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To browse Academia. Maths questions and solutions pdf reader to main content. Log In Sign Up. Download Free PDF. Ian Seepersad. Download PDF. A short summary of this paper. Section 1. Note that we were able to incorporate the parentheses by using the words maths questions and solutions pdf reader and.

This has been slightly reworded so that the tenses make more sense. Contrapositive: If I do not stay at home, then it will not snow tonight. Inverse: If it does not snow tonight, then I will not stay home. Contrapositive: Whenever I do not go to the beach, it is not a sunny summer day. Inverse: Whenever it is not a sunny day, I do not go to the beach. Contrapositive: If I do not sleep until noon, then I did not stay up late.

A truth table will need 2n rows if there are n variables. To construct the truth table maths questions and solutions pdf reader a compound proposition, we work from the inside. In each case, we will show the intermediate steps. For parts a and b we have the following table column three for part acolumn four for part b.

For parts a and b we have the following table column two for part acolumn four for part b. This time we have omitted the column explicitly showing the negation of q. It is irrelevant that the condition is now false. This cannot be a proposition, because it cannot have a truth value.

Indeed, if it were true, then it would be truly asserting that it is false, a contradiction; on the other hand if it were false, then its assertion that it is false must be false, so that it would be true�again a contradiction. Thus this string of letters, while appearing to be a proposition, is in fact meaningless.

This maths questions and solutions pdf reader a classical paradox. We will use the male pronoun in what follows, assuming that we are talking about males shaving their beards here, and assuming that all men have facial hair. If we restrict ourselves to beards and allow female barbers, then the barber could be female with no contradiction.

If such a barber existed, who would shave the barber? If the barber shaved himself, then he would be violating the rule that he shaves only those people who do not shave themselves. On the other hand, if he does not shave himself, then the rule says that he must shave. Neither is possible, so there can be no such barber. Note that we can make all the conclusion true by making a false, s true, and u false.

Thus the system is consistent. This system is consistent. This requires that both L and Q be true, by the two conditional statements that have B as their consequence. Note that there is just this one satisfying truth assignment. This is similar to Example 17, maths questions and solutions pdf reader universities in New Mexico.

If A is a knight, then his statement that both of them are knights is true, and both will be telling the truth. But that is impossible, because B is asserting otherwise that A is a knave. Thus we conclude that A is a knave and B is a knight. We can draw no conclusions.

A knight will declare himself to be a knight, telling the truth. A knave will lie and assert that he is a knight. If Smith and Jones are innocent and therefore telling the truththen we get an immediate contradiction, since Smith said that Jones was a friend of Cooper, but Jones said that he did not even know Cooper. If Jones and Williams are the innocent truth-tellers, then we again get a contradiction, since Jones says that he did not know Cooper and was out of town, but Williams says he saw Jones with Cooper presumably in town, and presumably if we was with him, then he knew.

Therefore it must be the case that Smith and Williams are telling the truth. Their statements do not contradict each. Therefore Jones is the murderer. Can none of them be guilty?

If so, then they are all telling the truth, but this is impossible, because as we just saw, some of the statements are contradictory.

Can more than one of them be guilty? If, for example, they are all guilty, then their statements give us no information. So that is certainly possible. This information is enough to determine the entire. Let each maths questions and solutions pdf reader stand for the statement maths questions and solutions pdf reader the person whose name begins with that letter is chatting. Note that we were able to convert all of these statements into conditional statements.

In what follows Maths Vector Questions And Solutions And Pdf we will sometimes make use of the contrapositives of these conditional statements as. First suppose that H is true. Then it follows that A and K are true, whence it follows that R and V are true. But R implies that V is false, so we get a contradiction. Therefore H must be false. From this it follows that K is true; whence V is true, and therefore R is false, as is A. We can now check that this assignment leads to a true value for each conditional statement.

There are four cases to consider. If Alice is the sole truth-teller, maths questions and solutions pdf reader Carlos did it; but this means that John is telling the truth, a contradiction. If John is the sole truth-teller, then Diana must be lying, so she did it, but then Carlos is telling the truth, a contradiction.

If Carlos is the sole truth-teller, then Diana did it, but that makes John truthful, again a contradiction. So the only possibility is that Maths questions and solutions pdf reader is the sole truth-teller. This means that John is lying when he denied it, so he did it. Note that in this case both Alice and Carlos are indeed lying. Since Carlos and Diana are making contradictory statements, the liar must be one of them we could have used this approach in part a as.

Therefore Alice is telling the truth, so Carlos did it. Note that John and Diana are telling the truth as well here, and it is Carlos who is lying.

There are two Maths Questions And Solutions Pdf 60 cases. Therefore the two propositions are logically equivalent. We see that the fourth and seventh columns are identical. For part a we have the following table. We argue directly by showing that if the hypothesis is true, then so is the conclusion.

An alternative approach, which we show only for part ais to use the equivalences listed in the section and work symbolically. Then p is false. To do this, we need only show that if p is true, then r is true.

Suppose p is true. It now follows from the second part of the hypothesis that r is true, as desired. Then p is true, and since the second part of the hypothesis is true, we conclude that q is also true, as desired. If p is true, then the second part of the hypothesis tells us that r is true; similarly, if q is true, then the third part of the hypothesis tells us that r is true.

Thus in either case we conclude that r is true. This is not a tautology. It is saying that knowing that the hypothesis of an conditional statement is false allows us to conclude that the conclusion is also false, and we know that this is maths questions and solutions pdf reader valid reasoning.

Since this is possible only if the conclusion if false, we want to let q be true; and since we want the hypothesis to be true, we must also let p be false. It is easy to check that if, indeed, p is maths questions and solutions pdf reader and q is true, then the conditional statement is false. Therefore it is not a tautology. The second is true if and only if either p and q are both true, or p and q are both maths questions and solutions pdf reader. Clearly these two conditions are saying the same thing.

We determine exactly which rows of the truth table will have T as their entries. The conditional statement will be true if p is false, or if q in one case or r in the other case is true, i. Since the two propositions are true in exactly the same situations, maths questions and solutions pdf reader are logically equivalent. But these are equivalent by the commutative and associative laws.

An conditional statement in which the conclusion is true or the hypothesis is false is true, and that completes the argument.

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