Given that 2 is irrational, prove that 5 + 3 2 is an irrational number.

Given that 2 is irrational. We have to prove that 5 + 3 2 is an irrational number. As we know that, A rational number is a number that is of the form p q where p and q are integers and q is not equal to 0 , and The product of any irrational number with a rational number is irrational. So, since 3 is a rational number, 3 2 is irrational. Also, The sum of a rational number with an irrational number is irrational. So, since 5 is a rational number, 5 + 3 2 is irrational. Thus, we proved that 5 + 3 2 is an irrational number.

Find HCF and LCM of 404 and 96 and verify that HCF × LCM = Product of the two given numbers .

Given numbers are 404 and 96 . Prime factorizing the given numbers, we get: 2 404 2 202 101 ∴ 404 = 2 × 2 × 101 and 2 96 2 48 2 24 2 12 2 6 3 ∴ 96 = 2 × 2 × 2 × 2 × 2 × 3 ∴ HCF of ( 404 , 96 ) = 2 × 2 = 4 and LCM ( 404 , 96 ) = 2 × 2 × 2 × 2 × 2 × 3 × 101 = 9696 Verification HCF × LCM = 4 × 9696 = 38784 . . . ( i ) and product of the given numbers = 404 × 96 = 38784 … (ii) From equations ( i ) and ( i i ) , we get HCF × LCM = Product of the given numbers. 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.

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.

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.

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.