Question

Prove that 3 +2√3 is irrational​

Answer 1

Solution :-

$let \: 3 \: + 2 \sqrt{3} \: is \: rational \: \\ therefore \: we \: can \: find \: two \: \\ co \: prime \: integers \: a \: and \: b \: \\ (b \: not \: equal \: to \: 0) \\ 3 \: + \: 2 \sqrt{3} = \: a \div b \: \\ \\ 2 \sqrt{3} \: = \: a \div b \: - 3 \\ \sqrt{3} \: = \: 1 \div 2(a \: \div \: b \: - \: 3) \\ since \: a \: and \: b \: are \: integers \: \: \\ 1 \div 2(a \div b \: - \: 3) \: will \: also \\ \: be \: a \: rational \: and \: therefore \: \\ \sqrt{3} \: is \: also \: rational \: \\ this \: contradicts \: the \: fact \: that \\ \sqrt{3} \: is \: irrational \: . \: hence \: our \\ \: assumption \: \: that \: 3 \: + \: 2 \sqrt{3} \\ is \: rational \: is \: false \: . \: therefore \: \\ 3 \: + \: 2 \sqrt{3} \: is \: irrational$

Answer 2

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Given:

• 3+2√3

To Find:

• 3+2√3 is an irrational number

Solution :

First, shall we prove that √3 is an irrational number.

So,

Let us assume that√3 is an rational number, a and b are co prime numbers and b ≠ 0.

√3 = a/b

b√5 = a

Square on both sides:

(b√3)² = a²

3b²= a² -------(★)

b²=a²/3

•°• 3 is an prime number

If 3 divides a² then 3 divides a also

3 is a factor of a ------(1)

now,

assume that a = 3c in (★)

3b²=(3c)²

3b²=9c²

c²=b²/3

b² is divisible by 3

•°• 3 is a factor of b -------(2)

From (1) and (2)

a and b are not co prime number because HCF is 1.

So, a and b is not a rational number, our assumption is wrong.

So, √3 is irrational number.

now,

Proof:

Let us assume that 3 + 2 √3 is an rational number

3+2√3 =a/b So, a and b are co prime numbers.

3+2√3=a/b

2√3 = a/b-3

√3 =a-3b/2b

a-3b/2b is a rational number

we know that,

√3 is an irrational number.

Our assumption is wrong

Hence 3+2√3 is an irrational number.

@BrainlyUniverse