REAL NUMBERS | Irrational Numbers | Ex-3

Prove that \( \sqrt[3]{6} \) is Not a Rational Number

Question:

Prove that \( \sqrt[3]{6} \) is an irrational number.

Solution:

We will use the method of contradiction.

Step 1: Assume the contrary

Suppose that \( \sqrt[3]{6} \) is a rational number.

Then it can be expressed as a fraction in lowest terms:

\( \sqrt[3]{6} = \frac{p}{q} \), where \( p \) and \( q \) are integers with no common factors other than 1, and q ≠ 0 .

Step 2: Cube both sides

\( \left( \frac{p}{q} \right)^3 = 6 \)

\( \frac{p^3}{q^3} = 6 \)

\( p^3 = 6q^3 \)

This means that \( p^3 \) is divisible by 2 and 3, so \( p \) must also be divisible by both 2 and 3 (since the cube of a number divisible by a prime must have that prime as a factor).

So we can write \( p = 6k \) for some integer \( k \).

Step 3: Substitute back into the equation

\( (6k)^3 = 6q^3 \)

\( 216k^3 = 6q^3 \)

\( 36k^3 = q^3 \)

This implies \( q^3 \) is divisible by 6, so \( q \) is also divisible by both 2 and 3.

However, this implies that both \( p \) and \( q \) share common factors (2 and 3), which goes against the initial assumption that \( \frac{p}{q} \) is in its simplest form.

Conclusion:

This contradiction shows that our assumption was wrong. Hence, \( \sqrt[3]{6} \) is not a rational number. That is, it is irrational.