Prove that the following are irrationals : (i) (ii) (iii)

Question

Homi · Accepted Answer

Understand the Question

To prove that a number is irrational, we use the method of contradiction:
1. Assume the given number is rational and can be written in the form $\dfrac{a}{b}$ , where $a$ and $b$ are co-prime integers and $b \neq 0$ .
2. Rearrange the equation to isolate the known irrational square root (e.g., $\sqrt{2}$ or $\sqrt{5}$ ) on one side.
3. Since integers are closed under basic arithmetic operations, the expression of integers on the other side is rational, which forces the irrational square root to be rational.
4. This contradiction proves our initial assumption was false, hence the number must be irrational.

(i) Prove that $\dfrac{1}{\sqrt{2}}$ is irrational.

Step 1 · Assume Rationality and Derive Contradiction

Let us assume, to the contrary, that $\dfrac{1}{\sqrt{2}}$ is rational.

Then there exist co-prime integers $a$ and $b$ ( $b \neq 0$ ) such that: $\dfrac{1}{\sqrt{2}} = \dfrac{a}{b}$

Taking the reciprocal on both sides: $\sqrt{2} = \dfrac{b}{a}$

Since $a$ and $b$ are integers with $a \neq 0$ , $\dfrac{b}{a}$ is a rational number.

This implies that $\sqrt{2}$ is rational.

However, this contradicts the fact that $\sqrt{2}$ is irrational.

Therefore, our assumption is false, and $\dfrac{1}{\sqrt{2}}$ is irrational.

Answer

(i) $\dfrac{1}{\sqrt{2}}$ is irrational.

(ii) Prove that $7\sqrt{5}$ is irrational.

Step 1 · Assume Rationality and Derive Contradiction

Let us assume, to the contrary, that $7\sqrt{5}$ is rational.

Then there exist co-prime integers $a$ and $b$ ( $b \neq 0$ ) such that: $7\sqrt{5} = \dfrac{a}{b}$

Isolating $\sqrt{5}$ : $\sqrt{5} = \dfrac{a}{7b}$

Since $a$ , $b$ , and $7$ are integers with $b \neq 0$ , $\dfrac{a}{7b}$ is a rational number.

This implies that $\sqrt{5}$ is rational.

However, this contradicts the fact that $\sqrt{5}$ is irrational.

Therefore, our assumption is false, and $7\sqrt{5}$ is irrational.

Answer

(ii) $7\sqrt{5}$ is irrational.

(iii) Prove that $6 + \sqrt{2}$ is irrational.

Step 1 · Assume Rationality and Derive Contradiction

Let us assume, to the contrary, that $6 + \sqrt{2}$ is rational.

Then there exist co-prime integers $a$ and $b$ ( $b \neq 0$ ) such that: $6 + \sqrt{2} = \dfrac{a}{b}$

Isolating $\sqrt{2}$ : $\sqrt{2} = \dfrac{a}{b} - 6 = \dfrac{a - 6b}{b}$

Since $a$ , $b$ , and $6$ are integers with $b \neq 0$ , $\dfrac{a - 6b}{b}$ is a rational number.

This implies that $\sqrt{2}$ is rational.

However, this contradicts the fact that $\sqrt{2}$ is irrational.

Therefore, our assumption is false, and $6 + \sqrt{2}$ is irrational.

Answer

(iii) $6 + \sqrt{2}$ is irrational.

Common Mistakes

Skipping Co-prime Condition: Forgetting to state that $a$ and $b$ are co-prime integers with $b \neq 0$ .
Proving $\sqrt{2}$ or $\sqrt{5}$ from Scratch: You can directly state that $\sqrt{2}$ and $\sqrt{5}$ are known irrationals when proving composite expressions like $6 + \sqrt{2}$ or $7\sqrt{5}$ , unless explicitly asked to prove them first.
Algebraic Sign Errors: Making sign mistakes while rearranging terms, such as writing $\sqrt{2} = \dfrac{6b - a}{b}$ instead of $\dfrac{a - 6b}{b}$ .

Question 3