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Q3(i):
Prove that the following are irrationals : (i) $\frac{1}{\sqrt{2}}$
Solution :
Given: A number $\frac{1}{\sqrt{2}}$.
To Prove: $\frac{1}{\sqrt{2}}$ is an irrational number.
Step 1: Assumption for Proof by Contradiction
Let us assume, to the contrary, that $\frac{1}{\sqrt{2}}$ is a rational number. By the definition of rational numbers, if a number is rational, 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., their greatest common divisor $\text{gcd}(p, q) = 1$).
Therefore, we can write:
$\frac{1}{\sqrt{2}} = \frac{a}{b}$
where $a$ and $b$ are integers, $b \neq 0$, and $\text{gcd}(a, b) = 1$.
Step 2: Algebraic Manipulation
To isolate the radical term, we rearrange the equation:
$\frac{1}{\sqrt{2}} = \frac{a}{b}$
Taking the reciprocal of both sides:
$\sqrt{2} = \frac{b}{a}$
Step 3: Analyzing the Rationality
Since $a$ and $b$ are integers, the quotient $\frac{b}{a}$ must be a rational number by the definition of rational numbers (the ratio of two integers is rational).
[Since $a, b \in \mathbb{Z}$ and $a \neq 0$, $\frac{b}{a} \in \mathbb{Q}$]
This implies that $\sqrt{2}$ is a rational number.
Step 4: Identifying the Contradiction
We know from the fundamental theorems of real numbers that $\sqrt{2}$ is an irrational number. Our assumption that $\frac{1}{\sqrt{2}}$ is rational has led to the conclusion that $\sqrt{2}$ is rational, which contradicts the established mathematical fact that $\sqrt{2}$ is irrational.
[Since the assumption leads to a contradiction, the initial assumption must be false.]
Step 5: Conclusion
Because the assumption that $\frac{1}{\sqrt{2}}$ is rational leads to a contradiction, we must conclude that $\frac{1}{\sqrt{2}}$ is irrational.
Final Answer: Since the assumption that $\frac{1}{\sqrt{2}}$ is rational leads to a contradiction, it is proven that $\frac{1}{\sqrt{2}}$ is an irrational number.
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