Download discrete mathematics and its applications 7th

Thus both sides of the logical equivalence are true hence equivalent. Now suppose that A is false. If P x is true for all x , then the left-hand side is true. On the other hand, if P x is false for some x, then both sides are false. Therefore again the two sides are logically equivalent. If P x is true for at least one x, then the left-hand side is true. On the other hand, if P x is false for all x , then both sides are false. If A is false, then both sides of the equivalence are true, because a conditional statement with a false hypothesis is true.

If A is false, then both sides of the equivalence are true, because a conditional statement with a false hypothesis is true and we are assuming that the domain is nonempty. It is saying that one of the two predicates, P or Q , is universally true; whereas the second proposition is simply saying that for every x either P x or Q x holds, but which it is may well depend on x.

As a simple counterexample, let P x be the statement that x is odd, and let Q x be the statement that x is even. Let the domain of discourse be the positive integers. The second proposition is true, since every positive integer is either odd or even. P x is true, so we form the disjunction of these three cases. So the response is no. So the response is yes. Following the idea and syntax of Example 28, we have the following rule: The unsatisfactory excuse guaranteed by part b cannot be a clear explanation by part a. If x is one of my poultry, then he is a duck by part c , hence not willing to waltz part a.

Or, more simply, a nonnegative number minus a negative number is positive which is true. The answers to this exercise are not unique; there are many ways of expressing the same propositions sym- bolically.

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Note that C x, y and C y, x say the same thing. Our domain of discourse for persons here consists of people in this class. We need to make up a predicate in each case. We let P s, c, m be the statement that student s has class standing c and is majoring in m. The variable s ranges over students in the class, the variable c ranges over the four class standings, and the variable m ranges over all possible majors. It is true from the given information. This is false, since there are some mathematics majors. This is true, since there is a sophomore majoring in computer science.

This is false, since there is a freshman mathematics major.

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This is false. It cannot be that m is mathematics, since there is no senior mathematics major, and it cannot be that m is computer science, since there is no freshman computer science major. Nor, of course, can m be any other major. The best explanation is to assert that a certain universal conditional statement is not true. We need to use the transformations shown in Table 2 of Section 1. The logical expression is asserting that the domain consists of at most two members. It is saying that whenever you have two unequal objects, any object has to be one of those two.

Note that this is vacuously true for domains with one element. Therefore any domain having one or two members will make it true such as the female members of the United States Supreme Court in , and any domain with more than two members will make it false such as all members of the United States Supreme Court in In each case we need to specify some predicates and identify the domain of discourse. In English, everybody in this class has either chatted with no one else or has chatted with two or more others.

In English, some student in this class has sent e-mail to exactly two other students in this class. In English, for every student in this class, there is some exercise that he or she has not solved. Word order in English sometimes makes for a little ambiguity. In English, some student has solved at least one exercise in every section of this book. This x provides a counterexample. The domain here is all real numbers. This statement says that there is a number that is less than or equal to all squares.

We need to show that each of these propositions implies the other. By our hypothesis, one of two things must be true. Either P is universally true, or Q is universally true. Next we need to prove the converse. Otherwise, P x0 must be false for some x0 in the domain of discourse.

Since P x0 is false, it must be the case that Q y is true for each y.

[Solution] Discrete Mathematics and It's Application by Kenneth H. Rosen (7th Edition)~Instructbd

Logic and Proofs c First we rewrite this using Table 7 in Section 1. This is modus tollens. Modus tollens is valid. This is, according to Table 1, disjunctive syllogism. See Table 1 for the other parts of this exercise as well. We want to conclude r. We set up the proof in two columns, with reasons, as in Example 6.

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Note that it is valid to replace subexpressions by other expressions logically equivalent to them. Step Reason 1. Alternatively, we could apply modus tollens. Another application of modus tollens then tells us that I did not play hockey. We could say using existential generalization that, for example, there exists a non-six-legged creature that eats a six-legged creature, and that there exists a non-insect that eats an insect. Now modus tollens tells us that Homer is not a student.

