Question

prove that 5-2√3 is irrational number​

Answer 1

ATQ, We have to show that 5 - √3 is an irrational number.

Let's assume that √3 is an irrational number. Which means that √3 can be expressed in the form p/q where q ≠ 0, and p and q are co-primes.

(Co-primes are numbers that are only by the number itself and 1)

⇒ √3 = p/q

Squaring on both sides we get;

⇒ (√3)² = p²/q²

⇒ 3q² = p²

⇒ 3 divides p²

∴ 3 divides p also. → Statement (1)

(If p is a prime number and divides a², then p divides 'a' as well where 'a' is a positive integer)

Let us take p to be equal to 3c.

Substitute p = 3c in 3q² = p².

⇒ 3q² = (3c)²

⇒ 3q² = 9c²

⇒ q² = 3c²

⇒ 3 divides q²

∴ 3 divides q also. → Statement (1)

On viewing statement 1 & 2, we can see that this contradicts our statement that p & q are co-primes. This further contradicts our statement that √3 is an irrational number.

∴ √3 is a rational number.

Now we'll prove 5 - 2√3 is an irrational number​.

Let's assume that 5 - 2√3 is an irrational number. Which means that 5 - 2√3 can be expressed in the form p/q where q ≠ 0, and p and q are co-primes.

⇒ 5 - 2√3 = p/q

⇒ 5 - (p/q) = 2√3

⇒  (5q - p)/q = 2√3

⇒  (5q - p)/2q = √3

Here, √3 is an irrational number, and  (5q - p)/2q is a rational number. But we know that Irrational no: ≠ Rational no: .This contradicts our assumption that 5 - 2√3 is a rational number.

∴ 5 - 2√3 is an irrational number.

(You can solve the question in two ways, one by proving √3 is irrational first, or you can directly mention that √3 is irrational if it's given in the question. Since it hasn't been mentioned in the question, we've begun by proving √3 is irrational.)

Answer 2

first,

need to prove √3 is a irrational number.

so,

Let us assume to the contrary that √3 is a rational number.

It can be expressed in the form of p/q

where p and q are co-primes and q≠ 0.

⇒ √3 = p/q

⇒ 3 = p2/q2 (Squaring on both the sides)

⇒ 3q2 = p2………………………………..(1)

This means that 3 divides p2. This means that 3 divides p because each factor should appear two times for the square to exist.

So we have p = 3r

where r is some integer.

⇒ p2 = 9r2………………………………..(2)

from equation (1) and (2)

⇒ 3q2 = 9r2

⇒ q2 = 3r2

Where q2 is multiply of 3 and also q is multiple of 3.

Then p, q have a common factor of 3. This runs contrary to their being co-primes. Consequently, p / q is not a rational number. This demonstrates that √3 is an irrational number.

Now,

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

5-2√3=p/q(where p and q are co prime number)

2√3=p/q+5

√3=p+5q/q

now p, 5,2 and q are integers.

p+5q/q is a rational number

WKT:

√3 is a irrational number.

so our assumptions is wrong,

5-2√3 is a irrational number