Question

Prove that √3 or √5 is irrational.​

Answer 1

♥HEY♥

☆QUESTION 1 - PROVE THAT ROOT 3 IS IRRATIONAL :-

➣Let us assume that root 3 is rational.

Root 3 = a/b where a and b are integers and coprimes.

Root 3 * b = a

Square LHS and RHS

3b² = a²

b² = a²/3

Therefore 3 divides a² and 3 divides a.

Now take ,

a = 3c

Square ,

a² = 9c²

3b² = 9c²

b²/3 = c²

Therefore 3 divides b² and b.

$\huge\mathfrak\red{ = = = = = = = = = = = }$

☆QUESTION 2 - PROVE THAT ROOT 5 IS IRRATIONAL:-

➣GIVEN: √5

We need to prove that √5 is irrational

Proof:

Let us assume that √5 is a rational number.

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

⇒ √5 = p/q

On squaring both the sides we get,

⇒5 = p²/q²

⇒5q² = p² —————–(i)

p²/5 = q²

So 5 divides p

p is a multiple of 5

⇒ p = 5m

⇒ p² = 25m² ————-(ii)

From equations (i) and (ii), we get,

5q² = 25m²

⇒ q² = 5m²

⇒ q² is a multiple of 5

⇒ q is a multiple of 5

Hence, p,q have a common factor 5. This contradicts our assumption that they are co-primes. Therefore, p/q is not a rational number

√5 is an irrational number.

HENCE PROVED

THANKS❤

Answer 2

Let us assume on the contrary that √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

where, a and b, are co-prime i.e. their HCF is 1Now,

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

Squaring Both sides

$3 = \frac{a {}^{2} }{b {}^{2} }$

$b {}^{2} = \frac{a {}^{2} }{3}$

$\pink{now \: if \: a {}^{2} \: is \: divided \: by \: 3}$

$\pink{then \: a \: is \: also \: divisible \: by \: 3}$

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Now let 3c= a -------------(eq 1)

where c is any integer

therefore putting eq 1 in

$b { }^{2} = \frac{a} {3}^{2}$

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

putting the value of A

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

cutting 9 by 3 resulting in 3

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

$\pink{now \: if \: b {}^{2} \: is \: \: divided \: by \: 3}$

$\pink{then \: b \: is \: also \: divisibe \: by \: 3}$

we observe that a and b have at least 3 as a common factor. But, this contradicts the fact that a and b are co-prime. This means that our assumption is not correct.

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Hence, √3 is an irrational number.

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