Question

prove that√3 is an irrational number​

Answer 1

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

Answer 2

Question:-

Prove $\sqrt{3}$ is an irrational Number.

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Answer:-

Let us assume $\sqrt{3}$ be a rational number and its simplest form be $\frac{a}{b}$ where a and b are coprimes.

So,

$\large\implies$ $\large\sqrt{3}$ = $\large\frac{a}{b}$

$\large\implies$ $\large{a²=3b²}$

Thus, a² is a multiple of 3

$\large\implies$ a is a multiple of 3 __$(1)$

Let $a=3m$ for some integer m

Then, $b²=3m²$

Thus, b² is multiple of 3

$\large\implies$ b is a multiple of 3 __$(2)$

From $(1)$ and $(2)$ $3$ is a common factor of a and b. This contradicts the fact that a and b are co-primes.

Hence $\sqrt{3}$ is an irrational Number

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