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Q3(iii):
Prove that the following are irrationals : (iii) $6 + \sqrt{2}$
Solution :
Given: A real number $6 + \sqrt{2}$.
To Prove: $6 + \sqrt{2}$ is an irrational number.
Step 1: Assumption for Contradiction
To prove that $6 + \sqrt{2}$ is irrational, we begin by assuming the contrary. Let us assume that $6 + \sqrt{2}$ is a rational number. [By the method of contradiction].
Step 2: Definition of Rational Numbers
If $6 + \sqrt{2}$ is a rational number, then it can be expressed in the form $\frac{p}{q}$, where $p$ and $q$ are integers, $q \neq 0$, and $p$ and $q$ are coprime (i.e., they have no common factors other than 1).
Thus, we can write:
$6 + \sqrt{2} = \frac{p}{q}$
Step 3: Algebraic Rearrangement
We isolate the irrational part ($\sqrt{2}$) on one side of the equation to analyze its nature:
$\sqrt{2} = \frac{p}{q} - 6$
$\sqrt{2} = \frac{p - 6q}{q}$
Step 4: Analyzing the Right-Hand Side (RHS)
Since $p$ and $q$ are integers, $6$ is an integer, and the difference of two integers ($p - 6q$) is also an integer. Similarly, $q$ is an integer. Therefore, $\frac{p - 6q}{q}$ is a quotient of two integers, which satisfies the definition of a rational number. [Since the set of rational numbers is closed under subtraction and division].
Step 5: Analyzing the Left-Hand Side (LHS) and Identifying Contradiction
From Step 4, we have concluded that the RHS ($\frac{p - 6q}{q}$) is rational. This implies that the LHS ($\sqrt{2}$) must also be rational.
However, we know from the fundamental theorem of real numbers that $\sqrt{2}$ is an irrational number. [A well-established mathematical fact].
Step 6: Conclusion
Our assumption that $6 + \sqrt{2}$ is rational has led to a contradiction, as it implies $\sqrt{2}$ is rational, which is false. Therefore, our initial assumption must be incorrect.
Hence, it is proved that $6 + \sqrt{2}$ is an irrational number.
Final Answer: Since the assumption that $6 + \sqrt{2}$ is rational leads to a contradiction, $6 + \sqrt{2}$ is an irrational number.
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