0
Add your answer:
Earn +20 pts
How would you show that every positive integer can be written as the product of a power of 2 and an odd?
If P is a positive integer, then let 2n be the largest power of two that divides P. Then P = Q2n, where Q is the quotient of this division. Clearly Q is odd - for otherwise, 2 would divide Q, which would mean 2n + 1 also divides P, a contradiction.
What is the sum or difference of p and q?
The sum of p and q means (p+q). The difference of p and q means (p-q).
What is the proof of the modus ponens not by the truth table?
1)p->q 2)not p or q 3)p 4)not p and p or q 5)contrudiction or q 6)q
What is p lots of q?
p*q
What is converse inverse and contrapositive?
if the statement is : if p then q converse: if q then p inverse: if not p then not q contrapositive: if not q then not
How would you show that every positive integer can be written as the product of a power of 2 and an odd?
If P is a positive integer, then let 2n be the largest power of two that divides P. Then P = Q2n, where Q is the quotient of this division. Clearly Q is odd - for otherwise, 2 would divide Q, which would mean 2n + 1 also divides P, a contradiction.
What symbol represents contradiction?
In Formal Logic proofs, the contradiction is represented with an inverted T (or upside-down T) as follows: ┴ The contradiction symbol can be introduced at any time a logical contradiction is encounterd, for example, all of the the following contradictory logical statements (using different symbols) can be replaced with the contradiction symbol: The ball is completely blue and the ball not completely blue. P ^ ¬P P & ~P P & !P P AND NOT P
If p q and q r then p r. Converse statement B.A syllogism C.Contrapositive statement D.Inverse statement?
Converse: If p r then p q and q rContrapositive: If not p r then not (p q and q r) = If not p r then not p q or not q r Inverse: If not p q and q r then not p r = If not p q or not q r then not p r
What is the sum or difference of p and q?
The sum of p and q means (p+q). The difference of p and q means (p-q).
What is the truth table for p arrow q?
Not sure I can do a table here but: P True, Q True then P -> Q True P True, Q False then P -> Q False P False, Q True then P -> Q True P False, Q False then P -> Q True It is the same as not(P) OR Q
What is qĀ²-pĀ² divided by q-p?
q + p
If p is 50 of q then what percent of p is q?
If p = 50 of q then q is 2% of p.
Is the square root of 6 an irrational number?
Yes it is. The proof is as follows:We prove the statement by contradiction i.e. Assume that sqrt(6) is a rational number.Then there exist positive integers p and q with gcd(p,q) = 1 such that p/q = sqrt(6).Square both sides: p^2 / q^2 = 6,p^2 = 6q^2.Now as 2 is a divisor of the right-hand side (RHS), it implies that 2 is also a divisor of the left-hand side (LHS).This is only possible if 2 is a factor of p.Let p =2k. Then k is a positive integer as well.Thus, 4k^2 = 6q^2,2k^2 = 3q^2.As 2 is a factor of the LHS, 2 is also a factor of the RHS.But this is only possible if 2 is a factor of q.=> gcd(p,q) >= 2. Contradiction!Thus sqrt(6) is irrational.yes it is
How do you write the negation of if and then?
If p then q is represented as p -> q Negation of "if p then q" is represented as ~(p -> q)
How do you show rootover 3 is not rational by contradiction method?
The proof is by the method of reductio ad absurdum. We start by assuming that sqrt(3) is rational. That means that it can be expressed in the form p/q where p and q are co-prime integers. Thus sqrt(3) = p/q. This can be simplified to 3*q^2 = p^2 Now 3 divides the left hand side (LHS) so it must divide the right hand side (RHS). That is, 3 must divide p^2 and since 3 is a prime, 3 must divide p. That is p = 3*r for some integer r. Then substituting for p gives, 3*q^2 = (3*r)^2 = 49*r^2 Dividing both sides by 3 gives q^2 = 3*r^2. But now 3 divides the RHS so it must divide the LHS. That is, 3 must divide q^2 and since 3 is a prime, 3 must divide q. But then we have 3 dividing p as well as q which contradict the requirement that p and q are co-prime. The contradiction implies that sqrt(3) cannot be rational.
Simplify pp - q - q q - p?
p-q
What does p over q mean in algebra?
P! / q!(p-q)!
