Determine whether or not the following arguments are valid. If they are valid, t

Question

Determine whether or not the following arguments are valid. If they are valid, then state the rules of inference used to prove validity. If they are invalid, outline precisely why they are invalid. If it is raining then I bring my umbrella to work. I bring my umbrella. Therefore it must be raining. Everyone who has a PC plays computer games. Everyone student taking COMP1501 next semester plays computer games. Therefore every student taking COMP1501 next semester has a PC. Prove that Squareroot 4 + Squareroot 5 is an irrational number. Prove, by indirect proof, that if n is an integer and n^5 + 7 is odd, then n is even. Show all your work.

Explanation / Answer

1) If it is raining then I bring my umbrella to work

P --> Q

In particular, modus ponens tells us that if a conditional statement and the
hypothesis of this conditional statement are both true, then the conclusion must
be true.

In this case it is Tautology which means the knowledge base is P -->Q
then the only model is TTT .

I bring my umbrella therefore it must be raining

A sentence or KB is satisfiable, if it has a model .
Propositional Satisfiability is the problem of deciding
if a sentence is satisfiable

b)

1)C(x) :Everyone who has a PC plays computer games.
2) D(x) :Everyone student taking COMP1501 next semester has a PC .

Then the premises are x(D(x)C(x)) and D

Step Reason
1. x(D(x)C(x)) PC has games
2. D(games)C(games) Universal instantiation from (1)
3. D(games) Premise
4. C(games) Modus ponens from (2) and (3)

2)2 and 5 is irrational.

2 is irrational.
Proof: We prove by means of contradiction. Assume
2 is a rational
number. So, there exists a0 and b0
integers with b0 6= 0 with 2 = a0/b0

We select such integers a and b with the additional property that a and b
have no common factors, i.e. the fraction a/b is in lowest terms (this is
always possible to obtain, for we can keep dividing by common factors).
So, 2 = a/b so 2 = a2/b2. This implies 2b2 = a2
By the definition of even, we know that a2 is even. Next we use a theorem that states that
if a2 is even then a is even (prove it as an exercise). Now, since a is even,
we know that there exists c such that a = 2c. Substituting in the formula
(1) above, we get that 2b2 = (2c)2 = 4c2
. Dividing both sides by 2 we get
b2 = 2c2
. By the definition of even, we see that b is even. Therefore, we
got that a is even and b is even, and so 2 is a common factor of a and b.
But we also had that a and b had no common factors. We just reached a
contradiction!

Similar iin the case of 5 .

3) n is a integer and n5 + 7 is odd then n is even .

Consider : Let x be any integer. Then x2 + x is even.

Proof:
p : x is even; q : x is odd; r : x2 + x is even.

Verify premise 1. If x is even, then x = 2n, for
some integer n. Hence, x2 + x = (2n)2+2n =
4n2 + 2n, which is even.

Verify premise 2. If x is odd, then x = 2n + 1, for
some n. Hence, x2 + x = (2n + 1)2 + (2n + 1) =
(4n2 + 4n + 1) + (2n + 1) = 4n2 + 6n + 2, which
is even.

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