
Prove that $5-2\sqrt{3}$ is an irrational number.
Answer
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Hint: First assume that $5-2\sqrt{3}$ is a rational number and represent it in the form of $\dfrac{p}{q}$. Then arrange the values in the LHS equal to $\sqrt{3}$ and you will find a rational number in RHS equal to $\sqrt{3}$. That will contradict your assumption that $5-2\sqrt{3}$ is a rational number because $\sqrt{3}$ is an irrational number.
Complete step-by-step answer:
Let us assume that the given expression, $5-2\sqrt{3}$ is a rational number.
As, $5-2\sqrt{3}$ is a rational number.
So, $5-2\sqrt{3}$ can be written in the form of $\dfrac{p}{q}$ where p and q are integers, having no common factor except 1 and q≠0.
i.e. $5-2\sqrt{3}$= $\dfrac{p}{q}$ (p and q are co-prime and q≠0)
$\begin{align}
& \Rightarrow -2\sqrt{3}=\dfrac{p}{q}-5 \\
& \Rightarrow -2\sqrt{3}=\dfrac{p-5q}{q} \\
& \Rightarrow -\left( \dfrac{p-5q}{2q} \right)=\sqrt{3} \\
\end{align}$ $\begin{align}
& \Rightarrow -2\sqrt{3}=\dfrac{p}{q}-5 \\
& \Rightarrow -2\sqrt{3}=\dfrac{p-5q}{q} \\
& \Rightarrow \sqrt{3}=-\left( \dfrac{p-5q}{2q} \right) \\
\end{align}$
Since, $-\left( \dfrac{p-5q}{2q} \right)$ is a rational number as p and q are integers.
That means, $-\left( \dfrac{p-5q}{2q} \right)=\sqrt{3}$, should also be a rational number.
That gives, $\sqrt{3}$ is a rational number.
But we know that $\sqrt{3}$ is an irrational number.
We know that,
Irrational ≠ rational
So, our assumption is wrong.
We contradict the statement that $5-2\sqrt{3}$ is rational.
Therefore, $5-2\sqrt{3}$ is an irrational number.
Note: In this question, students do not need to prove $\sqrt{3}$ is an irrational number, students can use this information without proving it to prove the given expression $5-2\sqrt{3}$ is an irrational number.
Complete step-by-step answer:
Let us assume that the given expression, $5-2\sqrt{3}$ is a rational number.
As, $5-2\sqrt{3}$ is a rational number.
So, $5-2\sqrt{3}$ can be written in the form of $\dfrac{p}{q}$ where p and q are integers, having no common factor except 1 and q≠0.
i.e. $5-2\sqrt{3}$= $\dfrac{p}{q}$ (p and q are co-prime and q≠0)
$\begin{align}
& \Rightarrow -2\sqrt{3}=\dfrac{p}{q}-5 \\
& \Rightarrow -2\sqrt{3}=\dfrac{p-5q}{q} \\
& \Rightarrow -\left( \dfrac{p-5q}{2q} \right)=\sqrt{3} \\
\end{align}$ $\begin{align}
& \Rightarrow -2\sqrt{3}=\dfrac{p}{q}-5 \\
& \Rightarrow -2\sqrt{3}=\dfrac{p-5q}{q} \\
& \Rightarrow \sqrt{3}=-\left( \dfrac{p-5q}{2q} \right) \\
\end{align}$
Since, $-\left( \dfrac{p-5q}{2q} \right)$ is a rational number as p and q are integers.
That means, $-\left( \dfrac{p-5q}{2q} \right)=\sqrt{3}$, should also be a rational number.
That gives, $\sqrt{3}$ is a rational number.
But we know that $\sqrt{3}$ is an irrational number.
We know that,
Irrational ≠ rational
So, our assumption is wrong.
We contradict the statement that $5-2\sqrt{3}$ is rational.
Therefore, $5-2\sqrt{3}$ is an irrational number.
Note: In this question, students do not need to prove $\sqrt{3}$ is an irrational number, students can use this information without proving it to prove the given expression $5-2\sqrt{3}$ is an irrational number.
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