If a and b are two unequal rational numbers, show that a + b 2 is a rational number lying between a and b .

Given that a and b are two unequal rational numbers. Let a is less than b . So, we get, a < b ⇒ a + b < b + b [Add b to both sides] ⇒ a + b < 2 b ⇒ a + b 2 < 2 b 2 [Divide both sides by 2 ] ⇒ a + b 2 < b -----(i) Again, a < b ⇒ a + a < a + b [Add a to both sides] ⇒ 2 a < a + b ⇒ 2 a 2 < a + b 2 [Divide both sides by 2 ] ⇒ a < a + b 2 -----(ii) Combining (i) and (ii), we get, a < a + b 2 < b . Now, since a and b are rational numbers, a + b is also a rational number [The sum of two rational numbers is always a rational number.] Also, 2 is an integer, which is also a rational number. Hence, a + b 2 is a rational number [The quotient of two rational numbers is a rational number] Therefore, a + b 2 is a rational number lying between a and b . [Proved]

Find five rational numbers between 1 4 and 1 3 .

We are given to find five rational numbers between 1 4 and 1 3 . We know that a rational number lying between two rational numbers a and b is given by 1 2 a + b . So, one rational number lying between 1 4 and 1 3 is 1 2 1 4 + 1 3 = 1 2 3 + 4 12 = 1 2 × 7 12 = 7 24 . One rational number lying between 1 4 and 7 24 is 1 2 1 4 + 7 24 = 1 2 6 + 7 24 = 1 2 × 13 24 = 13 48 . One rational number lying between 1 4 and 13 48 is 1 2 1 4 + 13 48 = 1 2 12 + 13 48 = 1 2 × 25 48 = 25 96 . One rational number lying between 7 24 and 1 3 is 1 2 7 24 + 1 3 = 1 2 7 + 8 24 = 1 2 × 15 24 = 15 48 . One rational number lying between 15 48 and 1 3 is 1 2 15 48 + 1 3 = 1 2 15 + 16 48 = 1 2 × 31 48 = 31 96 . Therefore, five rational numbers between 1 4 and 1 3 are 25 96 , 13 48 , 7 24 , 15 48 and 31 96 .

Find six rational numbers between 1 3 and 2 3 .

We are given to find six rational numbers between 1 3 and 2 3 . Here, we observe that the denominators of the given two rational numbers are the same. We shall multiply both the numerators and the denominators of the given two rational numbers by 7 [The required number of rational numbers between the given rational numbers is six. So, we shall multiply both the numerators and the denominators of the given two rational numbers by a number which is 6 + 1 = 7 or more] Thus, we get, 1 3 = 1 × 7 3 × 7 = 7 21 and 2 3 = 2 × 7 3 × 7 = 14 21 . So, six rational numbers between 1 3 [or 7 21 ] and 2 3 [or 14 21 ] are 8 21 , 9 21 , 10 21 , 11 21 , 12 21 and 13 21 or 8 21 , 3 7 , 10 21 , 11 21 , 4 7 and 13 21 . Therefore, the required six rational numbers between 1 3 and 2 3 are 8 21 , 3 7 , 10 21 , 11 21 , 4 7 and 13 21 .

Prove that 5 is an irrational number.

Let's assume 5 is a rational number. As per the definition of rational numbers, we can express 5 = p q , {where p , q are co-primes and q ≠ 0 }. On squaring both sides, we get, 5 2 = p 2 q 2 ⇒ 5 = p 2 q 2 ⇒ p 2 = 5 q 2 . . . . . . i Which means that, p 2 is a multiple of 5 , ⇒ 5 divides p Now, let p = 5 m ⇒ p 2 = 25 m 2 Now putting this value of p 2 in equation i , 25 m 2 = 5 q 2 ⇒ 5 m 2 = q 2 ⇒ 5 divides q 2 ⇒ 5 divides q This means that, 5 is a common divisor of p and q , which contradicts the assumption done in the first step. { p , q are co-primes}. Therefore, 5 is an irrational number.

Prove that 7 is not a rational number.

Let's assume 7 is a rational number. As per the definition of rational numbers, we can express 7 = p q , {where p , q are co-primes and q ≠ 0 }. On squaring both sides, we get, 7 2 = p 2 q 2 ⇒ 7 = p 2 q 2 ⇒ p 2 = 7 q 2 . . . . . . i Which means that, p 2 is a multiple of 7 , ⇒ 7 divides p Now, let p = 7 m ⇒ p 2 = 49 m 2 Now putting this value of p 2 in equation i , 49 m 2 = 7 q 2 ⇒ 7 m 2 = q 2 ⇒ 7 divides q 2 ⇒ 7 divides q This means that, 7 is a common divisor of p and q , which contradicts the assumption done in the first step. { p , q are co-primes}. Therefore, 7 is an irrational number.