The real numbers are built up in stages, each solving a problem the previous set could not: N={1,2,3,…} (counting) ⊂W={0,1,2,…} (adds zero) ⊂Z={…,−1,0,1,…} (adds negatives, for debts) ⊂Q={m/n:m,n∈Z,n=0} (adds ratios, for even division) ⊂R (fills in every remaining point of the number line, including the irrationals).
A rational number's decimal expansion always terminates or eventually repeats; conversely, any terminating or repeating decimal is rational. Numbers whose decimals neither terminate nor repeat are irrational, denoted Q′, with R=Q∪Q′ and Q∩Q′=∅.
Proving a number irrational (by contradiction). To show 2∈/Q: assume 2=m/n in lowest terms; squaring gives m2=2n2, so m is even; writing m=2k gives n2=2k2, so n is even too -- contradicting that m,n share no common factor. The identical technique proves 3, 5, π, etc. irrational.
Density. Between any two distinct rationals there is always another rational (e.g. their average); yet the rationals still leave gaps on the number line -- the diagonal of a unit square, 2, is a point no rational reaches.
Combining irrationals. Sums, differences, and products of irrational numbers are NOT automatically irrational: (1+2)−2=1∈Q, and 2×8=4∈Q -- the irrational parts can cancel.
Note
N⊂W⊂Z⊂Q⊂R, and the real-number field/order axioms (closure, associativity, distributivity, trichotomy, and 'multiplying an inequality by a negative number flips it') underlie every algebraic manipulation in this chapter.
Test each number for being a perfect square, a terminating/repeating decimal, or a whole number.
✓Final answer
7∈R−Q; −41∈Q; 0∈Z,Q; 3.14∈Q; 4∈N,Z,Q; 722∈Q.
Step 1.7: since 7 is not a perfect square, 7 is irrational, so 7∈R−Q.
Step 2.−41: this is a ratio of integers with nonzero denominator, so −41∈Q.
Step 3.0: an integer (and hence rational), so 0∈Z and 0∈Q.
Step 4.3.14: a terminating decimal, hence a ratio 100314, so 3.14∈Q.
Step 5.4: a positive whole number used for counting, so 4∈N; it is also in Z and Q.
Step 6.722: already a ratio of integers with nonzero denominator, so 722∈Q (note: this is only an approximation to π, not π itself, so it stays rational).