The negation of the statement "If a number is | vedclass

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(D) Let $p$,$q$,and $r$ be the following statements:$p$: $A$ number is divisible by $15$.$q$: $A$ number is divisible by $5$.$r$: $A$ number is divisi

Vedclass · Accepted Answer

$A$ number is divisible by $15$ and it is not divisible by $5$ and $3$.

Vedclass · Answer

(D) Let $p$,$q$,and $r$ be the following statements:$p$: $A$ number is divisible by $15$.$q$: $A$ number is divisible by $5$.$r$: $A$ number is divisible by $3$.The given statement is of the form $p \rightarrow (q \wedge r)$.The negation of an implication $p \rightarrow S$ is $p \wedge (\sim S)$.Therefore,the negation is $p \wedge \sim(q \wedge r)$.Using De Morgan's Law,$\sim(q \wedge r) \equiv (\sim q \vee \sim r)$.Thus,the negation is $p \wedge (\sim q \vee \sim r)$,which means: "$A$ number is divisible by $15$ and it is not divisible by $5$ or it is not divisible by $3$."However,looking at the logical structure provided in the options,the standard negation of "If $p$ then $q$ and $r$" is "$p$ and not ($q$ and $r$)",which is "$p$ and (not $q$ or not $r$)".Given the options provided,option $D$ is the intended answer representing the logical negation $p \wedge \sim(q \wedge r)$.

The negation of the statement "If a number is divisible by $15$,then it is divisible by $5$ and $3$" is:

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