Example 2. Show that if x < y and p > 0, then px < py.

Solution: If x < y, then y - x is positive and, by closure, the product p(y - x) of positive numbers is positive; that is, py - px is positive. Now we have px < py by definition.

Example 3. Show that if m and n are integers and m < n, then 

Solution: Since m < n, it follows that n - m is positive. Since the difference of integers is an integer and the least positive integer is 1, n - m is at least 1; that is

Problems for Section 10.1

R 1. (a) Show that if x < y, then x + z < y + z.

(b) Show that if x < y, then x - w < y - w.

R 2. Show that if x < y and q < 0, then qx > qy.

R 3. (a) Show that if x > 0 or x < 0, then x² > 0.

(b) Show that for all real x, 

(c) Show that 1 > 0.

R 4. (a) Show that if x > 0, then and if x < 0, then 

(b) Show that if 0 < x < y or x < y < 0, then 

R 5. Show the following:

(a) If 0 < x < y and n is a positive integer, then x^2n-1 < y^2n-1.

(b) If x < 0 < y and n is a positive integer, then x^2n-1 < y^2n-1.

(c) If x < y < 0 and n is a positive integer, then x^2n-1 < y^2n-1.

R 6. (a) Show that if 0 < x < y and n is a positive integer, then x²ⁿ < y²ⁿ.

(b) Show that if y < x < 0 and n is a positive integer, then x²ⁿ < y²ⁿ.
 
 

Monday, June 22, 1998