Question

prove that 2√3 is an irrational
--------
5​

Answer 1

We can prove it by contradictory method..

We assume that 2 + √3 is a rational number.

=> 2 + √3 = p/q , where p & q are integers, ‘q’ not = 0.

=> √3 = (p/q) - 2

=> √3 = (p - 2q)/ q ………… (1)

=> here, LHS √3 is an irrational number.

But RHS is a rational number.. Reason- the difference of 2 integers is always an integer. So the numerator (p- 2q) is an integer.

& the denominator ‘q’ is an integer.&‘q’ not = 0

This way, all conditions of a rational number are satisfied.

=> RHS (p- 2q)/q is a rational number.

But , LHS is an irrational.

=> LHS of….. (1) is not = RHS.

=> Our assumption, that 2 + √3 is a rational number, is incorrect..

=> 2 + √3 is an irrational number

ItzDopeGirl❣

Answer 2

$\huge\bf\red{Answer,}$

let 2√3 is a rational number.

so,

→2√3=r

→√3=r/2

now,

here is clear √3 is a irrational number and r/2 is also.

our contradiction is wrong

→2√3 is a irrational number (proved)

OR

let 2+√3=a/b is rational number.

on squaring both side.

we get,

→2+3+2√6=a2/b2

→5+2√6=a2/b2 is a rational number.

so,

→2√6= a2/b2- 5

since,

→2√6 is a irrational number.

→a2/b2-5 is a rational number.

therefore,

our contradiction is wrong (2+√3)is a irrational number.

:)