Question

prove that 3+2 root 5 is irrational where root 5 is irrational​

Answer 1

Step-by-step explanation:

Given Question:-

Prove that 3+2√5 is am irrational number where √5 is an irrational number

Proof:-

Let us assume that 3+2√5 is a rational number

So it must be in the form of p/q

Let 3+2√5 = a/b

Where a and b are co primes

3 = (a/b)-2√5

On squaring both sides then

3^2 =[(a/b) - 2√5]^2

We know that (a-b)^2 = a^2 -2ab +b^2

=>9 = (a/b)^2 -2(a/b)(2√5) +(2√5)^2

=>9 = (a^2/b^2) - 4√5a/b + 20

=>9 -20 = (a^2/b^2) - 4√5 a/b

=>-11 = (a^2/b^2) - 4√5 a/b

=>-11 -(a^2/b^2) = - 4√5a/b

=>-[11+(a^2/b^2)] = -4√5 a/b

=>11 +(a^2/b^2) = 4√5a/b

=>(11b^2+a^2)/b^2 =4√5a/b

=>(11b^2+a^2)/b^2×(b/a) = 4√5

=> (11b^2+a^2)/ab = 4√5

=>(11b^2+a^2)/4ab = √5

=>√5 = (11b^2+a^2)/4ab

=>√5 is in the form of p/q

=>It is a rational number

But given that √5 is an irrational number

So, This is contradiction to our assumption that

3+2√5 is a rational number

Therefore, It is not a rational number

3+2√5 is an irrational number.

Hence, Proved

(or)

Let us assume that 3+2√5 is a rational number

So it must be in the form of p/q

Let 3+2√5 = a/b

Where a and b are co primes

=>2√5 = (a/b)-3

=>2√5 = (a-3b)/b

=>√5 = (a-3b)/2b

=>√5 is in the form of p/q

=>It is a rational number

But given that √5 is an irrational number

So, This is contradiction to our assumption that

3+2√5 is a rational number

Therefore, It is not a rational number

3+2√5 is an irrational number.

Hence, Proved

Answer:-

3+2√5 is an irrational number

Used Method:-

Method of Contradiction or Indirect method

Used formula:-

• (a-b)^2 = a^2 -2ab +b^2