Prove that 5 is an irrational number.

Let us assume, to the contrary that 5 is a rational number. Then, there exist positive integers a & b such that 5 = a b Where a & b are co-prime. So, b 5 = a On squaring both sides, we get 5 b 2 = a 2 ⇒ 5 d i v i d e s a 2 ⇒ 5 d i v i d e d a ......(i) When a prime number p divides a 2 , then p divides a where a is a positive integer. ⇒ a = 5 c ( f o r s o m e i n t e g e r C ) ⇒ a 2 = 25 c 2 ⇒ 5 b 2 = 25 c 2 ⇒ 5 d i v i d e s b 2 ⇒ 5 d i v i d e s b .....(ii) We claim that 5 is a common factor of both a & b , But this contradicts the factor that a & b has no common factor than 1 as they are co-prime. This means our assumption is wrong. Hence, 5 is an irrational number.

Prove that 3 is an irrational number.

Let 3 be a rational number ∴ 3 = p q , where p , q are integers and q ≠ 0 and p and q are coprimes. Coprime number is a set of numbers which have no common divisor other than 1 p = 3 q Taking square on both sides, we get p 2 = 3 q 2 p 2 = 3 q 2 q 2 = p 2 3 we get p is the divisor of 3 ∴ p 3 = a ⇒ p = 3 a Taking square on both sides, we get p 2 = 9 a 2 Here, p 2 = 3 q 2 ⇒ 3 q 2 = 9 a 2 ⇒ q 2 = 9 a 2 3 ⇒ a 2 = q 2 3 ⇒ q is the factor of 3 . This contradicts out assumption. Hence, 3 is an irrational number.

Prove that 2 is an irrational number

We need to prove, 2 is an irrational number. Let 2 is a rational number. ∴ 2 can be expressed in the form p q , where p and q are both integers and co-primes, i.e., There is no common factor in p and q except 1 q ≠ 0 . ∴ 2 = p q ⇒ p = 2 q ⇒ p 2 = 2 q 2 ⇒ p 2 is an even number ⇒ p is an even number. So, let p = 2 m , where m is an integer ⇒ p 2 = 2 m 2 ⇒ 2 q 2 = 4 m 2 ∵ p 2 = 2 q 2 ⇒ q 2 = 2 m 2 ⇒ q 2 is an even number ⇒ q is an even number. Thus, p and q are both even numbers. So, there exists 2 as the common factor of both p and q . But, we have let that there is no common factor in p and q , except 1 . This is a contradiction. ∴ 2 is not a rational number. Hence, 2 is an irrational number.

Find the H.C.F of 1260 and 7344 using Euclid's algorithm.

According to Euclid division algorithm, any positive integer a can be expressed as a = b q + r , where q is quotient, b is divisor and r is remainder, 0 ≤ r < b Consider the largest integer out of the given two integers and then express the integer in the form of b q + r 7344 = 1260 × 5 + 1044 Now express 1260 in the form of a = b q + r , here a = 1260 and b = 1044 1260 = 1044 × 1 + 216 Now express 1044 in the form of a = b q + r , here a = 1044 and b = 216 1044 = 216 × 4 + 180 Now express 216 in the form of a = b q + r , here a = 216 and b = 180 216 = 180 × 1 + 36 Now express 180 in the form of a = b q + r , here a = 180 and b = 36 180 = 36 × 5 + 0 When the remainder is 0 , the divisor at this stage will be the H.C.F of 1260 and 7344 Hence, the H.C.F is 36

Prove that one of every three consecutive positive integers is divisible by 3 .

Let three consecutive positive integers be a , a + 1 and a + 2 When a number is divided by 3 , we can get 0 or 1 or 2 as remainders. So numbers can take the form of 3 x , 3 x + 1 and 3 x + 2 If a = 3 x then a is divisible by 3 but a + 1 and a + 2 are not divisible by 3 . If a = 3 x + 1 , then a + 2 = 3 x + 1 + 2 = 3 x + 3 = 3 ( x + 1 ) is divisible by 3 but a and a + 1 are not divisible by 3 . If n = 3 x + 2 then, a + 1 = 3 x + 2 + 1 = 3 x + 3 = 3 ( x + 1 ) is divisible by 3 but a and a + 2 are not divisible by 3 . So, we can say that one of the numbers among a , a + 1 and a + 2 is divisible by 3 .

