question_answer

Prove that \[\sqrt{5}\] is an irrational number. Hence show that \[3+2\sqrt{5}\] is also an irrational number.

Answer:

Let \[\sqrt{5}\] be a rational number.

So,                               \[\sqrt{5}=\frac{p}{q}\]

On squaring both sides

\[5=\frac{{{p}^{2}}}{{{q}^{2}}}\]

\[{{q}^{2}}=\frac{{{p}^{2}}}{5}\]

\[\Rightarrow \]\[5\] is a factor of \[{{p}^{2}}\]

\[\Rightarrow \]\[5\] is a factor of \[p\]

Now, again let                \[p=5c\].

So,                               \[\sqrt{5}=\frac{5c}{q}\]

On squaring both sides

\[5=\frac{25{{c}^{2}}}{{{q}^{2}}}\]

\[{{q}^{2}}=5{{c}^{2}}\]

\[{{c}^{2}}=\frac{{{q}^{2}}}{5}\]

\[\Rightarrow \]\[5\] is factor of \[{{q}^{2}}\]

\[\Rightarrow \]\[5\] is a factor of \[q\].

Here \[5\] is a common factor of p, q which contradicts the fact that p, q are co-prime.

Hence our assumption is wrong, \[\sqrt{5}\] is an irrational number.

Now we have to show that \[3+2\sqrt{5}\] is an irrational number. So let us assume

\[3+2\sqrt{5}\] is a rational number.

\[\Rightarrow \]               \[3+2\sqrt{5}=\frac{p}{q}\]

\[\Rightarrow \]               \[2\sqrt{5}=\frac{p}{q}-3\]

\[\Rightarrow \]               \[2\sqrt{5}=\frac{p-3q}{q}\]

\[\Rightarrow \]               \[\sqrt{5}=\frac{p-3q}{2q}\]

\[\frac{p-3q}{2q}\] is in the rational form of \[\frac{p}{q}\] so \[\sqrt{5}\] is a rational number but we have already proved that \[\sqrt{5}\] is an irrational number so contradiction arises because we supposed wrong that \[3+2\sqrt{5}\] is  rational number. So we can say that \[3+2\sqrt{5}\] is an irrational number.                         Hence Proved