Q3(ii):

Prove that the following are irrationals : (ii) $7\sqrt{5}$

Given: A number $7\sqrt{5}$.

To Prove: That $7\sqrt{5}$ is an irrational number.

Step 1: Assumption for the Method of Contradiction
To prove that $7\sqrt{5}$ is irrational, we assume the contrary. Let us assume that $7\sqrt{5}$ is a rational number.

Step 2: Definition of Rational Numbers
If $7\sqrt{5}$ is a rational number, then by definition, it can be expressed in the form $\frac{a}{b}$, where $a$ and $b$ are integers, $b \neq 0$, and $a$ and $b$ are co-prime (i.e., their highest common factor, $HCF(a, b) = 1$).

So, we can write:
$7\sqrt{5} = \frac{a}{b}$

Step 3: Algebraic Manipulation
We isolate the square root term on one side of the equation to analyze its nature.
Divide both sides of the equation by $7$:
$\sqrt{5} = \frac{a}{7b}$

Step 4: Analyzing the Rationality of the Expression
Since $a$ and $b$ are integers, $7$ is also an integer. Therefore, the expression $\frac{a}{7b}$ is a ratio of two integers. By the definition of rational numbers, any number that can be expressed as a quotient of two integers (where the denominator is non-zero) is a rational number.

Thus, $\frac{a}{7b}$ is a rational number.

Step 5: Identifying the Contradiction
From the equation in Step 3, we have:
$\sqrt{5} = \text{a rational number}$

However, it is a well-established mathematical fact (proven via the Fundamental Theorem of Arithmetic) that $\sqrt{5}$ is an irrational number.

[Since the assumption that $7\sqrt{5}$ is rational leads to the conclusion that $\sqrt{5}$ is rational, which contradicts the known fact that $\sqrt{5}$ is irrational, our initial assumption must be false.]

Step 6: Conclusion
Since the assumption that $7\sqrt{5}$ is rational leads to a contradiction, we conclude that $7\sqrt{5}$ must be irrational.

Final Answer: Since the assumption that $7\sqrt{5}$ is rational leads to a contradiction, it is proven that $7\sqrt{5}$ is an irrational number.