Question

Prove that 3 - 2 √11 is irrational

Answer 1

$\large\bf\underline {To \: prove:-}$

• 3 - 2 √11 is irrational

$\huge\bf\underline{Solution:-}$

Let us assume that 3-2√11 is rational.

So,

$\rm \longmapsto \: 3 - 2 \sqrt{11} = \frac{a}{b}$

where a and b are co - prime and b ≠ 0

$\rm \longmapsto \: - 2 \sqrt{11} = \frac{a}{b} - 3 \\ \\ \rm \longmapsto \: - 2 \sqrt{11} = \frac{a - 3b}{b} \\ \\ \rm \longmapsto \: \sqrt{11} = \frac{a - 3b}{ - 2b} \\ \\ \rm \longmapsto \: \sqrt{11} = \frac{ - (3b - a)}{ - 2b} \\ \\ \rm \longmapsto \: \sqrt{11} = \frac{3b - a}{2b}$

So, 3b-a/2b is rational as it is shown in the form of p/q where p and q are intigers and q ≠ 0.

then, a and be are intigers, so 3b - a/2b is rational.

thus √11 is also rational .

but as we know that√11 is rational .

So this contradiction is arissen because of our wrong assumption then,

3 - 2 √11 is irrational.

hence Proved

Answer 2

★ sOLUTIOn ★

Let us assume that 3-2√11 is a Rational number.

And we know that ,

if the number is rational so we can write it in the fractional from OR in the form of p/q.

where ( q≠0 and p , q are any integers )

Now,

$\longrightarrow 3-2\sqrt11=\frac{p}{q}\\\longrightarrow 2\sqrt11=\frac{p}{q}-3\\ \longrightarrow 2\sqrt=\frac{p-3q}{q}\\ \longrightarrow \sqrt11=\frac{p-3q}{2q}$

p and q both are integer so$\longrightarrow \frac {p-3q}{2q}$

is a rational number.

and √11 is a irrational number.

• Note :- A rational number can never be equal to irrational number

Therefore,

our contradiction is wrong 3-2√11 is a irrational number.