Mathematical Solutions Assignment

Ans1
a.
P q r ~p ~p v q p n r Result
0 0 0 1 1 0 0
0 0 1 1 1 0 0
0 1 0 1 1 0 0
0 1 1 1 1 0 0
1 0 0 0 0 0 1
1 0 1 0 0 1 1
1 1 0 0 1 0 0
1 1 1 0 1 1 1
b.
P q r ~q ~r p v ~q p n r q v r Conclusi
on
0 0 0 1 1 1 1 0 0
0 0 1 1 0 1 1 1 1
0 1 0 0 1 0 1 1 0
0 1 1 0 0 0 1 1 1
1 0 0 1 1 1 1 0 0
1 0 1 1 0 1 0 1 1
1 1 0 0 1 1 1 1 0
1 1 1 0 0 1 0 1 1
Invalid
c.
= p -> q = ~p v q
= ~(p -> q) = p n ~q
= (q n ~p) V (p n q )
= (~p n q) v (p n q)
= (~p v p) n q
= q
d.
premises p -> r, ~q -> ~r, ~s, q -> s
conclusion ~p
Step Reason
1. q -> s Hypothesis
2. ~s Hypothesis
3. ~q Modus tolles 1 and 2
4. ~q -> ~r Hypothesis

5. ~r Modul tollen using 3 and
4
6. P -> r Hypothesis
7. ~p Conclusion using modus
tollen 5 and 6
Ans 2.
a. Symbolically, R can be expressed as ∀x ∈ R, x2 ≥ 0, and thus its negation
is
∼ (∀x ∈ R, x2 ≥ 0) = ∃ x ∈ R, ∼ (x2 ≥ 0) = ∃ x ∈ R, x2 < 0. In words, this is
∼ R : There exists a real number whose square is negative.
Observe that R is true and ∼ R is false. You may be able to get ∼ R
immediately, without using Equation (7.3) as we did above. If so, that is
good; if not, you will probably be there soon.
If a statement has multiple quantifiers, negating it will involve several
iterations of Equations (7.3) and (7.4). Consider the following:
S : For every real number x there is a real number y for which y3 = x.
This statement asserts any real number x has a cube root y, so it’s true.
Symbolically S can be expressed as
∀x ∈ R,∃ y ∈ R, y3 = x.
Let’s work out the negation of this statement.
∼ (∀x ∈ R,∃ y ∈ R, y3 = x) = ∃ x ∈ R,∼ (∃ y ∈ R, y3 = x)
= ∃ x ∈ R,∀ y ∈ R, ∼ (y3= x)
= ∃ x ∈ R,∀ y ∈ R, y3= x.
Therefore the negation is the following (false) statement.
∼ S : There is a real number x for which y3= x for all real numbers y.
b. ∀x ∈ R, ∀y ∈ Z, ∃z ∈ N, (xy >z^y)
c. Take x = -1 and y = 2, xy<0 and x^4 =1 y^4=16
1+16 = 17, xy = -2
17/-2 not divisible
17/-3 not divisible
d. .
D Left Right Left -> Right
-7 T F F
-5 T F F
-2 T F F
-1 F F T
0 F T T
1 F F T
2 T F F
3 T T T
4 F F T
5 T F F
6 F T T
7 T F F
10 F F T
11 F F T
14 F F T
19 F F T

20 F F T
21 F T T
24 F T T
Ans 3.
a. Lets take if n=20 and m =10 then nm^2 = 2000 it means only possible
case for nm^2<=1985 means deacrease n or m if n=20 and m = 9 the
nm^2 = 1620 and if n = 19 and m = 10 then nm^2 = 1900. Hence either
n<=19 or m<=9 for nm^2 <=1985.
b. According to Fermat’s Theorem,
R[(7^16)] = 1
Also, R[(7^16k)] = 1, where 16k is a multiple of 16.
So, 54 = 16*3 + 6
So R[(7^54)/17] = R[(7^48)*R[(7^6)/17] = R[(7^6)/17
Now, R[(7^6)/17 = R{[(117649]/17} =9
c.
We know a-c divides a-c
And a-c divides ad+cb
So,
a-c | a-c
= a-c |(a-c)(d-b)
=a-c|(ad – ab – cd + cb)
=a-c|(ab + cd - ad - cb )
Q.E.D. a- c | (ab + cd)
d
Let us assume that √13 is rational no and equals to p/q are Co primes
√13=p/q
squaring both sides
√13*2=p*2/q*2
13=p*2/q*2
13q*2=p*2
13/p*2