Tag: proofs in mathematics

Questions Related to proofs in mathematics

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

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

  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.

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Correct Option: A

Either $p$ or $q$ is equivalent to:

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

  1. $p \vee q$

  2. $(p \vee \sim q) \vee (q \wedge \sim p)$

  3. $(p \vee \sim q) \wedge (q \vee \sim p)$

  4. none

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Correct Option: A

Equivalent statement of ''If $x\in Q$, then $x\in T$'' is
$x\in Q$ is necessary for $x\in l$
$x\in l$ is sufficient for $x\in Q$
$z\in Q$ or $x\in l$
$x\in Q$ but $x\in l$ 

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

  1. $I$ & $II$

  2. $I,\ II$ & $IV$

  3. $III$

  4. $All$

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Correct Option: B

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

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

  1. $S \rightarrow \sim Q$

  2. $\sim Q \rightarrow S$

  3. $\sim S \rightarrow \sim Q$

  4. $Q \rightarrow \sim S$

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Correct Option: B

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

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

  1. Equivalent to $p\wedge q$

  2. Tautology

  3. Fallacy

  4. Neither tautology nor fallacy

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Correct Option: A

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 $\displaystyle m\left ( n-k \right )-k\left ( k-1 \right )+k-m= 0.$

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

  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.

Show answer
Correct Option: A
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.

The dual of the statement $\sim p \wedge [\sim q \wedge (p \vee q) \wedge \sim  r]$ is:

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

  1. $\sim p \vee [\sim q \vee (p \vee q) \vee \sim r]$

  2. $ p \vee [q \vee (\sim p \wedge \sim q) \vee r]$

  3. $ \sim p \vee [\sim q \vee (p\wedge q) \vee \sim r]$

  4. $ \sim p \vee [\sim q \wedge (p\wedge q) \wedge \sim r]$

Show answer
Correct Option: C
Explanation:

The dual of the statement $\sim p \wedge [\sim q \wedge (p \vee q) \wedge \sim  r]$ is

$\equiv  \sim p \vee [\sim q \vee (p\wedge q) \vee  \sim r]$

Note: For dual of a statement just replace $\vee$ by $\wedge$ and vice versa.

Which of the following is equivalent to $(p \wedge q)$?

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

  1. $p \rightarrow \sim q $

  2. $ \sim (\sim p \wedge \sim q)$

  3. $ \sim ( p \rightarrow \sim q)$

  4. None of these

Show answer
Correct Option: C
Explanation:

$p \wedge q \equiv   \sim (\sim p \vee \sim q) \equiv   \sim (p \to \sim q)$

Which of the following is equivalent to $( p \wedge q)$?

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

  1. $p \rightarrow \sim q$

  2. $\sim (\sim p \wedge \sim q)$

  3. $\sim (p \rightarrow \sim q)$

  4. None of these.

Show answer
Correct Option: C
Explanation:

$p \wedge q \equiv   \sim (\sim p \vee \sim q) \equiv   \sim (p \to \sim q)$

The equivalent statement of (p $\leftrightarrow$ q) is

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

  1. $(p \wedge q) \vee (p \vee q)$

  2. $(p \rightarrow q) \vee (q \rightarrow p)$

  3. $(\sim p \vee q) \vee (p \vee \sim q)$

  4. $(\sim p \vee q) \wedge (p \vee \sim q)$

Show answer
Correct Option: D
Explanation:

$p\rightarrow q \equiv (\sim p\vee q)$

$q\rightarrow p \equiv (\sim q \vee p)$
$\therefore$
$p\leftrightarrow q \equiv (p\rightarrow q)\wedge(q\rightarrow p)$
$\Rightarrow p\leftrightarrow q\equiv (\sim p\vee q)\wedge(p\vee\sim q)$