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The Timpson Lyceum · College Algebra

Lecture 6 (College Algebra)

Overview

This lecture builds the working vocabulary of the real number system and the basic rules for manipulating numbers and fractions. It defines what it means for a number to be rational or irrational, names the standard number sets and how they nest inside one another, and gives precise definitions of additive inverse, subtraction, division, and absolute value. From there it develops the sign rules for multiplication and addition, the procedure for collecting like terms, the construction of the real number line, and the meaning of the inequality symbols. The second half states the rules for fraction equality, multiplication, addition, simplification, and division, and proves several of them. Mastering these definitions is the groundwork for every algebraic manipulation that follows.

Key Concepts

Rational number. A real number xx is rational if there exist integers p,qp, q with q≠0q \neq 0 such that x=pq.x = \frac{p}{q}. Any integer nn is rational because n=n1n = \tfrac{n}{1}; for example 17=17117 = \tfrac{17}{1} and 0=010 = \tfrac{0}{1}. A number can be written as a ratio of integers in many ways (e.g. 139=2618\tfrac{13}{9} = \tfrac{26}{18}).

Irrational number. A real number is irrational if it is not rational — it cannot be written as a quotient of two integers. Examples: π\pi, 2\sqrt{2}, 23\tfrac{\sqrt{2}}{\sqrt{3}}.

The number sets and their nesting.

Every natural is an integer, every integer is rational, and every rational is real: N⊂Z⊂Q⊂R.\mathbb{N} \subset \mathbb{Z} \subset \mathbb{Q} \subset \mathbb{R}. The symbol ⊂\subset ("is a subset of") means every element of the smaller set also belongs to the larger one; this can be pictured with nested circles (a Venn-style diagram).

Additive inverse. Given x∈Rx \in \mathbb{R}, the number yy is the additive inverse of xx if x+y=0x + y = 0 (where 00 is the additive identity). There is exactly one such yy, written −x-x. So −x-x is shorthand for "the number you add to xx to get 00." If x=−17x = -17, then −x=−(−17)=17-x = -(-17) = 17.

Division. To divide bb by cc (with c≠0c \neq 0) means to find the number xx with b=cxb = cx. Here bb is the dividend, cc the divisor, and xx the quotient. Writing x=bcx = \tfrac{b}{c} is exactly the same statement as b=cxb = cx, so you can move freely between the two forms. A negative divided by a negative is positive (e.g. −25−5=5\tfrac{-25}{-5} = 5).

Subtraction. To subtract cc from bb means to find xx with b=x+cb = x + c; equivalently x=b+(−c)x = b + (-c). Thus subtracting a number is the same as adding its additive inverse.

Absolute value. The absolute value of xx is ∣x∣={xif x≥0,−xif x<0.|x| = \begin{cases} x & \text{if } x \ge 0, \\ -x & \text{if } x < 0. \end{cases} When xx is negative, −x-x is the additive inverse of a negative number, hence positive — it is not a negative result. For example ∣−19∣=−(−19)=19|-19| = -(-19) = 19.

Sign rules for multiplication. When multiplying several factors, multiply the magnitudes and then count the negative factors: an odd number of negative factors gives a negative product, an even number gives a positive product. For instance (−1)(−3)(−7)(−2)(−1)(-1)(-3)(-7)(-2)(-1) has five negatives, so the product 4242 comes out negative: −42-42.

Adding signed numbers. Same signs: add the magnitudes and keep the common sign, e.g. (−4)+(−12)=−16(-4) + (-12) = -16. Different signs: subtract the smaller magnitude from the larger and keep the sign of the larger, e.g. (−12)+4=−8(-12) + 4 = -8.

Zero-product property. ab=0ab = 0 if and only if a=0a = 0 or b=0b = 0. Two nonzero numbers can never multiply to zero.

Factoring out a negative. Placing a minus sign in front of a grouping changes the sign of every term inside (only at that outermost level — it does not recurse into inner groupings). For example −12+4+3=−(12−4−3)-12 + 4 + 3 = -(12 - 4 - 3).

Coefficients and collecting like terms. The coefficient of a term is the number multiplying its variable part (a missing coefficient is an understood 11). To collect like terms, add the numeral coefficients and multiply by the common factor: −9ab+ab=(−9+1)ab=−8ab-9ab + ab = (-9 + 1)ab = -8ab.

The real number line. Choose an arbitrary point as the origin and label it 00; choose an arbitrary length as the unit; then mark off units in both directions to place 1,2,3,…1, 2, 3, \dots to the right and −1,−2,−3,…-1, -2, -3, \dots to the left. Every real number corresponds to exactly one point on the line. A numerical magnitude is meaningless without a unit of reference.

Inequalities. a<ba < b (equivalently b>ab > a) means b−ab - a is positive. So 4<64 < 6 because 6−4=2>06 - 4 = 2 > 0, and −3>−7-3 > -7 because −3−(−7)=4>0-3 - (-7) = 4 > 0. Consequences: x<0x < 0 exactly when xx is negative, and x>0x > 0 exactly when xx is positive. The relation x≤yx \le y means x<yx < y or x=yx = y. Note that 00 is neither positive nor negative, so 0<00 < 0 is false but 0≤00 \le 0 is true. On the line, the number farther to the right is the greater one.

Equality of fractions. With b,d≠0b, d \neq 0, ab=cd  ⟺  ad=bc.\frac{a}{b} = \frac{c}{d} \iff ad = bc. For example 35=1220\tfrac{3}{5} = \tfrac{12}{20} because 3⋅20=5⋅12=603 \cdot 20 = 5 \cdot 12 = 60.

Multiplying and adding fractions. ab⋅cd=acbd,ab+cd=ad+bcbd.\frac{a}{b} \cdot \frac{c}{d} = \frac{ac}{bd}, \qquad \frac{a}{b} + \frac{c}{d} = \frac{ad + bc}{bd}.

Building up / reducing a fraction. Multiplying (or dividing) numerator and denominator by the same nonzero kk leaves the value unchanged: ab=kakb.\frac{a}{b} = \frac{ka}{kb}.

Dividing fractions. Dividing by a fraction is multiplying by its reciprocal: a/bc/d=ab⋅dc=adbc.\frac{a/b}{c/d} = \frac{a}{b} \cdot \frac{d}{c} = \frac{ad}{bc}.

Worked Example

Show that building up a fraction by a common factor preserves its value: for real a,b,ka, b, k with b≠0b \neq 0 and k≠0k \neq 0, ab=kakb.\frac{a}{b} = \frac{ka}{kb}.

By the cross-multiplication test, ab=kakb\tfrac{a}{b} = \tfrac{ka}{kb} holds precisely when a⋅(kb)=b⋅(ka)a \cdot (kb) = b \cdot (ka). Verify that equality using commutativity and associativity of multiplication:

a⋅(kb)=b⋅(ka)akb=bkaabk=abk.✓\begin{aligned} a \cdot (kb) &= b \cdot (ka) \\ a k b &= b k a \\ abk &= abk. \quad\checkmark \end{aligned}

Since a(kb)=b(ka)a(kb) = b(ka) is a true statement, the equality-of-fractions definition gives ab=kakb\tfrac{a}{b} = \tfrac{ka}{kb}.

As a numerical check with a=5a = 5, b=7b = 7, k=3k = 3:

57=?3⋅53⋅7=1521,5⋅21=105=7⋅15.✓\begin{aligned} \frac{5}{7} &\overset{?}{=} \frac{3\cdot 5}{3\cdot 7} = \frac{15}{21}, \\ 5 \cdot 21 &= 105 = 7 \cdot 15. \quad\checkmark \end{aligned}

Summary