Solution for Discrete Mathematics

Running head: SOLUTION FOR DISCRETE MATHEMATICS 1
Solution for Discrete Mathematics
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SOLUTION FOR DISCRETE MATHEMATICS 2
Question 1
a. The truth table for P ↓q
p q P ↓q
0 0 1
0 1 0
1 0 0
1 1 0
P ↓q is true if and only if neither q nor p are true meaning when both are false
b. The truth table for( p ↓q ¿ ↓ ( p ↓q ¿
p q ( p ↓q ¿ ( p ↓q ¿ ( p ↓q ¿ ↓ ( p ↓q ¿
0 0 1 1 0
0 1 0 0 1
1 0 0 0 1
1 1 0 0 1
It can be noted that the ( p ↓q ¿ ↓ ( p ↓q ¿ is nothing not only OR gate only
c. Negation using ↓ only
we have
p Negation of p
0 1
1 0
We can now consider ( p ↓ p ¿, ( p ↓ p ¿ will be true only when p is small

SOLUTION FOR DISCRETE MATHEMATICS 3
P P ( p ↓ p ¿
0 0 1
1 1 0
Thus ( p ↓ p ¿ is nothing but negation of p
‘AND’ using ↓ only
The truth table for ‘p AND q’ is as below
P q Output
0 0 0
0 1 0
1 0 0
1 1 1
Now consider ( p ↓ p ¿, ( q ↓ q ¿ and ( p ↓ p ¿ ↓ (q ↓ q ¿
P q ( p ↓ p ¿ (q↓ q ¿ ( p ↓ p ¿ ↓ (q↓ q ¿
0 0 1 1 0
0 1 1 0 0
1 0 0 1 0
1 1 0 0 1
It can be observed that the output for ( p ↓ p ¿ ↓ (q ↓ q ¿ is equal and same as ‘p AND q’
Therefore, AND gate can also be represented by applying ↓ only
Question 2
The symbols to be used in this solution
−negation ∃−their exist ∀−for alll ∨−' V ' ∧−' ⋀ '
a. No computer in the company is connect
∃ x ¿
b. There is computer connected

SOLUTION FOR DISCRETE MATHEMATICS 4
∃ x ¿
c. Computer either connected to network or with less 100TB of storage
∃ x ¿
Question 3
Determining the truth value of the preposition
a. ∃ x P (6 , x )
62 =x
36=x
Hence, its truth value is TRUE because there is an x (x = 36) for the value P (6, x) is
correct
b. ∃ x P ( x , 6 ) ⟹ x2=6
x=± √ 6 ≠interger
Therefore, Truth value is FALSE because we have no x for which P (x, 6) is true
c. ∀ x ∃ y ¿
Substituting any integer value for x, the integer for y is determined. Therefore, the truth
value is TRUE
d. ∃ y ∀ x P ( x , y ) ⟹ x2= y
Truth value is FALSE
Because there may be a y (y=36) for which x (x is 6) may be correct but not for all the values
Question 4
a. If n2 is multiple of 4 then n is multiple of 4
This is a wrong statement
Counter example
If n=6
n2 ⇒ 36 is multiple of 4
but n=6 is not multiple of 4
b. If n3 is multiple of 2 then n is multiple of 2
Proof