Tag: principle of mathematical induction

Questions Related to principle of mathematical induction

$\sim (p \vee q)$

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

  1. $p \wedge \sim q$

  2. $p \wedge q$

  3. $\sim p \wedge \sim q$

  4. $\sim p \vee \sim q$

Show answer
Correct Option: C
Explanation:

$\sim (p\vee q)$

By using De morgan's theorem,
$\sim p\wedge \sim q$

Statement I : if p is false statement and q is true statement, then $ \sim \,p\, \wedge \,q$ is true
Statement II : $ \sim \,p\, \wedge \,q$ is equivalent to $ \sim \left( {pV \sim \,q\,} \right)$

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

  1. Statement I is true and Statement II is the correct explanation for statement.

  2. Statement I is true and Statement II is true. Statement II is not the correct explanation for the Statement I.

  3. Statement I is true but Statement II is false.

  4. Statement I is false but Statement II is true

Show answer
Correct Option: A
Explanation:
p q $ \sim p$ $ \sim q$ (i)$ \sim \,p\, \wedge \,q$ $p\,\, \vee \, \sim \,q$ (ii)$ \sim \left( {p\,\, \vee \, \sim \,q} \right)$
T T F F F T F
T F F T F T F
F T T F T F T
F F T T F T F

p is false
q is true
$ \sim p\, \wedge \,q$ is true
from (i) & (ii)
$ \sim p\, \wedge \,q$ = $ \sim \left( {p\,\, \vee  \sim \,\,q} \right)$
Statement (i) & (ii) is correct 
$ \sim \left( {p\,\, \vee  \sim \,\,q} \right)\, = \, \sim \,p\, \wedge \, \sim \,\left( { \sim \,q} \right)$
$ = \, \sim \,p\, \wedge \,\,q$

$ \sim (p \vee q) \vee ( \sim p \wedge q)$ is logically euivalent to 

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

  1. $ \sim p$

  2. p

  3. q

  4. $ \sim q$

Show answer
Correct Option: C
Explanation:

${ \sim  }({ p }{ \vee  }{ q }){ \vee  }({ \sim  }{ p }{ \wedge  }{ q })\ ={ \sim  }({ U }){ \vee  }({ q }{ \wedge  }{ q })\ ={ \phi  }{ \vee  }({ q })\ ={ q }$


Where U is universal set , $\phi $ is nulll set and p and q are two disjoint sets 
Correct option is C

Cost of a diamond varies directly as the square of its weight.A diamond broke into four pieces with their weight in the ratio $1:2:3:4$ If the loss in the total value of the diamond was $Rs.\ 70000$. Find the prices of the original diamond.

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

  1. $10000\ Rs$

  2. $100000\ Rs$

  3. $1000000\ Rs$

  4. $2000000\ Rs$

Show answer
Correct Option: B
Explanation:

Let, the weights of the four pieces of the diamond $x,2x,3x,4x$ respectively.

$\therefore$ Total weight of the original diamond$=x+2x+3x+4x=10x$
Let, price of the original diamond$=k{ (100{ x }^{ 2 }) },where\quad k\quad is\quad constant$
$\therefore$ Cost of four pieces$=k({ x }^{ 2 }+4{ x }^{ 2 }+9{ x }^{ 2 }+16{ x }^{ 2 })\ =k(30{ x }^{ 2 })$
Loss in the total value of the diamond,
$k(100{ x }^{ 2 })-k(30{ x }^{ 2 })=70000\ \Rightarrow k(70{ x }^{ 2 })=70000\ \Rightarrow k{ x }^{ 2 }=Rs.1000$
Hence, price of the original diamond$=k(100{ x }^{ 2 })\ =1000\times 100=Rs.100000.$

Which of the following is always true ? 

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

  1. $\left( {p \to q} \right) \cong \left( { \sim q \to \sim p} \right)$

  2. $ \sim \left( {p \vee q} \right) \cong \left( { \sim p \vee \sim q} \right)$

  3. $ \sim \left( {p \to q} \right) \cong \left( {p \vee \sim q} \right)$

  4. $ \sim \left( {p \wedge q} \right) \cong \left( { \sim p \wedge \sim q} \right)$

Show answer
Correct Option: C
Explanation:

We know that 


$p\rightarrow q\equiv \sim p\wedge q$

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

$\sim (p\rightarrow q)\equiv p\vee \sim q$               (De morgan's law)

$C$ is correct

The Boolean expression $ \sim\ ( p \vee q ) \vee ( \sim\ p \wedge q ) $ is equivalent to:

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

  1. $p$

  2. $q$

  3. $ \sim q $

  4. $ \sim p $

Show answer
Correct Option: D
Explanation:
$\sim (p\vee q) \vee  (\sim p\wedge q)$
$=(\sim p \wedge \sim q) \vee (\sim p\wedge q)$
$=\sim p\wedge (\sim q\vee q)$
$=\sim p\wedge T$
$=\sim p$

$p \leftrightarrow q \equiv  \sim \left( {p\Delta  \sim q} \right)\Delta  \sim \left( {q\Delta  \sim p} \right)$

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

  1. True

  2. False

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Correct Option: A

Let $p$ and $q$ be two statements, then $ \sim ( \sim p \wedge q) \wedge (p \vee q)$ is logically equivalent to 

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

  1. $q$

  2. $p\vee q$

  3. $p$

  4. $p\vee \sim q$

Show answer
Correct Option: D

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

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

  1. $p \vee \sim q$

  2. $p \wedge \sim q$

  3. $ \sim p \vee q$

  4. $ \sim p \wedge q$

Show answer
Correct Option: B

$\left( { \sim p\Delta q} \right)V\left( { \sim p\Delta  \sim q} \right)V\left( { \sim p\Delta  \sim q} \right) \equiv  \sim pV \sim q$

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

  1. True

  2. False

Show answer
Correct Option: A