Solutions for Session 1, Part C

See solutions for Problems: C1 | C2 | C3 | C4 | C5 | C6| C7

Problem C1

 a. Yes, the set of counting numbers is closed for addition. b. Yes, the set of counting numbers is closed for multiplication.

 Problem C2 We must include 0 (to subtract things like 4 - 4) and negative integers (to subtract things like 23 - 831).

Problem C3

 a. We must include all fractional numbers of the form p/q, where p and q are integers (positive, negative, or zero counting numbers), with the restriction that q cannot be 0 (dividing by 0 is not defined). For example, we will need numbers like 5/2 and 82/7 and -1/2. b. Take a division problem like 53 = r. This is the equivalent to saying, "What number multiplied by 3 gives us 5?" The equation for this is 3 • r = 5. To solve this equation, we must isolate r on one side of it. Doing this requires dividing by 3 or multiplying 3 by its multiplicative inverse. The multiplicative inverse of 3 is usually written as 1/3. Multiplying both sides by 1/3 produces the following: 1/3 • (3 • r) = 1/3 • 5 (1/3 • 3) • r = 1/3 • 5 1 • r = 1/3 • 5 r = 5/3 c. No. If y is the multiplicative inverse of 0, then y • 0 = 1. But every real number multiplied by 0 equals 0, so y cannot be a real number -- and there is no multiplicative inverse for 0. That's why we can't divide by 0.

 Problem C4 Counting numbers are not dense. There is no counting number between 2 and 3. The integers are not dense either. However, we can always find a rational number between any two given rational numbers; for example, the average of any two fractions must always be a fraction between the two given fractions. Therefore, rational numbers are dense. One rational number between 2.5 and 2.6 is 2.55. One rational number between 2.55 and 2.6 is 2.555.

Problem C5

 a. Some major examples include raising a number to a power (exponentiation) and its inverse function (taking roots, such as square or cube roots), working with circles (and the number , approximately 3.141593), and solving equations with exponents (such as 2x = 3). b. Such operations produce irrational numbers, like , , or e (the base of natural logarithms; e is a mathematical constant approximately equal to 2.7183). Roots such as and are algebraic irrationals since they can be solutions to polynomial equations: numbers such as and e are called transcendental irrationals since they cannot be solutions to polynomial equations. Other equations, like x2 = -1, do not have a solution on the number line at all; this solution would be an imaginary number.

Problem C6

 a. No. Proving this is actually pretty difficult, but for the to be rational, we would have to be able to write it as p/q in reduced form, where p and q are integers that are relatively prime. This would mean that p and q are solutions to the equation p2 = 2q2, which cannot be solved if p and q can only be counting numbers. b. The length of the is the hypotenuse of a right triangle whose legs are 1 and 1 (i.e., x2 = 12 + 12 or x2 = 2 or x = ). So we know it has a "physical" distance and therefore can be located on the number line. This is about 1.414, but no decimal could ever express the exactly.

 Problem C7 Each is on the number line some specific distance from 0 (since each number is a constant). As with the , the distance cannot be expressed as a terminating or repeating decimal. is approximately 3.141593, while e is approximately 2.7183.

Problem C8

 a. The real numbers could be represented as the horizontal axis (similar to the number line). All real numbers, like 2, 1/2, -3, and e, would be on this line. b. The pure imaginary numbers, like 2i, (1/2)i, -3i, and ei could be represented as the vertical axis. The coordinates of a real number are (x,0), where x is the real part. The coordinates of a pure imaginary number are (0,yi), where yi is the imaginary part.