Question

Prove that \[\left( 2\sqrt{3}+\sqrt{5} \right)\] is an irrational number. Also check whether \[\left( 2\sqrt{3}+\sqrt{5} \right)\left( 2\sqrt{3}-\sqrt{5} \right)\] is rational or irrational.

Answer 1

Hint:Consider the expression \[\left( 2\sqrt{3}+\sqrt{5} \right)\] as rational, so it is equal to \[{}^{p}/{}_{q}\]. Thus simplify the terms and find if the expression thus obtained is rational or not. If not then it’s irrational.

Complete step-by-step answer:
We know that a rational number can be written as a ratio of two integers, which is of the \[{}^{p}/{}_{q}\], where \[q\ne 0\] and p and q are both integers. But irrational numbers cannot be written as a ratio of two integers. One of the famous examples of irrational numbers is \[\pi \], where the value of \[\pi =3.14159265.....\].
We have been given to prove that \[\left( 2\sqrt{3}+\sqrt{5} \right)\] is irrational. So let us assume the opposite that \[\left( 2\sqrt{3}+\sqrt{5} \right)\] is rational.
Thus let us put that \[\left( 2\sqrt{3}+\sqrt{5} \right)={}^{p}/{}_{q}......(1)\]
\[\therefore \dfrac{1}{2\sqrt{3}+\sqrt{5}}=\dfrac{q}{p}\].
Now let us take its conjugate. Multiply \[\left( 2\sqrt{3}-\sqrt{5} \right)\] in the numerator and denominator. We get,
\[\dfrac{\left( 2\sqrt{3}-\sqrt{5} \right)}{\left( 2\sqrt{3}+\sqrt{5} \right)\left( 2\sqrt{3}-\sqrt{5} \right)}=\dfrac{q}{p}\].
The denominator is of the form,
\[\begin{align}
  & \left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}} \\
 & \therefore \dfrac{2\sqrt{3}-\sqrt{5}}{{{\left( 2\sqrt{3} \right)}^{2}}-{{\left( \sqrt{5} \right)}^{2}}}=\dfrac{q}{p} \\
 & \Rightarrow \dfrac{2\sqrt{3}-\sqrt{5}}{12-5}=\dfrac{q}{p}.......(2) \\
\end{align}\]
Now let us add the expression equation (1) and (2).
\[\left( 2\sqrt{3}+\sqrt{5} \right)+\left( 2\sqrt{3}-\sqrt{5} \right)=\dfrac{p}{q}+\dfrac{7q}{p}\].
Simplifying the above expression, we get,
\[\begin{align}
  & 4\sqrt{3}=\dfrac{{{p}^{2}}+7{{q}^{2}}}{pq} \\
 & \Rightarrow \sqrt{3}=\dfrac{{{p}^{2}}+7{{q}^{2}}}{4pq} \\
\end{align}\]
Now from the above expression we can say that \[\left( \dfrac{{{p}^{2}}+7{{q}^{2}}}{4pq} \right)\] is a rational number, but \[\sqrt{3}\] is an irrational number.
Since rational \[\ne \] irrational
This is a contradiction. Our assumption is wrong. Thus we can say that \[\left( 2\sqrt{3}+\sqrt{5} \right)\] is an irrational number.
\[\therefore \left( 2\sqrt{3}+\sqrt{5} \right)\] is an irrational number.
Now we need to check whether \[\left( 2\sqrt{3}+\sqrt{5} \right)\left( 2\sqrt{3}-\sqrt{5} \right)\] is rational or irrational.
Let us take \[\dfrac{p}{q}=\left( 2\sqrt{3}+\sqrt{5} \right)\left( 2\sqrt{3}-\sqrt{5} \right)\].
This is of the form \[\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}\].
\[\begin{align}
  & \therefore {}^{p}/{}_{q}={{\left( 2\sqrt{3} \right)}^{2}}-{{\left( \sqrt{5} \right)}^{2}}=\left( 4\times 3 \right)-5=12-5=7 \\
 & \therefore {}^{p}/{}_{q}={}^{7}/{}_{1} \\
\end{align}\]
Here it is rational as \[\dfrac{p}{q}=\dfrac{7}{1}\].
Hence both p and q are rational.
\[\therefore \left( 2\sqrt{3}+\sqrt{5} \right)\left( 2\sqrt{3}-\sqrt{5} \right)\] is a rational number.
Hence we proved that \[\left( 2\sqrt{3}+\sqrt{5} \right)\] is an irrational number and we also found that \[\therefore \left( 2\sqrt{3}+\sqrt{5} \right)\left( 2\sqrt{3}-\sqrt{5} \right)\] is a rational number.

Note:It is important to consider that the expression is rational and equal to \[\dfrac{p}{q}\] where p and q are integers and \[q\ne 0\] and take their inverse to formulate another expression, be careful while rationalizing so that we won’t mix up the signs.