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 largenegative x, this expression is negative; therefore it's not the case that it'salways 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 tominus x cubed—the answer is yes. Given any x, let y be negative x. If you cube a positive number, then you get apositive number. If you cube a negative number, then you get a negative number. Inall 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 theanswer would be yes. Namely, for every y value, there exists an x value suchthat x plus y equals 0 (namely when x = −y). This statement is false due to thefact that if we were to solve for x in terms of y (namely x = −y), then it wouldn'twork because if x is not positive then we cannot have y squared equal to it sothis 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 considersome examples: Let's consider the case where x is negative. If that were thecase, 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 asolution that satisfies the given condition. This is true because if given an x, there isalways a y that satisfies this equation. In fact, if the given value for x is negative, anynon-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 satisfythis condition. If you are given a negative value for x, there are no non-negative y'sthat satisfy y=x^2. However, there are solutions to this equation in positive numbers(ie., limits). In mathematics and in everyday life, one often finds oneself having to negate astatement involving quantifiers. For example, you can negate a statement by simplyputting a negation symbol in front, but that is not always sufficient. One mustproduce 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 oranother one in which any negation symbols are as far inside the statement aspossible without resulting in an unduly cumbersome expression. Let's look at our first example of negating a quantified statement. Let A(x) be someproperty 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). Forexample, it's not the case that all motorists run red lights, it is equivalent to there is amotorist who does not run red lights. Well, in this case, it's pretty obvious that thesetwo are equivalent. We begin with the left-to-right implication. If it is not the case that for all x A(x), thenat least one x fails to satisfy A(x). In symbols, there exists an x such that A(x) isfalse. This is the left-to-right implication. Now we want to prove the right-to-leftimplication. 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, orin 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 thisreasoning 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 tof x, should be fresh in your mind after this example. Now we can return to thenegation 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 isbadly made. With this notation, the sentence becomes 4x and C, D of x Implies m ofx. For all cars, if the car is domestic, then it's badly made. The negation becomes ifthere exists an x in c such that dx does not imply and here we've actuallyabbreviated the following. Not the case DX implies M of X but instead of writing notdx 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 xwith the negation of the condition inside. Just compare what's going on here with theprevious example. In this case, it's talking about Xs which are in c—the only Xs we'reinterested in are cars. But if the only objects we're interested in are cars, thenegation 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 holdwhile m fails, so we looked at this when we did truth tables. So the negation of theoriginal sentence, "There exists an x in C such that Dx" is equivalent to "For every xin 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 carwhich is domestic, and it is not badly made. Now, with the notation and method for dealing with these negations, you should beable to follow it quite easily.