Question

Prove that √3 is an irrational number.

Answer 1

Let us assume the opposite,

ie., \(\sqrt{3}\) is rational

Hence, \(\sqrt{3}\) can be written in the form \(\frac{a}{b}\) where a and \(b(b \neq 0)\) are co.prime.

Hence,

\(\sqrt{3} =\frac{a}{b} \)

\(\sqrt{3} b =a\)

squaring both sides

\((\sqrt{3} b)^2 =a^2 \)

\(3 b^2 =a^2 \)

\(b^2 =\frac{a^2}{3}\)

Hence, 3 divides \(a^2\)

So, 3 shall divide a also ...(i)

we Can say,

\(\frac{a}{3}=C\) where C is some integer

So, a = 3c

Now we know that

\(3 b^2 =a^2 \)

\(\text {Putting } a=3 c\)

\(3 b^2=(3 c)^2 \)

\( 3 b^2=9 c^2 \)

\( b^2=9 c^2 \times \frac{1}{3} \)

\( b^2=3 c^2 \)

\( c^2=\frac{b^2}{3}\)

Hence, 3 divides \(b^2\)

So, 3 divides b also .....(ii)

By (i) and (ii)

3 divides both a & b.

Hence 3 is a factor of a and b. 

So, a & b have a factor 3.

Therefore, a & b are not co-prime.

\(\therefore \sqrt{3}\) is an irrational number.

Answer 2

Let \(\sqrt 3\) be a rational number.

\(\therefore \sqrt{3}=\frac{p}{q},\) where \(\mathrm{q} \neq 0\) and let \(\mathrm{p} \,\&\, \mathrm{q}\) be co-prime.

\(3 q^2=p^2 \Rightarrow p^2\) is divisible by \(3 \Rightarrow p\) is divisible by 3 .....(i)

\(\Rightarrow p=3 a,\) where 'a' is some integer

\(9 a^2=3 q^2 \Rightarrow q^2=3 a^2 \Rightarrow q^2\) is divisible by \(3 \Rightarrow q\) is divisible by 3 ......(ii)

(i) and (ii) leads to contradiction as 'p' and 'q' are co-prime.