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.

Represent the number 5 on the number line.

Let O represent 0 on the number line. Let A be a point on the number line such that O A = 2 units. Draw A B perpendicular to O A at A such that A B = 1 unit. Then, O B = A B 2 + A O 2 = 1 2 + 2 2 = 5 With O as centre and O B as radius draw an arc cutting the line at C . Therefore, C represents 5 on the number line.

Represent the number 7 on the number line.

Let O represent 0 on the number line. Let A be a point on the number line such that O A = 1 unit. Draw A B perpendicular to O A at A such that A B = 1 unit. Then, by Pythagoras theorem, we get, O B = A B 2 + A O 2 = 1 2 + 1 2 = 2 units. With O as centre and O B as radius draw an arc cutting the line at C . Then, O C = O B = 2 units. Again, draw C D perpendicular to the line O C at C such that C D = 1 unit. Then, O D = O C 2 + C D 2 = 2 2 + 1 2 = 3 units Draw an arc with O as centre and O D as radius, cutting the line at E . Then, O E = O D = 3 units. Draw E F perpendicular to the number line at E such that E F = 2 units and join O F . Now, O F = O E 2 + E F 2 = 3 2 + 2 2 = 7 units. With O as centre and O F as radius, draw an arc cutting the line at G . Therefore, O G represents 7 units on the number line.