One test: can it be written as one integer over another?
Which one is rational?
Which of these is a rational number?
227 is a ratio of two integers, so it is rational by definition — it merely approximates π, it is not π. √20 does not reduce to a whole number, π is irrational, and B follows a pattern but never repeats, which is what rationality requires.
pq
p and q integers, q ≠ 0
no such fraction exists
π, e, √2, √20
Every integer is rational — 7 is 71. So are all fractions, all mixed numbers, and every decimal that stops or repeats.
Together the two kinds make up the real numbers, and there is no overlap: a number is one or the other, never both.
| Decimal | Behaviour | Verdict |
|---|---|---|
| 0.375 | terminates | rational — 38 |
| 0.3333… | repeats | rational — 13 |
| 0.142857142857… | repeats in blocks of six | rational — 17 |
| 0.101001000100001… | a pattern, but never repeating | irrational |
| 3.14159265… | never stops, never repeats | irrational — π |
0.101001000100001… follows a rule — the run of zeros keeps growing — but no fixed block of digits cycles forever, so it cannot be written as a fraction. Exam items love this one because it looks orderly.
A radical sign does not make a number irrational. Simplify first, then judge — the question is whether the root comes out exact.
1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225 — and the cubes 1, 8, 27, 64, 125, 216. Anything else under the root sign is irrational.
22 ÷ 7 = 3.142857142857…
repeats, so rational
π = 3.14159265358979…
never repeats, irrational
they differ from the third decimal
227 is only an approximation
Being close to π does not make a number irrational, and no fraction or finite decimal will ever equal π. A question asking which of the listed numbers is irrational is often testing exactly this.
| Combination | Result | Example |
|---|---|---|
| rational + rational | always rational | ½ + ⅓ = ⅚ |
| rational + irrational | always irrational | 2 + √3 |
| irrational + irrational | either | √2 + (−√2) = 0 |
| irrational × irrational | either | √2 · √2 = 2 |
| irrational × non-zero rational | always irrational | 3√5 |
So a claim like “the sum of two irrationals is irrational” is false, and the counterexample is one line: √2 + (2 − √2) = 2.
| Set | Contains | New arrivals |
|---|---|---|
| natural | 1, 2, 3, … | counting numbers |
| whole | 0, 1, 2, 3, … | zero |
| integers | …, −2, −1, 0, 1, 2, … | negatives |
| rational | all the above, plus fractions | ⅔, −1.5, 0.3̄ |
| real | rational and irrational together | π, √2 |
Each set sits inside the next. So −4 is an integer and rational and real — “which of these describes the number” usually wants all the labels that fit.
A block that repeats from the decimal point goes over that many nines. If something comes before the repeat, split it off first.
0.5454… = 5499 = 611
two-digit block → 99
2.1666… = 2.1 + 0.0666…
only the 6 repeats
0.0666… = (110)(0.666…) = (110)(69) = 115
one place later, so ÷ 10
2110 + 115 = 6330 + 230 = 136
check: 13 ÷ 6 = 2.1666… ✓
0.54 = 54100 stops; 0.5454… = 5499 repeats. And 2.1666… is not 21699 — that would be 2.1818…, where the block 18 repeats. Read exactly which digits carry the dots.
Roots of decimals: √0.04 and √0.4.
√0.04 = √4100 = 210 = 0.2
4 and 100 are both squares: rational
√0.4 = √40100 = √40 ÷ 10
40 is not a square
= 2√10 ÷ 10 ≈ 0.632 — irrational
the root does not come out
What is different: write the decimal as a fraction over a power of 100 before judging. Reading √0.4 as 0.2 is the classic slip — 0.2² is 0.04.
A quotient of roots: √72 ÷ √2 and √72 ÷ √3.
√72 ÷ √2 = √36 = 6
rational — divide under one root
√72 ÷ √3 = √24 = 2√6
irrational — 24 is not a square
What is different: two irrationals divided can go either way, just like a product. Combine first, then judge.
A repeat of nines: write 0.999… as a fraction.
x = 0.999…, 10x = 9.999…
shift one place
9x = 9, so x = 1
or 99 by the nines rule
What is different: the repeating decimal is a whole number in disguise. It is rational — and an integer.
Which of these are rational? Work each to its simplest form.
| Expression | Simplified | Verdict |
|---|---|---|
| (√5 + 2)(√5 − 2) | 5 − 4 = 1 | rational — the roots cancel |
| (√5 + 2)² | 5 + 4√5 + 4 = 9 + 4√5 | irrational |
| √3 × √27 | √81 = 9 | rational |
| √3 + √27 | √3 + 3√3 = 4√3 | irrational |
| (7 − √5) + √5 | 7 | rational |
(√5 + 2)² is not 5 + 4 = 9. The middle term 2 × √5 × 2 = 4√5 survives, so the result is irrational. Only the conjugate pair (a + √b)(a − √b) = a² − b loses its root.
Expect “which is irrational”, a repeating decimal to convert, a false statement about sums or products, and surd expressions to classify. Simplify every root and product before deciding, and keep one counterexample ready: √2 × √2 = 2.
The problem
Give an irrational number between 4 and 5.
√17 ≈ 4.12 sits in range, and 17 is not a perfect square, so its root is irrational. B is exactly 4 and rational; A and C are the same rational number written two ways. A square root is irrational only when what is under it is not a perfect square.
