By G. H. & Wright, E. M. Hardy

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**Sample text**

In the ﬁrst, we say that for each x we can ﬁnd a y with x + y 0. That’s a statement you have known to be true for years, ever since you learned about −x. On the other hand, the second statement says that there exists a y such that for all x, we have x + y 0. That statement is false, because the same y would have to work for all x. What’s the moral of this story? That the order of the quantiﬁers is very important. 5. 2. 46 4. 1). There are many possible answers. We list some below: (a) {1, −1}; (b) {1}; (c) the x-axis in R2 ; (d) the xy-plane in R3 ; (e) {2n : n ∈ Z} {.

G) If x is a nonzero real number, then x2 0. (h) If x is a nonzero real number, then there exists a real number y such that x · y 1. (i) If x and y are even integers, then x + y is an even integer. 11. Find a different useful description of the following: (a) {x ∈ R : x2 2}; (b) {(x, y) ∈ R2 : x y}; (c) {x ∈ N : x ≤ 0}; (d) {x ∈ Z : x2 > 0}. 12. Write each of the following in set notation. (a) The set of all odd integers. (b) The set of all points in the xy-plane above the line y x. 50 4. Set Notation and Quantiﬁers (c) The set of all points in the xy-plane that are inside the circle of radius one.

3. Let x be an integer. If x2 is odd, then x is odd. First we need to understand the problem. What does it mean for a number x to be odd? It means that there is an integer n such that x 2n + 1. So we are assuming that x2 2n + 1 for some integer n and trying to show x 2m + 1 for some integer m. It’s hard to see where to go from here, we think. Remember that P´olya suggests restating the problem, so let’s try that. ” Then we see that we wish to prove that P → Q is true. ” We can do better than that, since an integer is either odd or even.