Tag: principle of mathematical induction

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Questions Related to principle of mathematical induction

Multiple choice business maths proofs in mathematics implications principle of mathematical induction different forms of theoretical statements

$(p \wedge q) \vee  \sim p$ is equivalent to 

  1. $\sim p \wedge q$
  2. $\sim p \vee q$
  3. $p \wedge q$
  4. $p \vee q$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

By distributive law, (p and q) or not p is equivalent to (p or not p) and (q or not p). Since (p or not p) is a tautology (True), the expression simplifies to (q or not p), which is not p or q.

Multiple choice business maths proofs in mathematics implications principle of mathematical induction different forms of theoretical statements

$\sim (p \wedge q)\Rightarrow (\sim p)\vee (\sim p \vee q)$ is equal to

  1. $\sim p \vee q$
  2. $\sim p \wedge q$
  3. $p\vee \sim q$
  4. $p\wedge \sim q$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The expression ~(p AND q) -> (~p OR (~p OR q)) simplifies to ~(p AND q) -> (~p OR q). This is (~p OR ~q) -> (~p OR q). This is equivalent to (~(~p OR ~q) OR (~p OR q)) = (p AND q) OR (~p OR q). This simplifies to (~p OR q).

Multiple choice business maths proofs in mathematics implications principle of mathematical induction different forms of theoretical statements

The equivalent of $(p \rightarrow \sim p) \vee (\sim p \rightarrow p)$ is 

  1. $p \vee \sim p$
  2. $T \rightarrow F$
  3. $T \leftrightarrow F$
  4. $p \wedge \sim p$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The expression (p -> ~p) is equivalent to ~p, and (~p -> p) is equivalent to p. Therefore, the disjunction (~p) v p is a tautology, which is equivalent to p v ~p.

Multiple choice business maths proofs in mathematics implications principle of mathematical induction different forms of theoretical statements

Identify which of the following statement is not equivalent to the others

  1. If $x$ is bass then $x$ is bad.

  2. Boss implies bad,

  3. Bad is necessary condition for bass.
  4. $x$ is boss iff $x$ is bad.
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Statements A, B, and C all translate to the conditional statement that boss implies bad, or bad is a necessary condition for boss. Option D states an iff (biconditional) condition, which is a stronger claim and thus not equivalent to the others.

Multiple choice business maths proofs in mathematics implications principle of mathematical induction different forms of theoretical statements

Let  $P , Q , R$  and  $S$  be statements and suppose that  $P \rightarrow Q \rightarrow R \rightarrow P.$  If  $\sim S \rightarrow R,$  then

  1. $S \rightarrow \sim Q$
  2. $\sim Q \rightarrow S$
  3. $\sim S \rightarrow \sim Q$
  4. $Q \rightarrow \sim S$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given P -> Q, Q -> R, R -> P, we have P <-> Q <-> R. If ~S -> R, then by contrapositive ~R -> S. Since R is equivalent to Q, ~Q -> S.

Multiple choice business maths proofs in mathematics implications principle of mathematical induction different forms of theoretical statements

$(p\rightarrow q)\leftrightarrow (q\vee \sim p)$ is - 

  1. Equivalent to $p\wedge q$

  2. Tautology

  3. Fallacy

  4. Neither tautology nor fallacy
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The implication p implies q is logically equivalent to not p or q. The given statement equates this with (q or not p), which is identical by the commutative law. Thus, both sides are always equal, making the bi-implication a tautology.

Multiple choice business maths proofs in mathematics implications principle of mathematical induction different forms of theoretical statements

Let S be a set of n persons such that:(i)any person is acquainted to exactly k other persons in s;(ii)any two persons that are acquainted have exactly $\displaystyle l $ common acquaintances in s;(iii)any two persons that are not acquainted have exactly m common acquaintances in S.Prove that $\displaystyle m\left ( n-k \right )-k\left ( k-1 \right )+k-m= 0.$

  1. $\displaystyle k\left ( k-1-l \right )= m\left ( n-k-1 \right )$ is equivalent to the desired one.
  2. $\displaystyle k\left ( k+1-l \right )= m\left ( n-k-1 \right )$ is equivalent to the desired one.
  3. $\displaystyle k\left ( k-1+l \right )= m\left ( n-k-1 \right )$ is equivalent to the desired zero.
  4. $\displaystyle k\left ( k+1-l \right )= m\left ( n-k+1 \right )$ is equivalent to the desired one.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let a be a fixed element os S. Let us count the triples (a, x, y) such that a, x are acquainted, x, y are acquainted and a, y are not acquainted. Because a isacquainted to exactly k other persons in S, x can be chosen in k ways and for fixeda and x, y can be chosen in $\displaystyle k-1-l $ ways. Thus the number of such triples is $\displaystyle k\left ( k-1-l  \right ).$ Let us count again, choosing y first. The number of persons not acquainted to a equals n-k-1, hence y can be chosen in n-k-1 ways. Because x is a commonacquaintance of a and y, it can be chosen in m ways, yielding a total of $\displaystyle m\left ( n-k-1 \right )$ triples. It is not difficult to see that the equality $\displaystyle k\left ( k-1-l  \right )= m\left ( n-k-1 \right )$ is equivalent to the desired one.