Question

Prove that
$prove \: that \ \: \sqrt{3 \: is \: an \: \: irrational \: number}$
please answer my question​

Answer 1

EXPLANATION.

Prove that √3 is an irrational number.

As we know that,

Let, we assume that √3 is a rational number.

It means, √3 is written in the form of p/q.

⇒ √3 = p/q.

⇒ q√3 = p.

Squaring on both sides of the equation, we get.

⇒ (q√3)² = (p)².

⇒ 3q² = p².

⇒ q² = p²/3.

It means 3 divide by p².

So, 3 divides p also. - - - - - (1).

Let, we assume that,

⇒ p/3 = r.

Where r is a integers.

⇒ p = 3r.

As we know that,

⇒ 3q² = p².

Put the value of p = 3r in the equation, we get.

⇒ 3q² = (3r)².

⇒ 3q² = 9r².

⇒ q² = 3r².

⇒ q²/3 = r².

It means 3 divides q².

So, 3 divides q also. - - - - - (2).

From equation (1) and (2), we get.

We can see that 3 divides p and q.

It means 3 is a factor of p and q.

But, p and q are not Co-prime.

Hence, our assumption is wrong.

∴ √3 is irrational number.

Answer 2

Answer:

Question :-

$\mapsto$ Prove that √3 is an irrational number.

To Prove :-

• √3 is an irrational number.

Solution :-

Let,

$\leadsto$ √3 is a rational number.

Hence, we can find two integers a and b. [ where b ≠ 0 ]

$\implies \sf \sqrt{3} =\: \dfrac{a}{b}$

$\pink{\bigstar}\: \: \bf{Squaring\: both\: side\: we\: get\: :-}$

$\implies \sf \bigg\lgroup \sqrt{3}\bigg\rgroup^2 =\: \bigg\lgroup \dfrac{a}{b}\bigg\rgroup^2$

$\implies \sf 3 =\: \dfrac{a^2}{b^2}$

$\pink{\bigstar}\: \: \bf{By\: doing\: cross\: multiplication\: we\: get\: :-}$

$\implies \sf \bold{\purple{3b^2 =\: a^2\: ------\: (Equation\: No\: 1)}}$

Hence, a² is divisible by 3.

So, a is also divisible by 3.

Again,

Let,

$\leadsto$ a = 3c

Now, by putting the value of a in the equation no 1 we get,

$\implies \sf 3b^2 =\: a^2$

$\implies \sf 3b^2 =\: (3c)^2$

$\implies \sf 3b^2 =\: 9c^2$

$\implies \sf b^2 =\: \dfrac{\cancel{9}c^2}{\cancel{3}}$

$\implies \sf b^2 =\: 3c^2$

Hence, b² is divisible by 3.

So, b is also divisible by 3.

So, a and b both are divisible by 3. Therefore, we can say that a and b have atleast 3 as common factor. These contradict the fact that a and b are co-prime numbers.

It is happen due to our incorrect assumption that √3 is a rational number.

∴ √3 is an irrational number.

Hence Proved.