Question

prove that 2-√3 is irrational number​

Answer 1

Answer:

Let 2-√3 be rational no.

then,

2-√3=$\frac{a}{b}$

where,

a & b are positive integers and b≠0

2-√3=$\frac{a}{b}$

=> -2-$\frac{b}{a}$=√3

by taking LCM,

=> $\frac{-2a-b}{a}$=√3

we know that √3 is irrational and,

$\frac{-2a-b}{a}$=$\frac{integer}{integer}$= rational

rational≠irrational

Therefore, Our supposition is wrong 2-√3 is not rational. It is irrational.

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Answer 2

Step-by-step explanation:

We need to prove that$2-\sqrt3$ is irrational.

Let us say that $2-\sqrt3$ is a rational number.

And we know that rational number can be expressed in the form of $\frac{p}{q}$ where $q\neq0$

So, we can write $2-\sqrt3=\frac{p}{q}\\ \\ Taking\ 2\ on\ the\ RHS\ we\ get,\\ -\sqrt3=\frac{p}{q}-2\\ \\ \implies \sqrt3=2-\frac{p}{q}\\ \\ \rm But\ we\ know\ that\ \sqrt3\ is\ irrational\ and\ cant\ be\ written\ in\ a\ fraction\ form.$

$\rm This\ contradiction\ arised\ because\ we\ considered\ 2-\sqrt3\ as\ rational.\\ \therefore \bf 2-\sqrt3\ is\ irrational.$