Working with Quantifiers
Let’s see if we can find an answer to these problems. 1) Is it true that for every real number x, x+1 is greater than or equal to x? Yes, it is. 2) Is it true there is a real number x? Well, if we were talking about the rational numbers, the answer would be no because the square root of 2 is irrational. However, for the real numbers, it’s true. 3) What about this one? Well, for very large negative x, x cubed is an extremely large negative number which will dominate all of these numbers. So for large negative x, this expression is negative; therefore it’s not the case that it’s always positive. So that one’s false—so what about this one? 4) Is it true that for all x there is a y such that x cubed plus y cubed equals 0? X cubed plus y cubed equals 0—this would mean y cubed must be equal to minus x cubed—the answer is yes. Introduction to Mathematical Thinking QUIZ Which of the following are true and which false? Variables range over real numbers. T (a) forall x (x+1≥x) T (b) exists x (x^2+1=3) F (c) forall x (x^3+17^2+6x+11≥0) T (d) forall x exists x (x^3+y^3=0) F (e) exists x forall y (x+y=0) T (f) forall x exists y (x ≥ 0 Rightarrow y^2=x) Given any x, let y be negative x. If you cube a positive number, then you get anpositive number. If you cube a negative number, then you get a negative number. In all cases, y cubed will equal 0. Thus, x = y = 0 is always true. 5) What about this statement? Is there an x such that for every y value, we get 0? The answer is no because if the quantifiers were switched around then the answer would be yes. Namely, for every y value, there exists an x value suchnthat x plus y equals 0 (namely when x = −y). This statement is false due to the fact that if we were to solve for x in terms of y (namely x = −y), then it wouldn’t work because if x is not positive then we cannot have y squared equal to it so this statement is false. 6) What about this one? Is it true that for all values of x there exists a value of y such that if x is non-negative then y squared equals it? Well, let’s consider some examples: Let’s consider the case where x is negative. If that were the case, then this conditional would be true because it has a false antecedent. However, if x is not negative, then there is a y whose square is x. In this case, we can find the square root of x. So, whatever value we are looking for a solution to this equation, it must be a solution that satisfies the given condition. This is true because if given an x, there is always a y that satisfies this equation. In fact, if the given value for x is negative, any non-negative y will satisfy this condition because “y” squared equals negative “x”. However, if the given value for x is not negative, then any non-negative y will satisfy this condition. If you are given a negative value for x, there are no non-negative y’s that satisfy y=x^2. However, there are solutions to this equation in positive numbersn(ie., limits). In mathematics and in everyday life, one often finds oneself having to negate a statement involving quantifiers. For example, you can negate a statement by simply putting a negation symbol in front, but that is not always sufficient. One must produce a positive assertion—one which says what is rather than what is not. Roughly speaking, a positive statement is one that contains no negation symbols or another one in which any negation symbols are as far inside the statement as possible without resulting in an unduly cumbersome expression. Let’s look at our first example of negating a quantified statement. Let A(x) be some property of x. For example, x is a real root of the equation x squared + 2x + 1 = 0. Not every x has property A(x), so it’s not true that at least three x have A(x). For example, it’s not the case that all motorists run red lights, it is equivalent to there is a motorist who does not run red lights. Well, in this case, it’s pretty obvious that these two are equivalent. let A(x) be some property of x. (x is a real root of the equation x squared + 2x+1 = 0.) I’ll show that neg [forall x A (x)] is equivalent to exists x [neg A (x)]. “It’s not the case that all motorist run red lights” Leftrightarrow “There is a motorist who does not run red lights” Rightarrow Assume neg [forall x A (x)]. If it is not the case that for all x, A(x), then at least are x must fail to satisfy A(x). So for at least are x, neg A(x) is true. In symbols, exists x[neg A(x)]. Leftarrow Assume exists x[neg A(x)]. Then there is an x for which A(x) is false. Then A(x) cannot be true for all x. In other words, forall x A(x) must be false. In symbols, neg [forall x A(x)]. Try this one: Show that neg [exists x A(x)] Leftrightarrow forall x[neg A(x)]. want on everyday example. We begin with the left-to-right implication. If it is not the case that for all x A(x), then at least one x fails to satisfy A(x). In symbols, there exists an x such that A(x) is false. This is the left-to-right implication. Now we want to prove the right-to-left implication. Assume there exists an x such that A(x) is false, or in everyday English, assume there is an x for which f(x)=false. In this case, f(x) must be false for all x, ornin symbols, for all x, f(x) must be false. We have now proven both implications from “exists an x such that A(x) is false” to ” it is not the case that for all x A(x). Assume that there exists some x such that x is not a member of x. If you found this reasoning hard to follow, it’s because you are unfamiliar with the situation. The example of aerial warfare, where not exist in x of a of x is equivalent to 4 x, not t of x, should be fresh in your mind after this example. Now we can return to the negation of all domestic costs that are badly made and consider it as an example. The set C of all cars, D of x means that x is domestic, and M of x means that x is badly made. With this notation, the sentence becomes 4x and C, D of x Implies m of x. For all cars, if the car is domestic, then it’s badly made. The negation becomes if there exists an x in c such that dx does not imply and here we’ve actually abbreviated the following. Not the case DX implies M of X but instead of writing not dx implies m of x, I use this simple notation dx does not implies m of x. When you check for all x such that x is in C, and you negate it, you get the set of all x with the negation of the condition inside. Just compare what’s going on here with the previous example. In this case, it’s talking about Xs which are in c—the only Xs we’re interested in are cars. But if the only objects we’re interested in are cars, the negation will only be talking about those cars. So m c tells us which kinds of objects we’re dealing with, and we negate that. Okay, now let’s look at this part. We know that dx does not imply m of x if d of x can hold while m fails, so we looked at this when we did truth tables. So the negation of the original sentence, “There exists an x in C such that Dx” is equivalent to “For every x in C there exists a d such that dx”. This part is the same as this part, D of x and not m of x. It means that there is a car which is domestic, and it is not badly made. Now, with the notation and method for dealing with these negations, you should benable to follow it quite easily. “All domestic cost that are badly made” let C be the set of all case, D(x)means x is domestic, M(x) means x is badly made. (forall x in C) [D (x) rightarrow П (x)]. Negation is (exists x in C) [D (x) nRightarrow П (x)]. Why not (exists x in C)? neg [D (x) Rightarrow П (x)]. We know that [D(x) Rightarrow M(x) is equivalent to [D (x) Rightarrow M (x)]. So neg (forall x in C) [D (x) Rightarrow M (x) is equivalent (exists x in C) [D (x) Rightarrow M (x)]. There is a domestic case that is not badly made Now, with the notation and method for dealing with these negations, you should be able to follow it quite easily.
