Answer
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Hint:
Here, we have to prove that the given number is an irrational number. An irrational number is a number which cannot be expressed as a fraction or the ratio of two integers. An irrational number can neither be terminating or recurring.
Complete step by step solution:
We have to prove that $\left( 5-2\sqrt{3} \right)$ is an irrational number.
To prove that the number $\left( 5-2\sqrt{3} \right)$ is an irrational number, we have to prove that the number $5-2\sqrt{3}$is not a rational number.
A rational number is a number which can be expressed as a fraction or the ratio of two integers as $\dfrac{p}{q}$.
Now, considering $5-2\sqrt{3}$ as a rational number.
Expressing the number as rational number, we get
\[ \Rightarrow 5 - 2\sqrt 3 = \dfrac{p}{q}\]
Rewriting the equation, we get
\[ \Rightarrow - 2\sqrt 3 = \dfrac{p}{q} - 5\]
By taking L.C.M, and adding the like terms, we get
\[ \Rightarrow - 2\sqrt 3 = \dfrac{p}{q} - 5 \times \dfrac{q}{q}\]
$\Rightarrow -2\sqrt{3}=\dfrac{p}{q}-\dfrac{5q}{q}$
$\Rightarrow -2\sqrt{3}=\dfrac{p-5q}{q}$
Rewriting the equation, we get
\[ \Rightarrow \sqrt 3 = \dfrac{{p - 5q}}{{ - 2q}}\]
We know that the surd is always an irrational number.
So, $\sqrt{3}$ is an irrational number whereas $\dfrac{p-5q}{-2q}$ is a rational number.
So, we have $L.H.S\ne R.H.S$.
So, our assumption that $5-2\sqrt{3}$ is a rational number is wrong.
Hence, $5-2\sqrt{3}$ is an irrational number.
Therefore, $5-2\sqrt{3}$ is an irrational number.
Hence, proved.
Note:
We should know the properties of irrational numbers such that addition of two irrational numbers may or may not be irrational. Subtraction of two irrational numbers may or may not be irrational. But the difference between a rational number and an irrational number is always irrational.
We know that surd is defined as the number which cannot be simplified to find the square root. Every rational number is not a surd but every irrational number is a surd. Here, we have used the concept of contradiction. Contradiction is a method of proving the statement true by showing the assumption to be false. This method is quite easy to prove that $5-2\sqrt{3}$ is an irrational number.
Here, we have to prove that the given number is an irrational number. An irrational number is a number which cannot be expressed as a fraction or the ratio of two integers. An irrational number can neither be terminating or recurring.
Complete step by step solution:
We have to prove that $\left( 5-2\sqrt{3} \right)$ is an irrational number.
To prove that the number $\left( 5-2\sqrt{3} \right)$ is an irrational number, we have to prove that the number $5-2\sqrt{3}$is not a rational number.
A rational number is a number which can be expressed as a fraction or the ratio of two integers as $\dfrac{p}{q}$.
Now, considering $5-2\sqrt{3}$ as a rational number.
Expressing the number as rational number, we get
\[ \Rightarrow 5 - 2\sqrt 3 = \dfrac{p}{q}\]
Rewriting the equation, we get
\[ \Rightarrow - 2\sqrt 3 = \dfrac{p}{q} - 5\]
By taking L.C.M, and adding the like terms, we get
\[ \Rightarrow - 2\sqrt 3 = \dfrac{p}{q} - 5 \times \dfrac{q}{q}\]
$\Rightarrow -2\sqrt{3}=\dfrac{p}{q}-\dfrac{5q}{q}$
$\Rightarrow -2\sqrt{3}=\dfrac{p-5q}{q}$
Rewriting the equation, we get
\[ \Rightarrow \sqrt 3 = \dfrac{{p - 5q}}{{ - 2q}}\]
We know that the surd is always an irrational number.
So, $\sqrt{3}$ is an irrational number whereas $\dfrac{p-5q}{-2q}$ is a rational number.
So, we have $L.H.S\ne R.H.S$.
So, our assumption that $5-2\sqrt{3}$ is a rational number is wrong.
Hence, $5-2\sqrt{3}$ is an irrational number.
Therefore, $5-2\sqrt{3}$ is an irrational number.
Hence, proved.
Note:
We should know the properties of irrational numbers such that addition of two irrational numbers may or may not be irrational. Subtraction of two irrational numbers may or may not be irrational. But the difference between a rational number and an irrational number is always irrational.
We know that surd is defined as the number which cannot be simplified to find the square root. Every rational number is not a surd but every irrational number is a surd. Here, we have used the concept of contradiction. Contradiction is a method of proving the statement true by showing the assumption to be false. This method is quite easy to prove that $5-2\sqrt{3}$ is an irrational number.
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