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Q1:
Prove that $\sqrt{5}$ is irrational.
Solution :
Given: The number $\sqrt{5}$.
To Prove: $\sqrt{5}$ is an irrational number.
Step 1: Assumption for Contradiction
Let us assume, to the contrary, that $\sqrt{5}$ is a rational number. By the definition of rational numbers, if $\sqrt{5}$ is rational, it can be expressed in the form $\frac{a}{b}$, where $a$ and $b$ are integers, $b \neq 0$, and $a$ and $b$ are coprime (i.e., their highest common factor, $\text{HCF}(a, b) = 1$).
Step 2: Algebraic Manipulation
We have:
$\sqrt{5} = \frac{a}{b}$
Squaring both sides, we get:
$(\sqrt{5})^2 = (\frac{a}{b})^2$
$5 = \frac{a^2}{b^2}$
$a^2 = 5b^2$ --- (Equation 1)
Step 3: Applying the Fundamental Theorem of Arithmetic
From Equation 1, it is evident that $a^2$ is divisible by $5$.
[Theorem: If a prime number $p$ divides $a^2$, then $p$ divides $a$, where $a$ is a positive integer.]
Therefore, $a$ must also be divisible by $5$.
We can write $a = 5k$ for some integer $k$.
Step 4: Substitution and Further Deduction
Substitute $a = 5k$ into Equation 1:
$(5k)^2 = 5b^2$
$25k^2 = 5b^2$
Divide both sides by $5$:
$b^2 = 5k^2$
This implies that $b^2$ is divisible by $5$.
Following the same theorem as in Step 3, if $b^2$ is divisible by $5$, then $b$ must also be divisible by $5$.
Step 5: Analysis of the Contradiction
From Step 3 and Step 4, we have concluded that both $a$ and $b$ have at least $5$ as a common factor. This contradicts our initial assumption that $a$ and $b$ are coprime (i.e., $\text{HCF}(a, b) = 1$).
Step 6: Conclusion
The contradiction has arisen because of our incorrect assumption that $\sqrt{5}$ is a rational number. Therefore, we conclude that $\sqrt{5}$ must be irrational.
Final Answer: Since the assumption that $\sqrt{5}$ is rational leads to a contradiction, it is proven that $\sqrt{5}$ is an irrational number.
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