Question

probe that root 3 is ireational number ​

Answer 1

Answer:

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)

It means that 3 divides p2 and also 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

We have two cases to consider now.

Case I

Suppose that r is even. Then r2 is even, and 3r2 is even which implies that q2 is even and so q is even, but this cannot happen. If both q and r are even then gcd(q,r)≥2 which is a contradiction.

Case II

Now suppose that r is odd. Then r2 is odd and 3r2 is odd which implies that q2 is odd and so q is odd. Since both q and r are odd, we can write q=2m−1 and r=2n−1 for some m,n∈N.

Therefore

q2=3r2

(2m−1)2=3(2n−1)2

4m2−4m+1=3(4n2−4n+1)

4m2−4m+1=12n2−12n+3

4m2−4m=12n2−12n+2

2m2−2m=6n2−6n+1

2(m2−m)=2(3n2−3n)+1

We note that the lefthand side of this equation is even, while the righthand side of this equation is odd, which is a contradiction. Therefore there exists no rational number r such that r2=3.

Hence the root of 3 is an irrational number.

Step-by-step explanation:

Hence Proved

Answer 2

Answer:

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Step-by-step explanation:

let √3 be rational.

let a/b = √3 where a and b are co-prime

(a/b)² = a²/b² = (√3)² = 3

a² = 3b²

3 is a factor a²

3 is a factor of a

let a = 3c

a² = 3b²

(3c)² = 9c² = 3b²

9c²/3 = 3c² = b²

b² = 3c²

3 is a factor of b²

3 is a factor of b

Thus, a and b have a common factor, 3.

Thus, a and b are not co-prime

Since the assumption was wrong

Therefore, √3 is irrational