 |
|
|
|
|
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. |
<< back to Problem C1
|
|
| |
 |
Problem C2 | |
|
We must include 0 (to subtract things like 4 - 4) and negative integers (to subtract things like 23 - 831).

<< back to Problem C2
|
|
|
| |
 |
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 5 3 = 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. |
<< back to Problem C3
|
|
|
| |
 |
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.
<< back to Problem C4
|
|
|
| |
 |
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.

|
<< back to Problem C5
|
|
|
| |
 |
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. |
<< back to Problem C6
|
|
|
| |
 |
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.
<< back to Problem C7
|
|
|
| |
 |
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. |
<< back to Problem C8
|
|
|