Question

Prove that 4+5√3 is irrational

Answer 1

$\Huge{\underline{\underline{\mathfrak{Answer \colon}}}}$

To prove that 4 + 5√3 is an irrational number

On the contrary,let us assume that 4 + 5√3 is a rational number

Thus,

$\sf{4 + 5 \sqrt{3} = \frac{p}{q} }$

where 'p' and 'q' are rationals q≠ 0

Now,

$\implies \: \sf{5 \sqrt{3} = \frac{p}{q} - 4} \\ \\ \implies \: \sf{5 \sqrt{3} = \frac{p - 4q}{q} } \\ \\ \implies \: \underline{ \boxed{ \sf{ \sqrt{3} = \frac{p - 4q}{5q} }}}$

But √3 is an irrational

Thus,

$\implies \: \sf{ \sqrt{3} \neq \: \frac{p - 4q}{5q} }$

Here,

LHS ≠ RHS

Thus,our assumption is wrong and 4 + 5√3 is an irrational

Hence,proved

Answer 2

To prove:

4+5√3 is an irrational number

Proof:

Rational number:

A number in the form of p/q where p,qεZ q≠0

and p,q are co-primes

Now let us assume that 4+5√3 is a rational number and let it be equated to p/q

$= > 4 + 5 \sqrt{3} = \dfrac{p}{q} \\ \\ = > 5 \sqrt{3} = ( \dfrac{p}{q}) - 4 \\ \\ = > \sqrt{3} = \dfrac{(( \dfrac{p}{q} ) - 4) }{5}$

As we can see that

LHS=irrational number

RHS=rational number

rational can not be equal to irrational

This is contradiction due to our wrong assumption

Therefore

4+5√3 is an irrational number