Mathematical Reasoning

Let p = you are born in India.

q = you are a citizen of India.

$ \therefore $ $ \sim $ p = you are not born in India.

$ \sim $ q = you are not a citizen of India.

We know Contrapositive of p $ \to $ q is ~q $ \to $ ~p

So contrapositive of statement will be :

“If you are not a citizen of India, then you are not born in India.”

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

$ \equiv $ $\left( {\left( {\left( {p \vee \left( {p \vee \sim q} \right)} \right)} \right) \wedge \left( {q \vee \left( {p \vee \sim q} \right)} \right)} \right) \wedge \left( { \sim p \wedge \sim q} \right)$

$ \equiv $ $\left( {\left( {p \vee \sim q} \right) \wedge \left( {q \vee \sim q \vee p} \right)} \right) \wedge \left( { \sim p \wedge \sim q} \right)$

$ \equiv $ $\left( {\left( {p \vee \sim q} \right) \wedge \left( {t \vee p} \right)} \right) \wedge \left( { \sim p \wedge \sim q} \right)$

$ \equiv $ $\left( {\left( {p \vee \sim q} \right) \wedge t} \right) \wedge \left( { \sim p \wedge \sim q} \right)$

$ \equiv $ $\left( {p \vee \sim q} \right) \wedge \left( { \sim p \wedge \sim q} \right)$

$ \equiv $ $\left( {p \wedge \sim p \wedge \sim q} \right) \vee \left( { \sim q \wedge \sim p \wedge \sim q} \right)$

$ \equiv $ $\left( {f \wedge \sim q} \right) \vee \left( { \sim q \wedge \sim p} \right)$

$ \equiv $ $f' \vee \left( { \sim q \wedge \sim p} \right)$

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

Let,

p : two numbers are not equal

q : squares of two numbers are not equal

Contrapositive of p $ \to $ q is $ \sim $q $ \to $ $ \sim $p.

$ \therefore $ $ \sim $q $ \to $ $ \sim $p means "If the squares of two numbers are equal, then the numbers are equal".

Given q is F and (p $ \wedge $ q) $ \leftrightarrow $ r is T

$ \Rightarrow $  p $ \wedge $ q is F which implies that r is F

$ \Rightarrow $  q is F and r is F

$ \Rightarrow $  (p $ \wedge $ r) is always F

$ \Rightarrow $  (p $ \wedge $ r) $ \to $ (p $ \vee $ r) is tautology.

It is obvious

P(n) : n² $-$ n + 41 is prime

P(5) = 61 which is prime

P(3) = 47 which is also prime

Given that, (p $ \oplus $ q) $\wedge$ (~ p $ \odot $ q) $ \equiv $ p $\wedge$ q.

From the truth table you can see this equivalence only hold when ordered pair $\left( { \oplus , \odot } \right)$ is = $\left( { \wedge , \vee } \right)$

p	q	~p	~q	~p $ \vee $ ~q	p $ \to $ (~p $\vee$ ~q)
T	T	F	F	F	F
T	F	F	T	T	T
F	T	T	F	T	T
F	F	T	T	T	T

From the truth table,

p $ \to $ (~p $\vee$ ~q) is false only when p and q both are true.

From the table you can see ~p and
~(p $ \vee $ q) $ \vee $ (~p $ \wedge $ q) are equivalent.

Statement p :
sin 120^o = cos 30^o = ${{\sqrt 3 } \over 2}$ $ \Rightarrow $ 2 sin 120^o = $\sqrt 3 $

So, $\sqrt {1 + \sin {{240}^o}} - \sqrt {1 - \sin {{240}^o}} $

$ = \sqrt {{{1 - \sqrt 3 } \over 2}} - \sqrt {{{1 + \sqrt 3 } \over 2}} \ne \sqrt 3 $

Statement q :
So,   A + B + C + D = 2$\pi $

$ \Rightarrow $ ${{A + C} \over 2} + {{B + D} \over 2} = \pi $

$ \Rightarrow $  cos$\left( {{{A + C} \over 2}} \right) + \cos \left( {{{B + D} \over 2}} \right)$

= cos $\left( {{{A + C} \over 2}} \right)$ $-$ cos$\left( {{{A + C} \over 2}} \right) = 0$

Therefore, statement p is false and statement q is true.

From the truth table you can see (p $ \wedge $ ~q) $ \wedge $ (p $ \wedge $ r) $\to$ ~p $\vee$ q is false only when values of (p, q, r) is (T, F, T).

Let,

p : two numbers are not equal

q : squares of two numbers are not equal

Contrapositive of p $ \to $ q is $ \sim $q $ \to $ $ \sim $p.

$ \therefore $ $ \sim $q $ \to $ $ \sim $p means "If the squares of two numbers are equal, then the numbers are equal".

We have

p	q	~p	p$\to$q	~p$\to$q	(~p$\to$q)$\to$q	(p$\to$q)$\to$((~p$\to$q)$\to$q)
T	F	F	F	T	F	T
T	T	F	T	T	T	T
F	F	T	T	F	T	T
F	T	T	T	T	T	T

$ \therefore $ It is tautology.