There are no conclusions to be drawn about Maggie. Universal instantiation and modus ponens therefore tell us that tofu does not taste good. The third sentence says that if you eat x, then x tastes good. No conclusions can be drawn about cheeseburgers from these statements. Therefore by modus ponens we know that I see elephants running down the road. In each case we set up the proof in two columns, with reasons, as in Example 6.

In what follows y represents an arbitrary person. After applying universal instantiation, it contains the fallacy of denying the hypothesis. We know that some s exists that makes S s, Max true, but we cannot conclude that Max is one such s. We will give an argument establishing the conclusion. We want to show that all hummingbirds are small. Let Tweety be an arbitrary hummingbird.


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We must show that Tweety is small. Therefore by universal modus ponens we can conclude that Tweety is richly colored. The third premise implies that if Tweety does not live on honey, then Tweety is not richly colored. Therefore by universal modus tollens we can now conclude that Tweety does live on honey. Finally, the second premise implies that if Tweety is a large bird, then Tweety does not live on honey. Therefore again by universal modus tollens we can now conclude that Tweety is not a large bird, i. Notice that we invoke universal generalization as the last step.

Thus we want to show that if P a is true for a particu- lar a, then R a is also true. The right-hand side is equivalent to F. Let us use the following letters to stand for the relevant propositions: As we noted above, the answer is yes, this conclusion is valid. This conditional statement fails in the case in which s is true and e is false. If we take d to be true as well, then both of our assumptions are true. Therefore this conclusion is not valid.

This does not follow from our assumptions. If we take d to be false, e to be true, and s to be false, then this proposition is false but our assumptions are true. We noted above that this validly follows from our assumptions. The only case in which this is false is when s is false and both e and d are true. Therefore, in all cases in which the assumptions hold, this statement holds as well, so it is a valid conclusion.

We must show that whenever we have two even integers, their sum is even.

Rosen Discrete mathematics Book Review - Discrete Mathematics and Its Applications

Suppose that a and b are two even integers. We must show that whenever we have an even integer, its negative is even. Suppose that a is an even integer. This is true. We give a proof by contradiction. By Exercise 26, the product is rational. We give a proof by contraposition. If it is not true than m is even or n is even, then m and n are both odd. By Exercise 6, this tells us that mn is odd, and our proof is complete.

Assume that n is odd. But this is obviously not true. Therefore our supposition was wrong, and the proof by contradiction is complete. Therefore the conditional statement is true. This is an example of a trivial proof, since we merely showed that the conclusion was true. Then we drew at most one of each color. This accounts for only two socks. But we are drawing three socks. Therefore our supposition that we did not get a pair of blue socks or a pair of black socks is incorrect, and our proof is complete.

Since we have chosen 25 days, at least three of them must fall in the same month. Since n is even, it can be written as 2k for some integer k. This is 2 times an integer, so it is even, as desired. So suppose that n is not even, i. This is 1 more than 2 times an integer, so it is odd. That completes the proof by contraposition. There are two things to prove. Now the only way that a product of two numbers can be zero is if one of them is zero. We write these in symbols: It is now clear that all three statements are equivalent. We give direct proofs that i implies ii , that ii implies iii , and that iii implies i.

These are therefore the only possible solutions, but we have no guarantee that they are solutions, since not all of our steps were reversible in particular, squaring both sides. Therefore we must substitute these values back into the original equation to determine whether they do indeed satisfy it. But these each follow with one or more intermediate steps: We claim that 7 is such a number in fact, it is the smallest such number.

The only squares that can be used to contribute to the sum are 0 , 1 , and 4. Thus 7 cannot be written as the sum of three squares. By Exercise 39, at least one of the sums must be greater than or equal to Example 1 showed that v implies i , and Example 8 showed that i implies v. The cubes that might go into the sum are 1 , 8 , 27 , 64 , , , , , and We must show that no two of these sum to a number on this list. Having exhausted the possibilities, we conclude that no cube less than is the sum of two cubes. There are three main cases, depending on which of the three numbers is smallest.


