Question

I will really appreciate it if you could answer all of them all for me. Thank you :)

1.

use a direct proof to SHOW the following:

The square of an even natural number is even.

The sum of an even and odd number is odd.

The sum of two even number is even.

The sum of two odd number is even.

2.

Examine below compound proposition:

[-p ^ ( p v q ) ] -> q.

(1) Complete truth table

(2) Explain if this IS or IS NOT a tautology, and why.

3.

Determine the satisfiability of the following compound proposition:

(p v -q) ^ (q v -r) ^ (r v -p) ^ (p v q v r) ^ (-p v -q v -r ).

4. Select ALL that is true about the following logical operation between 'p' and 'q.'

p ∧ q.

if, and only if, both p and q are T, the end result will be T (T/F)

only if both p and q are 1 will result in a 1 (T/F)

If only 1 of the variables is 1, the result CAN still be a 1 (T/F)

in 3 of the 4 outcomes, the outcome will be F. (T/F)

Answer 1

1 -
The square of an even natural number is even.
Let n be a even natural number , then n can be written as n = 2(k) for positive integer k.
Therefore n^2 can be written as n^2 = (2k)^2
= 4k^2
= 2(2k^2)
Because we can write n^2 in form of 2(something) , we can say that n^2 is even.

The sum of an even and odd number is odd.
Let n1 be a even number , then n1 can be written as = 2(k1) for some integer k1.
Let n2 be a odd number , then n2 can be written as = 2(k2)+1 , for some integer k2.
Adding n1 and n2,
n = n1 + n2
= 2k1 + (2k2 +1)
= 2(k1+k2) + 1
Thus , n can be written in the form of 2k+1 , we conclude that n is odd.

The sum of two even number is even.
Let n1 be a even number , then n1 can be written as = 2(k1) for some integer k1.
Let n2 be a even number , then n2 can be written as = 2(k2), for some integer k2.
Adding n1 and n2,
n = n1 + n2
= 2k1 + 2k2
= 2(k1+k2)
Thus , n can be written in the form of 2k , we conclude that n is even.

The sum of two odd number is even.
Let n1 be a odd number , then n1 can be written as = 2(k1)+1,for some integer k1.
Let n2 be a odd number , then n2 can be written as = 2(k2)+1, for some integer k2.
Adding n1 and n2,
n = n1 + n2
= (2k1 + 1) + (2k2 + 1)
= 2k1 + 2k2 + 2
= 2(k1+k2+1)
Thus , n can be written in the form of 2k , we conclude that n is even.

2 -
[-p ^ ( p v q ) ] -> q.
Using :

p	q	p->q
T	T	T
T	F	F
F	T	T
F	F	T

Here's the truth table :

p	q	-p	p v q	-p^(pvq)	[-p ^ ( p v q ) ] -> q.
F	F	T	F	F	T
F	T	T	T	T	T
T	F	F	T	F	T
T	T	F	T	F	T

A tautology is a formula which is "always true" and as we can see that this proposition remains always trur, this is a Tautology.

3 -
(p v -q) ^ (q v -r) ^ (r v -p) ^ (p v q v r) ^ (-p v -q v -r ).

A proposition is satisfiable if there is at least one true result in its truth table

Truth Table :

p	q	r	p v -q	q v -r	r v -p	p v q v r	-p v -q v -r	(p v -q) ^ (q v -r) ^ (r v -p) ^ (p v q v r) ^ (-p v -q v -r )
F	F	F	T	T	T	F	T	F
F	F	T	T	F	T	T	T	F
F	T	F	F	T	T	T	T	F
F	T	T	F	T	T	T	T	F
T	F	F	T	T	F	T	T	F
T	F	T	T	F	T	T	T	F
T	T	F	T	T	F	T	T	F
T	T	T	T	T	T	T	F	F

As there is no true condition , the given preposition in not satisfiable.

4 -
p ^ q
Truth Table :

p	q	p^q
T	T	T
T	F	F
F	T	F
F	F	F

if, and only if, both p and q are T, the end result will be T. => True
only if both p and q are 1 will result in a 1 . => True
If only 1 of the variables is 1, the result CAN still be a 1  => False
in 3 of the 4 outcomes, the outcome will be F.  => True