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# Prove that $2+\sqrt{5}$ is an irrational number.

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Hint: Prove that $\sqrt{5}$ is an irrational number by assuming that it is a rational number and then using contradiction to prove that it is an irrational number. Use the fact that the sum of a rational number and an irrational number is an irrational number to prove that $2+\sqrt{5}$ is an irrational number.

We have to prove that $2+\sqrt{5}$ is an irrational number. We will firstly prove that $\sqrt{5}$ is an irrational number. We will prove this by contradiction technique.
Letâ€™s assume that $\sqrt{5}$ is a rational number. We know that any rational number can be written in the form $\dfrac{a}{b}$ where $a$ and $b$ are co-prime numbers and $b\ne 0$.
Thus, we have $\sqrt{5}=\dfrac{a}{b}$. Squaring on both sides, we get $5=\dfrac{{{a}^{2}}}{{{b}^{2}}}$.
Rearranging the terms, we get ${{a}^{2}}=5{{b}^{2}}......\left( 1 \right)$.
We see that $5$ divides ${{a}^{2}}$. We know that if a prime number $p$ divides ${{a}^{2}}$, then $p$ must divide $a$ as well, where $a$ is a positive integer.
Thus, as $5$ divides ${{a}^{2}}$, $5$ must divide $a$ as well.
Hence, we can write $a=5c$ for some positive integer $c$.
Substituting the value $a=5c$ in equation $\left( 1 \right)$, we have $25{{c}^{2}}=5{{b}^{2}}\Rightarrow {{b}^{2}}=5{{c}^{2}}$.
As ${{b}^{2}}=5{{c}^{2}}$, we observe that $5$ divides ${{b}^{2}}$. So, $5$ divides $b$ as well using the fact that if a prime number $p$ divides ${{a}^{2}}$, then $p$ must divide $a$ as well, where $a$ is a positive integer.
Thus, $5$ divides both $a$ and $b$. But this contradicts our assumption that $a$ and $b$ are coprimes.
Hence, our assumption is wrong that $\sqrt{5}$ is a rational number.
We conclude that $\sqrt{5}$ is an irrational number.
We can write $2$ as $\dfrac{2}{1}$, thus observing that it is a rational number.
We know that a sum of a rational number and an irrational number is an irrational number.
Hence, we observe that $2+\sqrt{5}$ is an irrational number.

Note: Itâ€™s necessary to keep the definition of rational numbers in mind which states that any rational number can be written in the form $\dfrac{a}{b}$ where $a$ and $b$ are co-prime numbers and $b\ne 0$, while irrational numbers are all those real numbers which are not rational numbers. Also, itâ€™s necessary to use the theorem which states that if a prime number $p$ divides ${{a}^{2}}$, then $p$ must divide $a$ as well, where $a$ is a positive integer. We also observe that the square root of any prime number will be an irrational number.
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