Question

how that proving question of irrational number for eg prove that 2+root 3 molded in class 10​

Answer 1

Answer:

Let 2 + root 3 be rational in the form p/q where p and q are integers and co primes

2 + root 3 = p/q

Squaring both the sides

4 + 3 + 4 root 3 = p²/q²

7 + 4 root 3= p²/q²

root 3 = (p²/q² - 7) ÷ 4..... eq 1

Now let root 3 be rational in the form p/q where p and q are integers and co primes

root 3 = p/q

3 = p²/q²

3q² = p²

This means p² is a divisible by 3

This implies that p is also a divisible by 3

q² = p²/3

This means q² is divisible by 3 that means even q is divisible by 3

But according to our earlier assumption p/q are co primes(Meaning hcf is 1)

But we came to know that 3 is a multiple of both p and q

So we can conclude that our statement was wrong.

That means root 3 is irrational

in eq 1

root 3 = (p²/q² - 7) ÷ 4

As irrational is equal to a number we can say that rhs is also irrational and that our assumption was wrong so 2+ root 3 is irrational.

HOPE THIS HELPS YOU

Answer 2

$\huge \bold \red { \fcolorbox{green}{grey}{required \: answer}}$

to prove :- 2+√3 is an irrational number

first we have to prove that √3 is an irrational number..

proof:-

let √3 be a rational number and is of the form of p/q , where p and q are co-prime integers and q ≠ 0

$\bold{so \: \sqrt{3} = \frac{p}{q} } \\ \\ \implies \: p = \sqrt{3} q$

on squaring both sides..

${p}^{2} = 3 {q}^{2} \: \: \: ............(i)$

→ 3 divides p2

we know that..if a prime number is divides square of any positive integer..then prime number divides that positive integer..(theorem).....(ii)

→ 3 divides p

now again let, p = 3r

on squaring both side, p² = 9r²( p = 3q² from eq.i )

→ 3 divides q²

→ 3 divide q ( by eq.ii )

since , 3 divides p and q both ( they are prime numbers)

so our assumption is wrong..

→ √3 is an irrational number

now, we have to prove that 2+√3 is an irrational number

proof :-

let 2+√3 be a rational number and is of the form of p/q where p and q are Integers and q ≠ 0

$\bold{so \: 2 + \sqrt{3} = \frac{p}{q} } \\ \\ \implies \: \sqrt{3} = \frac{p}{q} - 2 \\ \\ \implies \: \sqrt{3} = \frac{p - 2q}{q}$

here; p, 2q and q are Integers

$\bold{so \: \frac{p - 2q}{q} \: is \: a \: rational \: number}$

but √3 is an irrational number ( proved above )

so that, our assumption is wrong..

→ 2+√3 is an irrational number

$\bold{ {\huge{☦︎}}\color{purple} hence , \: proved \: \red{ ✔︎✔︎}}$