Question

It's urgent

Prove that:
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Answer 1

hey mate!

here is your answer:

Proof:

let us assume that root 3 is rational and root 3=a/b where a and b are two co-primes

so, root 3=a/b
squaring both sides
we get, 3=a^2/b^2

3b^2=a^2

now, a is divisible by 3
hence a^2 is a is also divisible by 3

let a =3c for some integer c
so, a^2=9c^2

now, 3b^2=9c^2
b^2=3c^2

now, b is divisible by 3
hence b^2 is also divisible by 3

hence , both a and b have a common factor as 3

Thus our assumption that a and b are co- primes is wrong and hence our assumption that root 3 is rational is wrong

Thus , root 3 is irrational.


hope it helped ^_^

Answer 2

$\textbf{\underline{Question}} -$

Prove that , √3 is irrational number

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$\textbf{\underline{Solution}} -$

Assume that √3 is rational number

Then , there exist positive integers a and b

Now,

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

squaring both side

$3 = \frac{ {a}^{2} }{ {b}^{2} } \\ \\ = > 3 {b}^{2} = {a}^{2} - - - - (1)$

So,

If 3 is the factor of a²

Then, 3 is also a factor of a –––––––(2)

Now,

let a = 3c ( where c is any integer)

again, squaring both side

a² = (3c)²

a² = 9c²

putting the value of a² in eq(1)

$3 {b}^{2} = {a}^{2} \\ \\ 3 {b}^{2} = 9 {c}^{2} \\ \\ {b}^{2} = \frac{9 {c}^{2} }{3} \\ \\ {b}^{2} = 3 {c}^{2}$

So,

If 3 is the factor of b²

then 3 is also a factor of b

Since,

we observed 3 is the factor of a and b .

But this contradicts the fact that a and b are co-prime .

This means that our assumption is not correct.

$\textsf{Hence \:root \: 3 is \: an \: irrational \: number}$

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$\textbf{Thanks}$