prove that under root 3 is irrational​
sorry it's a math, s question

Answer:

Let us assume that √3 is a rational number.

then, as we know a rational number should be in the form of p/q

where p and q are co- prime number.

So,

√3 = p/q { where p and q are co- prime}

√3q = p

Now, by squaring both the side

we get,

(√3q)² = p²

3q² = p² ........ ( i )

So,

if 3 is the factor of p²

then, 3 is also a factor of p ..... ( ii )

=> Let p = 3m { where m is any integer }

squaring both sides

p² = (3m)²

p² = 9m²

putting the value of p² in equation ( i )

3q² = p²

3q² = 9m²

q² = 3m²

So,

if 3 is factor of q²

then, 3 is also factor of q

Since

3 is factor of p & q both

So, our assumption that p & q are co- prime is wrong

hence,. √3 is an irrational number

Explanation:

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.

Hence Proved

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