Question

Prove that $\sqrt{3}$ is an irrational number.

Hint:

This method of proof by contradiction works identically for any prime root, such as $\sqrt{2}$, $\sqrt{5}$, or $\sqrt{7}$.
Memorizing this logical structure is extremely helpful as it is a highly expected question in board exams!

Answer 1

Step 1: Understanding the Question:
We need to prove that $\sqrt{3}$ cannot be expressed as a ratio of two integers.
This is a standard theorem proved using the method of contradiction.

Step 2: Key Formula or Approach:
1. Assume $\sqrt{3}$ is a rational number.
2. Write $\sqrt{3} = \frac{p}{q}$, where $p$ and $q$ are co-prime integers and $q \neq 0$. Co-prime means they share no common factors other than 1.
3. Square both sides and rearrange to show that 3 divides both $p$ and $q$.
4. This contradicts the co-prime assumption, proving $\sqrt{3}$ must be irrational.
5. Use the lemma: "If a prime number $a$ divides $x^2$, then $a$ divides $x$."

Step 3: Detailed Explanation:

• Assume on the contrary that $\sqrt{3}$ is a rational number.
Therefore, we can find co-prime integers $p$ and $q$ ($q \neq 0$) such that:
\[ \sqrt{3} = \frac{p}{q} \]

• Square both sides of the equation:
\[ 3 = \frac{p^2}{q^2} \]
Rearrange the equation:
\[ p^2 = 3q^2 \] --- (Equation 1)

• Analyze divisibility for $p$:
Since $p^2 = 3q^2$, 3 divides $p^2$.
By the standard number theory lemma, if a prime number divides $p^2$, it must also divide $p$.
Therefore, 3 divides $p$.
So we can write:
\[ p = 3m \] for some integer m.

• Substitute $p = 3m$ back into Equation 1:
\[ (3m)^2 = 3q^2 \]
\[ 9m^2 = 3q^2 \]
Divide both sides by 3:
\[ q^2 = 3m^2 \] --- (Equation 2)

• Analyze divisibility for $q$:
Since $q^2 = 3m^2$, 3 divides $q^2$.
This implies that 3 must also divide $q$.

• Identify the contradiction:
From our derivations, we found that 3 is a common factor of both $p$ and $q$.
However, this directly contradicts our initial assumption that $p$ and $q$ are co-prime (having no common factors other than 1).
Therefore, our assumption that $\sqrt{3}$ is rational is incorrect.
Thus, $\sqrt{3}$ is an irrational number.


Step 4: Final Answer:
By contradiction, $\sqrt{3}$ is proved to be an irrational number.