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In the second case, b is smallest or tied for smallest. Since one of the three has to be smallest we have taken care of all the cases. The number 1 has this property, since the only positive integer not exceeding 1 is 1 itself, and therefore the sum is 1. This is a constructive proof.

Therefore these two consecutive integers cannot both be perfect squares. This is a nonconstructive proof—we do not know which of them meets the requirement. In fact, a computer algebra system will tell us that neither of them is a perfect square. Of these three numbers, at least two must have the same sign both positive or both negative , since there are only two signs.

It is conceivable that some of them are zero, but we view zero as positive for the purposes of this problem. The product of two with the same sign is nonnegative. In fact, a computer algebra system will tell us that all three are positive, so all three products are positive. We know from algebra that the following equations are equivalent: This shows, constructively, what the unique solution of the given equation is.

Given r , let a be the closest integer to r less than r , and let b be the closest integer to r greater than r. In the notation to be introduced in Section 2. We follow the hint. This is clearly always true, and our proof is complete. This is impossible with an odd number of bits. Clearly only the last two digits of n contribute to the last two digits of n2. So we can compute 02 , 12 , 22 , 32 ,.

From that point on, the list repeats in reverse order as we take the squares from to , and then it all repeats again as we take the squares from to Thus our list which contains 22 numbers is complete. Clearly there are no integer solutions to these equations, so there are no solutions to the original equation. One proof that 3 2 is irrational is similar to the proof that 2 is irrational, given in Example 10 in Section 1. Thus p3 is even. Now we play the same game with q. Since q 3 is even, q must be even.

We have now concluded that p and q are both even, that is, that 2 is a common divisor of p and q. The solution is not unique, but here is one way to measure out four gallons. Fill the 5-gallon jug from the 8-gallon jug, leaving the contents 3, 5, 0 , where we are using the ordered triple to record the amount of water in the 8-gallon jug, the 5-gallon jug, and the 3-gallon jug, respectively.

Pour the contents of the 3-gallon jug back into the 8-gallon jug, leaving 6, 2, 0. Without loss of generality, we number the squares from 1 to 25, starting in the top row and proceeding left to right in each row; and we assume that squares 5 upper right corner , 21 lower left corner , and 25 lower right corner are the missing ones.

We argue that there is no way to cover the remaining squares with dominoes. By symmetry we can assume that there is a domino placed in using the obvious notation. If square 3 is covered by , then the following dominoes are forced in turn: Therefore we must use along with If we use all of , , and , then we are again quickly forced into a sequence of placements that lead to a contradiction.

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Therefore without loss of generality, we can assume that we use , which then forces , , , , , , and , and we are stuck once again. This completes the proof by contradiction that no placement is possible. The barriers shown in the diagram split the board into one continuous closed path of 64 squares, each adjacent to the next for example, start at the upper left corner, go all the way to the right, then all the way down, then all the way to the left, and then weave your way back up to the starting point.

Because each square in the path is adjacent to its neighbors, the colors alternate. Therefore, if we remove one black square and one white square, this closed path decomposes into two paths, each of which starts in one color and ends in the other color and therefore has even length.

Clearly each such path can be covered by dominoes by starting at one end. This completes the proof. Supplementary Exercises 31 Therefore the same argument as was used in Example 22 shows that we cannot tile the board using straight triominoes if any one of those other 60 squares is removed. The following drawing rotated as necessary shows that we can tile the board using straight triominoes if one of those four squares is removed.

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Textbook Solutions. Looking for the textbook? R CH1. SE CH1. R CH2. SE CH2. R CH3. SE CH3. R CH4. SE CH4. R CH5. SE CH5. R CH6. SE CH6. R CH7. SE CH7. R CH8. SE CH8. R CH9. SE CH9. R CH SE CH Sample Solution. JavaScript Not Detected. The objective is to prove that multiplicative identity element of the real numbers is unique. By the multiplicative identity law, for every real number for every real number. Comment 0. Use then Note that, Thus, have same multiplicative identity element, this implies that multiplicative identity element of the real numbers is unique.

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