Tag: proofs in mathematics

Questions Related to proofs in mathematics

Write the converse and contrapositive of the statement
"If a dog is barking,then it will not bite"
$(i)$Converse of the statement:If a dog will bite then the dog is barking.
$(ii)$Contrapositive of the statement:If a dog will bite then the dog is not barking.

  1. $(i)$True $(ii)$False

  2. $(i)$True $(ii)$True

  3. $(i)$False $(ii)$False

  4. $(i)$False $(ii)$True


Correct Option: D
Explanation:

"If a dog is barking,then it will not bite"
$(i)$Converse of the statement:If a dog will not bite then the dog is barking.
$(ii)$Contrapositive of the statement:If a dog will bite then the dog is not barking.

Write the dual of the following statement:
(p$\vee$ q)$\wedge$ T

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

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

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

  4. (p$\vee$ q) $\vee$ T


Correct Option: B
Explanation:

To obtain the dual of a formula , replace ∧ with V, T with F and vice versa

So, B is correct option.

Which of the following is true about the converse and contrapositive of the statement
"If two triangles are congruent, then their areas are equal."
(i) Converse of the statement :
If the areas of the two triangles are equal, then the triangles are congruent.
(ii) Contrapositive of the statement:
If the areas of the two triangles are not equal, then the triangles are not congruent.

  1. (i) True (ii) False

  2. (i) False (ii) True

  3. (i) True (ii) True

  4. (i) False (ii) False


Correct Option: C
Explanation:

"If two triangles are congruent, then their areas are equal."
(i) Converse of the statement :
If the areas of the two triangles are equal, then the triangles are congruent.
(ii) Contrapositive of the statement:
If the areas of the two triangles are not equal, then the triangles are not congruent.

Identify the Law of Logic: $p \wedge q \equiv q \wedge p$

  1. Idempotent Law

  2. Commutative Law

  3. Associative Law

  4. Conditional Law


Correct Option: B
Explanation:
Given logic 
$p\wedge q\equiv q \wedge p$ 
 It is commutative law 
according to commutative law the order does not matter
we can also check it by truth table 

$p\wedge q$ is logically equivalent to

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

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

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

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


Correct Option: A
Explanation:
$p$ $q$ $p\wedge q$ $\sim(p\rightarrow\sim q)$
$T$ $T$ $T$ $T$
$T$ $F$ $T$ $T$
$F$ $F$ $T$ $T$
$F$ $T$ $T$ $T$

$(p\wedge q)\longrightarrow $$\sim(p\rightarrow\sim q)$ is a tautology.

The contrapositive of $p \to \left( { \sim q \to  \sim r} \right)$ is 

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

  2. $\left( {q \wedge \sim r} \right) \to \sim p$

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

  4. $p \wedge \left( {q \vee r} \right)$


Correct Option: A
Explanation:

Contraceptive of $a \rightarrow b$ is $\sim b \rightarrow \sim a$


Then

Contraceptive of $p \rightarrow (\sim q \rightarrow \sim r )$

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

$\equiv \sim (q \wedge \sim r) \rightarrow \sim p [a \rightarrow b \equiv \sim a \wedge b]$

$\equiv (\sim q \vee r) \rightarrow \sim p $  [Demorgas law]

$A$ is correct.

$\sim (p \vee q)$

  1. $p \wedge \sim q$

  2. $p \wedge q$

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

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


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)$

  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


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 

  1. $ \sim p$

  2. p

  3. q

  4. $ \sim q$


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.

  1. $10000\ Rs$

  2. $100000\ Rs$

  3. $1000000\ Rs$

  4. $2000000\ Rs$


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$=k(100{ x }^{ 2 })\ =1000\times 100=Rs.100000.$