Prove that 7 is an irrational number.

We need to prove, 7 is an irrational number. Consider 7 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 . ∴ 7 = p q ⇒ p = 7 q ⇒ p 2 = 7 q 2 ⇒ p 2 is an even number ⇒ p is an even number. So, let p = 7 m , where m is an integer ⇒ p 2 = 7 m 2 ⇒ 7 q 2 = 49 m 2 ∵ p 2 = 7 q 2 ⇒ q 2 = 7 m 2 ⇒ q 2 is an even number ⇒ q is an even number. Thus, p and q are both even numbers. So, there exists 7 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. ∴ 7 is not a rational number. Hence, 7 is an irrational number.

Express 0 . 375 in the form of p q .

Given decimal number is 0 . 375 A decimal number can be expressed in the form of a fraction, keeping all the digits removing the decimal point on the numerator and the denominator is a one followed by a zeroes equals to the number of digits after a decimal point. There are three digits after a decimal point. So, the denominator is the one followed by three zeroes. So, 0 . 375 = 375 1000 The simplified form of the above fraction is 0 . 375 = 375 1000 = 3 8

Prove that 2 is irrational.

By the proof of contradiction, let us assume 2 is a rational number. A rational number is a number that can be written in the form of p q , where p , q are integers and q ≠ 0 So, 2 can be written in the form of p q , where p and q are co-prime. 2 = p q 2 q = p Squaring on both sides, we get 2 q 2 = p 2 p 2 = q 2 2 ....(1) So, q 2 is divided by 2 If p is a prime number and p divides a 2 , then p divides a , where a is a positive integer. So, 2 shall divides q also.....(1) Consider q 2 = c , where c is some integer. ⇒ q = 2 c Substituting q = 2 c in equation (1), we get p 2 = 2 c 2 p 2 2 = c 2 Since p 2 is divided by 2 , 2 will divide p .....(2) From (1) and (2), we can say that 2 is a factor of p and q Therefore, p and q are not co-prime. Hence, our assumption is wrong. By contradiction 2 is irrational.

Prove that 2 is irrational.

By the proof of contradiction, let us assume 2 is a rational number. A rational number is a number that can be written in the form of p q , where p , q are integers and q ≠ 0 So, 2 can be written in the form of p q , where p and q are co-prime. 2 = p q 2 q = p Squaring on both sides, we get 2 q 2 = p 2 p 2 = q 2 2 ....(1) So, q 2 is divided by 2 If p is a prime number and p divides a 2 , then p divides a , where a is a positive integer. So, 2 shall divides q also.....(1) Consider q 2 = c , where c is some integer. ⇒ q = 2 c Substituting q = 2 c in equation (1), we get p 2 = 2 c 2 p 2 2 = c 2 Since p 2 is divided by 2 , 2 will divide p .....(2) From (1) and (2), we can say that 2 is a factor of p and q Therefore, p and q are not co-prime. Hence, our assumption is wrong. By contradiction 2 is irrational.

Express 0 . 375 in the form of p q .

Given decimal number is 0 . 375 A decimal number can be expressed in the form of a fraction, keeping all the digits removing the decimal point on the numerator and the denominator is a one followed by a zeroes equals to the number of digits after a decimal point. There are three digits after a decimal point. So, the denominator is the one followed by three zeroes. So, 0 . 375 = 375 1000 The simplified form of the above fraction is 0 . 375 = 375 1000 = 3 8

Prove that 7 is an irrational number.

We need to prove, 7 is an irrational number. Consider 7 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 . ∴ 7 = p q ⇒ p = 7 q ⇒ p 2 = 7 q 2 ⇒ p 2 is an even number ⇒ p is an even number. So, let p = 7 m , where m is an integer ⇒ p 2 = 7 m 2 ⇒ 7 q 2 = 49 m 2 ∵ p 2 = 7 q 2 ⇒ q 2 = 7 m 2 ⇒ q 2 is an even number ⇒ q is an even number. Thus, p and q are both even numbers. So, there exists 7 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. ∴ 7 is not a rational number. Hence, 7 is an irrational number.