Show that the square of any positive integer cannot be of the form 5 q + 2 or 5 q + 3 for any integer q .

We have to show that the square of any positive integer cannot be of the form 5 q + 2 or 5 q + 3 for any integer q . Let, a be any positive integer. Now, by Euclid's division lemma, a = b m + r where b = 5 (given) ∴ a = 5 m + r So, r can be any of 0 , 1 , 2 , 3 when r = 0 a = 5 m ⇒ a 2 = ( 5 m ) 2 = 25 m 2 ⇒ a 2 = 5 ( 5 m 2 ) = 5 q Here, q = 5 m 2 when r = 1 a = 5 m + 1 ⇒ a 2 = ( 5 m + 1 ) 2 = 25 m 2 + 10 m + 1 ⇒ a 2 = 5 ( 5 m 2 + 2 m ) + 1 = 5 q + 1 Here, q = 5 m 2 + 2 m when r = 2 a = 5 m + 2 ⇒ a 2 = ( 5 m + 2 ) 2 ⇒ a 2 = 25 m 2 + 20 m + 4 ⇒ a 2 = 5 ( 5 m 2 + 4 m ) + 4 ⇒ a 2 = 5 q + 4 Here, q = 5 m 2 + 4 m when r = 3 a = 5 m + 3 ⇒ a 2 = ( 5 m + 3 ) 2 = 25 m 2 + 30 m + 9 ⇒ a 2 = 5 ( 5 m 2 + 6 m + 1 ) + 4 ⇒ a 2 = 5 q + 4 Here, q = 5 m 2 + 6 m + 1 So, from the above solutions, we can conclude that square of any positive integer cannot be of the form 5 q + 2 or 5 q + 3 .

Prove that 2 is an irrational number.

We need to prove, 2 is an irrational number. Consider 2 is a rational number and it can be expressed in the form p q , where p and q are both integers and co-primes, i.e., There is no common factor in p and q except 1 q ≠ 0 . ∴ 2 = p q ⇒ p = 2 q ⇒ p 2 = 2 q 2 ⇒ p 2 is an even number ⇒ p is an even number. So, let p = 2 m , where m is an integer ⇒ p 2 = 2 m 2 ⇒ 2 q 2 = 4 m 2 ∵ p 2 = 2 q 2 ⇒ q 2 = 2 m 2 ⇒ q 2 is an even number ⇒ q is an even number. Thus, p and q are both even numbers. So, there exists 2 as the common factor of both p and q . But, we have let that there is no common factor in p and q , except 1 . This is a contradiction. ∴ 2 is not a rational number. Hence, 2 is an irrational number.

Show that every positive odd integer is of the form 4 q + 1 or 4 q + 3 , where q is some integer.

Let a be any positive integer and b = 4 . Applying division lemma on a and b , a = 4 q + r , where 0 ≤ r < 4 . When r = 0 , a = 4 q + 0 = 2 ( 2 q ) When r = 1 , a = 4 q + 1 When r = 2 , a = 4 q + 2 = 2 ( 2 q + 1 ) When r = 3 , a = 4 q + 3 Since, 4 q , 4 q + 2 are multiples of 2 , these are even integers. So, every odd positive integers are in the form of either ( 4 q + 1 ) or ( 4 q + 3 ) , where q is some integer. Hence, proved.

Find a rational number between 2 and 3 .

We know the decimal form of 2 and 3 as: 2 = 1 . 4142135 . . . . . 3 = 1 . 73205080 . . . . . . As we know that rational numbers are those decimal numbers which are either terminating or repeating. So, you can given any terminating or repeating decimal number between 1 . 4142135 . . . and 1 . 73205080 . . . as your answer. So, numbers between 2 and 3 are just greater than 1 . 41 and less than 1 . 73 . Thus answer can be any of the following rationals: 1 . 415 , 1 . 416 , 1 . 417 , 1 . 532 , 1 . 582 , 1 . 60 , 1 . 62 , 1 . 70 , 1 . 72 etc.