Question

Prove that 3 + 2√5 is irrational.

Answer 1

Hint: Firstly, assume that$3+2\sqrt{5}$ is rational. Now, it’s a rational number so equate it to$\dfrac{a}{b}$. Then cross multiply and after cross multiplication and simplification, you will find that$\sqrt{5}$ is equal to some rational expression which is a contradiction because$\sqrt{5}$ is an irrational number. So, we proved by contradiction.

Complete step-by-step answer:
Let us assume that $3+2\sqrt{5}$ is a rational number. So,$3+2\sqrt{5}$ is equal to $\dfrac{a}{b}$.
$3+2\sqrt{5}=\dfrac{a}{b}$
In the above expression, “a” & “b” are co prime (meaning HCF (a, b) is equal to 1).
Now, cross multiplying the above expression we get,
$\begin{align}
  & 3b+2\sqrt{5}b=a \\
 & \Rightarrow 2\sqrt{5}=\dfrac{a-3b}{b} \\
 & \Rightarrow \sqrt{5}=\dfrac{a-3b}{2b} \\
\end{align}$
Now, the expression $\dfrac{a-3b}{2b}$ is a rational number because the definition of a rational number states that it is a number which is in the form of $\dfrac{p}{q}$ where p and q are integers and $q\ne 0$.
And we know that $\sqrt{5}$ is an irrational number.
So, irrational number ≠ rational number
Hence, the assumption that we have made that $3+2\sqrt{5}$ is a rational number contradicts with the result that we are getting above.
Hence, $3+2\sqrt{5}$ is an irrational number.

Note: To prove $\sqrt{5}$ is an irrational number, the proof is similar to the one that we have done above by assuming$\sqrt{5}$ is a rational number and equate it to $\dfrac{a}{b}$ then cross multiply and squaring both the sides will give:
$5{{b}^{2}}={{a}^{2}}$
From the above expression we can say that ${{a}^{2}}$ is divisible by 5 and 5 is prime number so a must be divisible by 5. We can write a = 5q and q must be a positive integer.
a = 5q
Squaring on both the sides will give:
$\begin{align}
  & {{a}^{2}}={{\left( 5 \right)}^{2}}{{q}^{2}} \\
 & 5{{b}^{2}}={{a}^{2}}=5.5.{{q}^{2}} \\
 & {{b}^{2}}=5{{q}^{2}} \\
\end{align}$
$\begin{align}
  & \dfrac{{{b}^{2}}}{{{q}^{2}}}=5 \\
 & \dfrac{b}{q}=\sqrt{5} \\
\end{align}$
Now, a > b > q and the above expression is showing that (b, q) is the smallest integer pair whose quotient is $\sqrt{5}$ which is contradiction with the assumption that “a” and “b” are the only pair. Hence,$\sqrt{5}$ is an irrational number.