Tag: proofs in mathematics

Questions Related to proofs in mathematics

The equivalent statement of $(p \vee q) \wedge \sim p$ is?

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

  1. $\sim p \vee q$

  2. $ p \wedge \sim q$

  3. $\sim p \wedge q$

  4. $ p \vee q$

Show answer
Correct Option: C
Explanation:

$(p\vee q)\wedge \sim p$
$=(p\wedge \sim p)\vee (q\wedge \sim p)$ Distributive Law
$=F\vee (q\wedge \sim p)$ Negation Law
$=(q\wedge \sim p)$ Identity Law
$=(\sim p \wedge q)$ Commutative Law

$p\rightarrow q$ is equivalent to

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

  1. $\sim p\vee \sim q$

  2. $ p\vee \sim q$

  3. $\sim p\vee q$

  4. $\sim p\wedge q$

Show answer
Correct Option: C
Explanation:
$p$ $q$ $p\rightarrow q$ $\sim p\vee q$
$T$ $T$ $T$ $T$
$T$ $F$ $T$ $T$
$F$ $F$ $T$ $T$
$F$ $T$ $T$ $T$

$(p\rightarrow q)\longleftrightarrow (\sim p\vee q)$ is a tautology.

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

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

  1. $p \leftrightarrow q$

  2. $p \rightarrow q$

  3. $p \leftrightarrow \sim q$

  4. $\sim p \rightarrow q$

Show answer
Correct Option: A
Explanation:
given statement 
$(p\wedge q)\vee (\sim p\wedge \sim q)$
$(p\wedge q)\vee (\sim p\wedge \sim q)$
$(p\wedge q)\vee (\sim(p\wedge  q))$
$(p\wedge q)\wedge(p\wedge  q)$
$p\leftrightarrow q$

$p \leftrightarrow q \equiv ?$

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

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

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

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

  4. None of these

Show answer
Correct Option: D
Explanation:

$p\leftrightarrow q=(p\rightarrow q)\wedge (q \rightarrow p)$
We know that $(p\rightarrow q)=(\sim p \vee q)$
So,
$(p\rightarrow q)\wedge (q \rightarrow p)=(\sim p \vee q)\wedge (\sim q \vee p)$
Now, apply the De'morgan law state that $\sim(p\vee q)= (\sim p \wedge \sim q)$
Therefore,
$(\sim p \vee q)\wedge (\sim q \vee p)=\sim (p \wedge \sim q) \wedge \sim (q\wedge \sim p) $

Identify the Law of Logic
$\sim(p \vee q) \equiv \sim p \wedge \sim q$

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

  1. Conditional Law

  2. Demorgan's Law

  3. Absorption Law

  4. Identity Law

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

It is Demorgan's law 
according to the if we take transpose or negation of any quatity then all the relation get opposite

Identify the Law of Logic
$p \rightarrow q \equiv \sim p \vee q$

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

  1. Idempotent Law

  2. Conditional Law

  3. Involution Law

  4. Commutative Law

Show answer
Correct Option: B
Explanation:
|  $p$ |  $q$ |  $p\rightarrow q$ |  $\sim p$ |  $(\sim p)\vee q$ | | --- | --- | --- | --- | --- | |  $T$ |   $T$ |   $T$ |   $F$ |   $T$ | |   $T$ |   $F$ |   $F$ |   $F$ |   $F$ | |  $F$ |   $T$ |   $T$ |   $T$ |   $T$ | |   $F$ |   $F$ |   $T$ |   $T$ |   $T$ |
We can say that if $p$ ,then $q$ or $p$ implies $q$ .
'$\rightarrow$' is called a conditional operator.
So, the giving logical equivalence is the conditional law.

Let p and q be any two logical statements and $r : p \rightarrow (\sim p \vee q)$. If r has a truth value F, then the truth values of p and q are respectively

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

  1. F, F

  2. T, T

  3. F, T

  4. T, F

Show answer
Correct Option: D
Explanation:
p q $\sim$p $\sim$ p $\vee$ q r
T T F T T
F F T T T
T F F F F
F T T T T

$\therefore$ Clearly from above able, If r has a truth value F, then the truth values of p and core T and F respectively.

State, whether the is given the statement, is True or False.
$\sim [(p \vee \sim q) \rightarrow (p \wedge \sim q)] \equiv (p \vee \sim q) \wedge (\sim \vee q)$

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

  1. True

  2. False

Show answer
Correct Option: A
Explanation:
 $p$  $q$  $\sim q$  $(p\vee \sim q)$  $(p\wedge \sim q)$ $(p\vee \sim q)\rightarrow (p\wedge \sim q)$  $\sim[(p\vee \sim q)\rightarrow (p\wedge \sim q)]$
 $T$  $T$  $F$  $T$  $F$  $F$  $T$
 $T$  $F$  $T$  $T$  $T$  $T$   $F$
 $F$  $T$  $F$  $F$  $F$  $T$   $F$
 $F$  $F$  $T$  $T$  $F$  $F$ $T$
 $p$  $q$  $\sim q$  $(p\vee \sim q)$ $\sim p$  $(\sim p \vee q)$ $(p\vee \sim q)\wedge (\sim p \vee q)$
 $T$  $T$  $F$  $T$  $F$  $T$  $T$
 $T$  $F$  $T$  $T$  $F$  $F$  $F$
 $F$  $T$  $F$   $F$  $T$  $T$  $F$
 $F$  $F$  $T$  $T$  $T$  $T$  $T$

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

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

  1. $\sim q$

  2. $q$

  3. $\sim p$

  4. $p$

Show answer
Correct Option: C
Explanation:

$\sim (p\vee q)\vee (\sim p \wedge q)$
$=(\sim p \wedge \sim q) \vee (\sim p \wedge q)$ De'Morgan Law
$=(\sim p)\wedge (\sim q\vee q)$ Distributive Law
$=(\sim p)\wedge T$ Negation Law
$=\sim p$ Identity law

State whether the following statements is True or False?
$p \leftrightarrow q \equiv (p \wedge q) \vee (\sim p \wedge \sim q)$

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

  1. True

  2. False

Show answer
Correct Option: A
Explanation:
given statement 
$p\leftrightarrow q=(p\wedge q)\vee (\sim p\wedge \sim q)$
taking RHS
$(p\wedge q)\vee (\sim p\wedge \sim q)$
$(p\wedge q)\vee (\sim(p\wedge  q))$
$(p\wedge q)\wedge(p\wedge  q)$
$p\leftrightarrow q$
It is true