Question 56 of 58advanced🔍 AnalyzeLong Answer5 marks

Let a=712a = \frac{7}{12} and b=56b = \frac{5}{6}. Express both aa and bb in the form k1m\frac{k_1}{m} and k2m\frac{k_2}{m} where k1,k2k_1, k_2 and mm are integers and k2−k1>6k_2 - k_1 > 6. Using the same denominator mm, write exactly five distinct rational numbers lying between aa and bb keeping an integer numerator. Explain why the condition k2−k1>n+1k_2 - k_1 > n + 1 is necessary to find nn such rational numbers between the two rational numbers aa and bb using this method.

Correct Answer

(i) a=2136andb=3036a = \frac{21}{36} \quad \text{and} \quad b = \frac{30}{36}

where k1=21k_1 = 21, k2=30k_2 = 30, and m=36m = 36.

(ii) 2236,2336,2436,2536,2636\frac{22}{36}, \quad \frac{23}{36}, \quad \frac{24}{36}, \quad \frac{25}{36}, \quad \frac{26}{36}

(iii) The number of integers strictly between k1k_1 and k2k_2 is k2−k1−1k_2 - k_1 - 1. To find nn distinct rational numbers, we need at least nn integers between k1k_1 and k2k_2, so:

k2−k1−1≥nk_2 - k_1 - 1 \ge n

or equivalently:

k2−k1≥n+1k_2 - k_1 \ge n + 1

The condition k2−k1>n+1k_2 - k_1 > n + 1 ensures k2−k1−1>nk_2 - k_1 - 1 > n, providing strictly more than nn integers, which guarantees we can find at least nn distinct rational numbers with integer numerators.

Exercise: END-OF-CHAPTER EXERCISES | Q: *13 | (Page 25)
For More Understanding

Explanation

The question assesses the method of finding rational numbers between two given numbers by converting them to equivalent fractions with a common denominator. By ensuring the difference between the numerators is sufficiently large (greater than n+1n + 1 for nn rational numbers), we create enough integer values between the numerators to easily select the required number of distinct rational numbers.

Solution Steps

  1. Step 1: Find a common denominator.

    a=712andb=56=1012a = \frac{7}{12} \quad \text{and} \quad b = \frac{5}{6} = \frac{10}{12}

    Here, k2−k1=10−7=3k_2 - k_1 = 10 - 7 = 3, which is not >6> 6.

  2. Step 2: Multiply to satisfy the condition. Multiply numerator and denominator by 3:

    a=2136andb=3036a = \frac{21}{36} \quad \text{and} \quad b = \frac{30}{36}

    Now, k1=21k_1 = 21, k2=30k_2 = 30, m=36m = 36, and k2−k1=9>6k_2 - k_1 = 9 > 6.

  3. Step 3: List five rational numbers. The integers between 21 and 30 are 22, 23, 24, 25, 26, 27, 28, 29. Five distinct rational numbers are:

    2236,2336,2436,2536,2636\frac{22}{36}, \quad \frac{23}{36}, \quad \frac{24}{36}, \quad \frac{25}{36}, \quad \frac{26}{36}

  4. Step 4: Explain the condition. The count of integers strictly between k1k_1 and k2k_2 is k2−k1−1k_2 - k_1 - 1. To find nn rational numbers, we need:

    k2−k1−1≥nk_2 - k_1 - 1 \ge n

    or:

    k2−k1≥n+1k_2 - k_1 \ge n + 1

    The condition k2−k1>n+1k_2 - k_1 > n + 1 ensures k2−k1−1>nk_2 - k_1 - 1 > n, guaranteeing enough integers for nn distinct numerators.

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