Question

prove that root 3 is an irrational number hence show that 2root 3 minus 8 is also an irrtionl number

Answer 1

Required to prove :-

√3 is an irrational number

2√3 - 8 is also an irrational number

Conditions used :-

Here, conditions refer to the properties used .

The properties of rational numbers are ,

It should be expressed in the form of p/q .

where, p and q are integers

q ≠ 0 ( q is not equal to zero )

p, q are co - primes

Solution :-

Let's assume on the contradict that ,

√3 is a rational number .

So, equal the √3 with p by q where , p and q are integers , q ≠ 0 , p and q are co - primes .

Hence ,

$\red{\mathrm{ \sqrt{3} = \dfrac{p}{q}}}$

Now, do cross multiplication .

Hence,

√3q = p

Now perform squaring on both sides .

So,

$\tt{ {(\sqrt{3}q )}^{2} = {(p)}^{2}}$

3q² = p²

Recall the fundamental theorem of arithmetic .

According to which ,

If a divides q² , then a divides q also .

So,

3 divides q²

3 divides q also .

Now , let take q = 3k

Where k is an positive integer .

So,

√3q = 3k

Squaring on both sides

$\tt{ {(\sqrt{3}q)}^{2} = {(3k)}^{2}}$

3q² = 9k²

$\mathsf{ {q}^{2} = \dfrac{9 {k}^{2}}{3}}$

q² = 3k²

Now interchange the terms on both sides .

3k² = q²

Here,

3 divides q²

So,

3 divides q also.

From the above we can conclude that,

3 is the common factor of both p and q

But according to rational numbers properties , p and q should have common factor as 1 because p,q are co - primes .

So, this contradicton is due to the wrong assumption that,

√3 is a rational number .

So, our assumption is wrong .

Hence,

√3 is an irrational number .

$\rule{400}{6}$

Similarly,

Let's assume on the contradictory that 2√3 - 8 is a rational number

So, equal this number with a by b

( where a and b are integers , b ≠ 0 , a,b are co - primes )

So,

$\tt{ 2 \sqrt{3} - 8 = \dfrac{a}{b}}$

Now transpose - 8 to right side

$\tt{ 2 \sqrt{3} = \dfrac{a}{b} + 8 }$

$\tt{ 2 \sqrt{3} = \dfrac{a + 8b }{b}}$

Transpose 2 to the right side

we get,

$\tt{ \sqrt{3} = \dfrac{ \dfrac{a + 8b}{b}}{2}}$

$\tt{ \sqrt{3} = \dfrac{ a + 8b}{2b}}$

Here,

$\boxed{\tt{ \dfrac{a + 8b}{2b} \;is\; a \;rational \;number }}$

But from the above we can conclude that ,

$\boxed{\mathsf{ \sqrt{3} \; is \; an \; irrational\; number }}$

However,

An irrational number is not equal to a rational number

This contradicts the fact that our assumption was wrong .

So,

$\boxed{\huge{\orange{\tt{ \sqrt{3} \neq \dfrac{a + 8b}{2b}}}}}$

Hence,

2√3 - 8 is an irrational number .

$\rule{400}{6}$

Hence proved

Answer 2

Given ,

√3 is an irrational number

Let us assume that , 2√3 - 8 is an rational number

So , 2√3 - 8 can be written as in the form of p/q

$\Rightarrow \sf 2 \sqrt{3} - 8 = \frac{p}{q} \\ \\ \Rightarrow \sf 2 \sqrt{3} = \frac{p}{q} + 8 \\ \\ \Rightarrow \sf 2 \sqrt{3} = \frac{p + 8q}{q} \\ \\ \Rightarrow \sf \sqrt{3} = \frac{p + 8q }{2q}$

Here , √3 is an irrational number but p/2q is an rational number

Since , Irrational ≠ Rational

Thus , our assumption is wrong

$\therefore \sf \underline{ \bold{2 \sqrt{3} - 8 \: is \: an \: irrational \: number }}$