Prove that 3 2 is irrational.

Let, 3 2 be a rational number. Therefore, 3 2 = p q , where p and q are co-prime integers and q ≠ 0 . 2 = p 3 q Here, 3 , p and q are integers. So, p 3 q is a rational number. ∴ 2 is also a rational number. This is the contradiction to the fact that 2 is an irrational number. Therefore, our assumption that 3 2 is a rational number is wrong. Thus, 3 2 is an irrational number. Hence, proved.

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 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 7 is irrational.

Let 7 is rational and 7 = a b Where a , b are co-prime and b ≠ 0 ⇒ 7 = a 2 b 2 ⇒ 7 b 2 = a 2 . . . i ⇒ 7 is a factor of a 2 ⇒ 7 is also factor of a Let a = 7 c where c is some integer, then i becomes ⇒ 7 b 2 = 7 c 2 ⇒ 7 b 2 = 49 c 2 ⇒ 7 b 2 = 7 c 2 ⇒ 7 is a factor of b 2 ⇒ 7 is also factor of b 7 is common factor of a and b but a and b are co-prime ∴ our assumption is not true ⇒ 7 is irrational.

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.

Suppose p , q , r are positive rational numbers such that p + q + r is also rational. Then

p q , p r , q r are rational, but p , q , r are irrational

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.

Prove that 3 + 5 is an irrational number.

Let us assume that 3 + √ 5 is a rational number ⇒ 3 + √ 5 = a b where, b ≠ 0 and a , b are integers. ⇒ √ 5 = a b - 3 ⇒ √ 5 = a - 3 b b Since a , b are integers. Hence, a – 3 b is also integer. ⇒ a - 3 b b is rational which cannot be possible. ∵ a - 3 b b = 5 which is an irrational number. I t is contradicting our assumption. So, the assumption was wrong. Hence, 3 + √ 5 is an irrational number.

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 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.