Question

Prove that√3 is an irrational number and hence show that 5 - 2√3 is an irrational number...
please help​

Answer 1

Proof of root 3 is irrational is attached in the picture.

show that 5 - 2√3 is an irrational number

Proof:

Let us assume that 5 - 2 √3 is a rational number.

So, 5 - 2 √3 may be written as

5 - 2 √3 = p/q, where p and q are integers, having no common factor except 1 (co primes) and q ≠ 0.

⇒ 5 - p/q = 2 √3

⇒ √3 = 5q - p/2q

Since, 5q - p/2q is a rational number as p and q are integers.

Therefore, √3 is also a rational number, which contradicts our assumption.

This contradiction has arisen because of our incorrect assumption

Hence, 5 - 2 √3 is an irrational number.

Answer 2

Step-by-step explanation:

To Prove : √3 is irrational,

Assume that √3 is rational.

√3 = a/b (a and b are co prime numbers)

⇒ √3b = a

Squaring both the sides,

⇒ 3b² = a² ---- (I)

As a² = 3b², a will be divisible by 3.

⇒ a = 3c ---- (II)

3 is a factor or a.

Putting eq. II in eq. I,

⇒ 3b² = (3c)²

⇒ 3b² = 9c²

⇒ b² = 3c²

As b² = 3c², b will be divisible by 3. 3 is also b's factor.

Both a and b have 3 as common factor, which means they are not co-prime numbers.

This is a contradiction to our assumption. This was arisen due to wrong assumption.

Therefore, √3 is irrational.

__________________

To Prove : 5 - 2√3 is irrational

Assume that 5 - 2√3 is rational

5 - 2√3 = a/b (a and b are integers)

⇒ -2√3 = a/b - 5

⇒ -2√3 = (a - 5b)/b

⇒ √3 = (a - 5b)/-2b

⇒ √3 = (-a + 5b)/2b

(-a + 5b)/2b is rational as a and b are integers. This means √3 is rational too.

This is contradiction to the fact that √3 is irrational.

This contradiction was arisen due to wrong assumption.

Therefore, 5 - 2√3 is irrational.