



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. 
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Problem C2  
We must include 0 (to subtract things like 4  4) and negative integers (to subtract things like 23  831).
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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. 
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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.
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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 2^{x} = 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 x^{2} = 1, do not have a solution on the number line at all; this solution would be an imaginary number.

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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 p^{2} = 2q^{2}, 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., x^{2} = 1^{2} + 1^{2} or x^{2} = 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. 
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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.
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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. 
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