Contrapositive of p $ \to $ q is $ \sim $q $ \to $ $ \sim $p.

Here,

Let

   p = Side of a square is doubles.

   q = Area of square increases four times.

$ \therefore $   $ \sim $q $ \to $ $ \sim $p = If the area of a square does not increase four times, then its side is not doubled.

Contrapositive of P : If 7 is not divisible by 2, then 7 is not an odd number.

This statement is false.

$ \therefore $    V₁ = False (F)

Contrapositive of Q : If 7 is not an odd number ,then 7 is not a prime number.

This statement is true.

$ \therefore $   V₂ = True (T)

$ \therefore $   (V₁, V₂) = (F, T).

From the table you can see for both p $\vee$ q and (p $\wedge$ ~q) $\vee$ q $\vee$ (~p $\wedge$ q) truth table matches so they are equivalent.

From the table you can see (s $\wedge$ r) and negation of ~s $\vee$ (~r $\wedge$ s) have same truth value. So they are equivalent.

p	q	$ \sim p$	$ \sim q$	$p \leftrightarrow q$	$p \leftrightarrow \sim q$	$ \sim p \leftrightarrow q$	$ \sim (p \leftrightarrow \sim q)$
T	F	F	T	F	T	T	F
F	T	T	F	F	T	T	F
T	T	F	F	T	F	F	T
F	F	T	T	T	F	F	T

From the truth table we can say that given statement is equivalent to $p \leftrightarrow q$.

Statement - I : $(p \wedge \sim q) \wedge ( \sim p \wedge q)$

p	q	$ \sim p$	$ \sim q$	$p \wedge \sim q$	$ \sim p \wedge q$	$(p \wedge \sim q) \wedge ( \sim p \wedge q)$
T	T	F	F	F	F	F
T	F	F	T	T	F	F
F	T	T	F	F	T	F
F	F	T	T	F	F	F

Hence, statement-I is a fallacy.

Statement - II : $(p \to q) \leftrightarrow ( \sim q \to \sim p)$

p	q	$ \sim p$	$ \sim q$	$p \to q$	$ \sim q \to \sim p$	$(p \to q) \leftrightarrow ( \sim q \to \sim p)$
T	T	F	F	T	T	T
T	F	F	T	F	F	T
F	T	T	F	T	T	T
F	F	T	T	T	T	T

Hence, statement - II is a tautology.

Now, statement-II does not depend on statement-I.

Let, p : I become a teacher

q : I will open a school

$\therefore$ "If I become a teacher, then I will open a school", is p $\to$ q

So, negation of $(p \to q) = \sim (p \to q) = p \wedge \sim q$ = I will become a teacher and I will not open a school

"Suman is brilliant and dishonest" an be expressed as : ${P \wedge \sim R}$

So “Suman is brilliant and dishonest if and only if Suman is rich” can be expressed as :

$\left( {P \wedge \sim R} \right) \leftrightarrow Q$

Now negation of this = $ \sim \left[ {\left( {P \wedge \sim R} \right) \leftrightarrow Q} \right]$

You should know, $p \leftrightarrow q \equiv q \leftrightarrow p$

So, $ \sim \left[ {\left( {P \wedge \sim R} \right) \leftrightarrow Q} \right] \equiv \sim \left[ {Q \leftrightarrow \left( {P \wedge \sim R} \right)} \right]$

Given that S is a non-empty subset of R.

$\bullet$ P : There is a rational number x $\in$ S such that x > 0.

Now, we need to find the negation of P. Clearly, P is equivalent to saying that "There is a positive rational number in S.

So, its negation ($\sim$ P) is "There is no positive rational number in S".

Thus, for $\sim$ P : There exists no positive rational number in S.

$\bullet$ $\Leftrightarrow$ $\sim$ P : Every rational number x $\in$ S satisfies x $\le$ 0.

Therefore, S1 is true and S2 is false.

p : x is an irrational number

q : y is a transcendental number

r : x is a rational number, if y is a transcendental number

$\Rightarrow$ r : $\sim$ p $\leftrightarrow$ q

S₁ : r $\equiv$ q $\vee$ p

and S₂ : r $\equiv$ $\sim$ (p $\leftrightarrow$ $\sim$ q)

p	q	$ \sim p$	$ \sim q$	r $ \sim p \leftrightarrow q$	Statement I $q \vee p$	$(p \leftrightarrow \sim q)$	Statement II $ \sim (p \leftrightarrow \sim q)$
T	T	F	F	F	T	F	T
T	F	F	T	T	T	T	F
F	T	T	F	T	T	T	F
F	F	T	T	F	F	F	T

It is clear from the table that r is not equivalent to either of the statements.

Therefore, both statements are false.

q	p	$q \to p$	$p \to (q \to p)$	$(p \vee q)$	$p \to (p \vee q)$
T	T	T	T	T	T
T	F	F	T	T	T
F	T	T	T	T	T
F	F	T	T	F	T

$\therefore$ Statement p $\to$ (q $\to$ p) is equivalent to p $\to$ (p $\vee$ q).