Question

Prove that root 2 is an irrational no.
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Answer 1

Prove that root 2 is an irrational number

Given √2

To prove: √2 is an irrational number.

Proof:

Let us assume that √2 is a rational number.

So it can be expressed in the form p/q where p, q are co-prime integers and q≠0

√2 = p/q

Here p and q are coprime numbers and q ≠ 0

Solving

√2 = p/q

On squaring both the side we get,

=>2 = (p/q)2

=> 2q2 = p2……………………………..(1)

p2/2 = q2

So 2 divides p and p is a multiple of 2.

⇒ p = 2m

⇒ p² = 4m² ………………………………..(2)

From equations (1) and (2), we get,

2q² = 4m²

⇒ q² = 2m²

⇒ q² is a multiple of 2

⇒ q is a multiple of 2

Hence, p, q have a common factor 2. This contradicts our assumption that they are co-primes. Therefore, p/q is not a rational number

√2 is an irrational number.

Hope will be helpful ☺️

Answer 2

$\bold{Let's \: assume \: that \: \sqrt{2} \: is \: rational}$

$: \mapsto \: \: \boxed{ \sf{ \sqrt{2} = \frac{a}{b} }} \: ; \sf b≠0$

$\leadsto \green{\sf \: Also \: a \: and \: b \: are \: Co-prime}$

$\: \: \: \: \: \: \: \: \: \: \displaystyle \sf \sqrt{2} \: b = a$

$\tt \: Squaring \: both \: Sides$

$\ \underline{ \fbox{ { \orange{ ★} \: \sf2b {}^{2} = a { }^{2} ...(1)} }}$

Therefore, a² is divisible by 2

Also, a is divisible by 2

Again, $\bf Let , a= 2c$

$\mapsto\sf2b {}^{2} = (2c) {}^{2} \\ \longrightarrow \: \cancel{ \sf2} \bf \: b {}^{2} = \cancel{14} \: c {}^{2} \\ \longrightarrow \sf \: b {}^{2} = 2 \: c {}^{2}$

$\underline{\fbox{ {\tt \orange★ \: 2c {}^{2} = b {}^{2} } }}$

Therefore, b² is divisible by 2

Also, b is divisible by 2

$\leadsto \green{ \sf so \: a \: and \: b \: are \: divisible \: by \: 2}$

Therefore we can say that a and b have atleast

2 as a Common factor.

This Contradicts fact that a and b are Co- prime.

$\\ \green{ \bf \: It \: happens \: due \: to \: our \: }\\ \bf \red{ incorrect \: assumption \: }$

$\tt {∴ \sqrt{2} \: is \: a \: irrational \: }$

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