2}$ is even.
Therefore,the converse is: If $n^{2}$ is even,then $n$ is even.
Given statement: If $n$ is even,then $n^{2}$ is even.
Therefore,the converse is: If $n^{2}$ is even,then $n$ is even.
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Similar Questions
Which of the following statement patterns is a tautology?
$S_1 \equiv (\sim q \wedge p) \wedge q$
$S_2 \equiv [p \wedge (p$ $\rightarrow q)]$ $\rightarrow q$
$S_3 \equiv (p \wedge q) \wedge (\sim p \vee \sim q)$
$S_4 \equiv (p \wedge q) \rightarrow r$
$S_1 \equiv (\sim q \wedge p) \wedge q$
$S_2 \equiv [p \wedge (p$ $\rightarrow q)]$ $\rightarrow q$
$S_3 \equiv (p \wedge q) \wedge (\sim p \vee \sim q)$
$S_4 \equiv (p \wedge q) \rightarrow r$
State the converse and contrapositive of the following statement:
$r:$ If it is hot outside,then you feel thirsty.
$r:$ If it is hot outside,then you feel thirsty.
Easy
View SolutionWrite the following statement in the form "if-then":
$A$ quadrilateral is a parallelogram if its diagonals bisect each other.
$A$ quadrilateral is a parallelogram if its diagonals bisect each other.
Easy
View SolutionThe negation of the compound statement $\sim p \vee (p \vee (\sim q))$ is
In a Boolean Algebra $B$,for all $x, y \in B$,$x \wedge (x \vee y)$ is equal to
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