Question

prove that ^3 is a irrational number

Answer 1

We have to prove that the square root of 3 is an irrational number.

Solution :

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.

:$\implies$√3 = p/q

:$\implies$3 = p2/q2 (Squaring on both the sides)

:$\implies$3q2 = p2………………………………..(i)

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.

:$\implies$ p2 = 9r2………………………………..(ii)

from equation (i) and (ii)

:$\implies$3q2 = 9r2

:$\implies$ 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

Answer:

$Let \: us \: assume \: on \: \\ the \: contrary \: that \sqrt{3} \\ is \: a \: rational \: number.$

Then, there exist positive integers a and b such that

$\sqrt{3} = \frac{a}{b} \: where, \:  a and b, \: are \: \\ co-prime \: i.e. \: their  \: HCF \:  is  \: 1$

Now,

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

$⇒3 = \frac{a {}^{2} }{b {}^{2} }$

$⇒3b {}^{2} = a {}^{2}$

$⇒3 \: divides \: a {}^{2}$

Let a = 3c

$3b {}^{2} = (3c) {}^{2}$

$3b {}^{2} = 9c {}^{2}$

$3c {}^{2} = b {}^{2}$

$3 \: divides \: b {}^{2}$

As 3 divides both a and c it contradicts the fact that it is rational.

$so \: \sqrt{3} \: is \: irrational$