Let a = 712 and b = 56. Express both a and b in the form k1m and k2m where k1, k2 and m are integers and k2 − k1 > 6. Using the same denominator m, write exactly five distinct rational numbers lying between a and b keeping an integer numerator. Explain why the condition k2 − k1 > n + 1 is necessary to find n such rational numbers between the two rational numbers a and b using this method.
Step-by-step solution
Idea: Fractions with the same denominator are ordered by their numerators. So the rational numbers jm between k1m and k2m correspond to the whole numbers j strictly between k1 and k2. Choose m big enough to make the gap wide.
- LCM of 12 and 6 is 12: a = 712, b = 1012, gap 3 (too small). m = 24 gives 1424, 2024, gap 6 (not > 6). m = 36 gives a = 2136 and b = 3036: k1 = 21, k2 = 30, k2 − k1 = 9 > 6 ✓.1 mark
- Numerators strictly between 21 and 30: 22, 23, …, 29. Five of them give 2236, 2336, 2436, 2536, 2636 (that is 1118, 2336, 23, 2536, 1318).1 mark
- Why a condition is needed: the numbers this method can produce are k1 + 1m, k1 + 2m, …, k2 − 1m: exactly k2 − k1 − 1 of them. To find n of them we need k2 − k1 − 1 ≥ n.1 mark
- The condition k2 − k1 > n + 1 makes k2 − k1 − 1 > n, so there are always more than enough integer numerators; if the gap is too small (as with m = 12, gap 3), we must enlarge m. Note: strictly speaking, k2 − k1 ≥ n + 1 is the exact requirement: with m = 24 the gap is 6 = n + 1 and 1524, …, 1924 are exactly five numbers. So k2 − k1 > n + 1 is a safe (sufficient) condition rather than the least one.1 mark
Check: 2136 = 712 ✓, 3036 = 56 ✓; 0.583 < 0.611 < 0.639 < 0.667 < 0.694 < 0.722 < 0.833 ✓.
Answer to write in the exam
m = 36: a = 712 = 2136, b = 56 = 3036
k1 = 21, k2 = 30, k2 − k1 = 9 > 6
Five rational numbers: 2236, 2336, 2436, 2536, 2636
Integers strictly between k1 and k2: k2 − k1 − 1 in number
n numbers needed ⇒ k2 − k1 − 1 ≥ n
∴ k2 − k1 > n + 1 guarantees n such numbers (the least requirement is k2 − k1 ≥ n + 1).
Common mistakes that cost marks
- Using m = 12 and looking for five numbers among 812 and 912 only: there are just 2.
- Counting the integers from k1 to k2 including the ends (k2 − k1 + 1); the end points are a and b themselves, not between them.
- Writing 56 = 2536 (wrong: 56 = 3036).
How this can come in the exam
With a = 38 = 924 and b = 23 = 1624, how many rational numbers j24 with integer j lie strictly between a and b?
- 5
- 6
- 7
- 8
Show answer
(B) 6
j = 10, 11, …, 15: 16 − 9 − 1 = 6 numbers.
Write 14 and 13 with a common denominator m so that at least four fractions jm lie between them, and list four.
Show answer
With m = 12 the numerators are 3 and 4: no room (1 mark). m = 60 gives 1560 and 2060, with 20 − 15 − 1 = 4 numerators between (1 mark): 1660, 1760, 1860, 1960 (1 mark).Try one yourself
Write 12 and 23 with a common denominator that leaves room for three fractions between them, and list them.
Show answer
Denominator 24: 1224 and 1624; between: 1324, 1424 = 712, 1524 = 58.
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