Question

prove irrratinal 3+2√3
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Answer 1

Letus assume that 3 + 2√5 is a rational number.

Soit can be written in the form a/b

3 + 2√5 = a/b

Here a and b are coprime numbers and b ≠ 0

Solving3 + 2√5 = a/b we get,

=>2√5 = a/b – 3

=>2√5 = (a-3b)/b

=>√5 = (a-3b)/2b

This shows (a-3b)/2b is a rational number. But we know that But √5 is an irrational number.

so it contradictsour assumption.

Our assumption of3 + 2√5 is a rational number is incorrect.

3 + 2√5 is an irrational number

Hence proved

Answer 2

Prove that 3 + 2√3 is irrational.

We have to prove 3 + 2√3 is irrational

Let us assume the opposite,

i.e., 3 + 2√3 is rational

Hence, 3 + 2√3 can be written in the form

where a and b (b+0) are co-prime (no common factor other than 1)

Hence, 3 + 2√3 =$\dfrac{a}{b}$

=2√3=$\dfrac{a}{b}$-3

=2√3=$\dfrac{a-3b}{b}$

√3=$\dfrac{1}{2}$×$\dfrac{a-3b}{b}$

√3=$\dfrac{a-3b}{2b}$

here √3 is irrational and $\dfrac{a-3b}{2b}$is rational

but √3 is irrational

since,

rational is not equal to irrational

this is a contradiction

therefore,our assumption is wrong

hence 3+2√3 is irrational

hence proved✓