Tag: proofs in mathematics

Quizzes (2) Questions (121)

Questions Related to proofs in mathematics

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

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

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

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

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

  1. $p \leftrightarrow q$
  2. $p \rightarrow q$
  3. $p \leftrightarrow \sim q$
  4. $\sim p \rightarrow q$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
given statement 
$(p\wedge q)\vee (\sim p\wedge \sim q)$
$(p\wedge q)\vee (\sim p\wedge \sim q)$
$(p\wedge q)\vee (\sim(p\wedge  q))$
$(p\wedge q)\wedge(p\wedge  q)$
$p\leftrightarrow q$

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

$p \leftrightarrow q \equiv ?$

  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
Reveal answer Fill a bubble to check yourself
D Correct answer
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) $

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

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

  1. Conditional Law

  2. Demorgan's Law

  3. Absorption Law

  4. Identity Law
Reveal answer Fill a bubble to check yourself
B Correct answer
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

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

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

  1. Idempotent Law

  2. Conditional Law

  3. Involution Law

  4. Commutative Law
Reveal answer Fill a bubble to check yourself
B Correct answer
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.
Multiple choice maths proofs in mathematics implications principle of mathematical induction different forms of theoretical statements

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

  1. F, F

  2. T, T

  3. F, T

  4. T, F
Reveal answer Fill a bubble to check yourself
D Correct answer
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.

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

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

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
 $p$  $q$  $\sim q$  $(p\vee \sim q)$  $(p\wedge \sim q)$ $(p\vee \sim q)\rightarrow (p\wedge \sim q)$  $\sim[(p\vee \sim q)\rightarrow (p\wedge \sim q)]$
 $T$  $T$  $F$  $T$  $F$  $F$  $T$
 $T$  $F$  $T$  $T$  $T$  $T$   $F$
 $F$  $T$  $F$  $F$  $F$  $T$   $F$
 $F$  $F$  $T$  $T$  $F$  $F$ $T$
 $p$  $q$  $\sim q$  $(p\vee \sim q)$ $\sim p$  $(\sim p \vee q)$ $(p\vee \sim q)\wedge (\sim p \vee q)$
 $T$  $T$  $F$  $T$  $F$  $T$  $T$
 $T$  $F$  $T$  $T$  $F$  $F$  $F$
 $F$  $T$  $F$   $F$  $T$  $T$  $F$
 $F$  $F$  $T$  $T$  $T$  $T$  $T$
Multiple choice maths proofs in mathematics implications principle of mathematical induction different forms of theoretical statements

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

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
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