Question

Prove that 3 + 2√3 is an irrational number​

Answer 1

Answer:

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

So, it can be written in the form a/b

3 + 2√5 = a/b

Here a and b are coprime numbers and b ≠ 0

Solving 3 + 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 √5 is an irrational number.

So, it contradicts our assumption. Our assumption of 3 + 2√5 is a rational number is incorrect.

3 + 2√5 is an irrational number

Hence proved

Answer 2

Understanding the Problem

① $\sqrt{3}$ is a real number.

② A number that is real and isn't rational is irrational.

③ Rational numbers are closed under operations.

Solution

Method A

Let us suppose there is a pair of co-prime numbers $a,\ b$ that satisfies $\sqrt{3}=\dfrac{a}{b}$, on the contrary.

By definition,

$\implies 3=\left(\dfrac{a}{b} \right)^{2}$

$\implies a^{2}=3b^{2}$

By the fundamental theorem of arithmetic, $a$ has $3$ as a factor.

$\implies \boxed{a=3k_{1}}$

Returning to the previous equation,

$\implies 9k^{2}_{1}=3b^{2}$

$\implies b^{2}=3k^{2}_{1}$

By the fundamental theorem of arithmetic, $b$ has $3$ as a factor.

$\implies \boxed{b=3k_{2}}$

We see both numbers are divisible by 3. Hence, the two numbers are never co-primes.

This is a contradiction since we supposed $a$ and $b$ as co-prime rational numbers.

Hence, the assumption is wrong, $\sqrt{3}$ is an irrational number.

Again, assume on the contrary that $(\text{irrational})+(\text{rational})=(\text{rational})$.

$\implies (\text{rational})-(\text{rational})=(\text{irrational})$

This is impossible as rational numbers are closed under operations. So, the assumption is wrong.

$\implies (\text{rational})+(\text{irrational})=(\text{irrational})$

$\implies 3+2\sqrt{3} =(\text{irrational})\ \blacksquare$

Method B

Let $x=3+2\sqrt{3}$.

$\implies x-3=2\sqrt{3}$

$\implies (x-3)^{2}=(2\sqrt{3} )^{2}$

$\implies x^{2}-6x+9=12$

$\implies x^{2}-6x-3=0$

This equation has $x=3+2\sqrt{3}$ as a solution. By the rational root theorem, the possible roots are between $\pm1,\ \pm3$. But none of the roots are between these, hence the roots are irrational. $\blacksquare$