Question

$prove \: that \: 3 + 2 \sqrt{5} \: is \: irrational$
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Answer 1

Let us assume, to the contrary, that 3 + 2√5 is a rational number.

Now,

3 + 2√5 = p/q [ Where p and q are co - primes and q is not equal to zero ]

So,

⇒ p/q - 3 = 2√5

⇒ √5 = p/2q -3/2

Since, p and q are integers, we get p/2q -3/2 is rational number and so, √5 is rational number.

But this contradicts the fact that √5 is irrational number.

This shows that our assumption is incorrect.

So, we concluded that 3 + 2√5 is an irrational number.

Answer 2

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

$\implies \bold{3+2\sqrt{5}= \frac{p}{q} \;\;[where\;p\;and\;q\;are\;co-primes,q\neq 0]}\\\\\implies \bold{2\sqrt{5}=\frac{p}{q}-3 }\\\\\implies \bold{2\sqrt{5}=\frac{(p-3q)}{q} }\\\\\implies \bold{\sqrt{5}=\frac{(p-3q)}{2q}}$

So our assumption is wrong that 3+2√5 is rational number . Since we know that √5 is an irrational number.

So 3+2√5 is an irrational number.