Question

Show that 5-√3 is irrational, where √3 is irrational number

Answer 1

$\sf It's\ given\ that\ \sqrt{3}\ is\ an\ irrational\ number.\\\\Let's\ assume\ that\ 5\ -\ \sqrt{3}\ is\ rational.\\\\5\ -\ \sqrt{3} = \frac{a}{b},\ where\ a\ and\ b\ are\ co-prime\ integers\ and\ b\ \neq 0.\\\\5\ -\ \sqrt{3} = \frac{a}{b}\\\\5\ -\ \frac{a}{b}\ =\ \sqrt{3}\\\\\frac{5b\ -\ a}{b}\ =\ \sqrt{3}\\\\Since\ a\ and\ b\ are\ integers,\ \frac{5b\ -\ a}{b}\ is\ rational.\\\\This\ implies\ that\ \sqrt{3}\ is\ rational\ too!\\\\ This\ contradicts\ the\ fact\ that\ \sqrt{3}\ is\ irrational.\\\\Therefore,\ our\ assumption\ is\ wrong.\\\\Hence,\ 5\ -\ \sqrt{3}\ is\ irrational.$

Answer 2

Answer:

=> 5 - √3

Solution:  

let us assum that 5-√3 is rational number so we can find two integers a , b. Where a and b are two co - primes number.  

= 5-√3 = a/b

= √3= 5- a/b

=> a and b are integers so (5 - a/b ) is rational

But √3 is irrational ( we know that and it is given)

So it arise contradiction due to our wrong assumption that 5 - √3 is rational number.

Hence, 5 -√3 is irrational number.