1. The Real-Number System
Section R.1
The Real-Number System

2. R.1 The Real-Number System
Identify various kinds of real numbers.
Use interval notation to write a set of numbers.
Identify the properties of real numbers.
Find the absolute value of a real number.

3. Rational Numbers
Numbers that can be expressed in the form p/q, where p and q are integers and q  0. Decimal notation for rational numbers either terminates (ends) or repeats.
Examples:
a) 0 d)
b) 7 e) c)

4. Irrational Numbers
The real numbers that are not rational are irrational numbers. Decimal notation for irrational numbers neither terminates nor repeats.
Examples: a) b) c)

6. Examples Write interval notation for each set and graph the set.
a) {x|5 < x < 2}
Solution: {x|5 < x < 2} = (5, 2)
( )
b) {x|4 < x  3}
Solution: {x|4 < x  3} = (4, 3]
( ]

7. Properties of the Real Numbers
Commutative property of addition and multiplication
a + b = b + a and ab = ba
Associative property
a + (b + c) = (a + b) + c and
a(bc) = (ab)c
Additive identity property
a + 0 = 0 + a = a
Additive inverse property
a + a = a + (a) = 0

8. More Properties of Real Numbers
Multiplicative identity property
a • 1 = 1 • a = a
Multiplicative inverse property
Distributive property
a(b + c) = ab + ac

9. Examples State the property being illustrated in each sentence.
a) Commutative Property of Multiplication
b) Associative property of addition
c) Additive inverse property
d) Multiplicative identity property
e) Distributive property

10. Absolute Value
The absolute value of a number a, denoted |a|, is its distance from 0 on the number line. For example,
|−5| = 5

11. Absolute Value For any real number a,
When a is nonnegative, the absolute value of a is a.
When a is negative, the absolute value of a is the opposite, or additive inverse, of a.
Thus, |a| is never negative; that is, for any real number a, |a| ≥ 0.

12. Distance Between Two Points on the Number Line
For any real numbers a and b, the distance between a and b is |a  b|, or equivalently |b  a|.
Example:
Find the distance between 2 and 3.
Solution:
The distance is |2  3| = |5| = 5,
or |3 (2)| = |3 + 2| = |5| = 5.