Question

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

Hint:

This proof template is identical for any prime root, such as $\sqrt{2}$, $\sqrt{5}$, or $\sqrt{7}$.
Memorizing the logical flow of this contradiction proof is highly valuable as it is a very common question in board exams.

Answer 1

Step 1: Understanding the Question:
The topic is Real Numbers, specifically proving the irrationality of a number.
An irrational number is a real number that cannot be expressed as a ratio of two integers.
To prove that $\sqrt{3}$ is irrational, we will use the method of proof by contradiction.

Step 2: Key Formula or Approach:
We assume the opposite of what we want to prove: that $\sqrt{3}$ is rational.
A rational number can be expressed as $\frac{a}{b}$, where $a$ and $b$ are co-prime integers and $b \neq 0$.
Co-prime integers mean that $a$ and $b$ share no common factors other than 1.
We will also use the theorem: if $p$ is a prime number and $p$ divides $a^2$, then $p$ divides $a$.

Step 3: Detailed Explanation:

• Let us assume, on the contrary, that $\sqrt{3}$ is a rational number.
Therefore, we can write:
\[ \sqrt{3} = \frac{a}{b} \]
where $a$ and $b$ are integers such that $b \neq 0$, and $a$ and $b$ are co-prime (they have no common factors other than 1).

• Rearranging the terms, we get:
\[ a = b\sqrt{3} \]

• Squaring both sides of the equation:
\[ a^2 = 3b^2 \quad \text{--- (Equation 1)} \]

• From Equation 1, we can see that $3$ divides $a^2$ (since $a^2$ is a multiple of 3).
By theorem, if a prime number divides $a^2$, it must also divide $a$.
Therefore, $3$ divides $a$.

• Since 3 divides $a$, we can write $a$ in terms of some integer $c$:
\[ a = 3c \]
where $c$ is an integer.

• Substitute $a = 3c$ into Equation 1:
\[ (3c)^2 = 3b^2 \]
\[ 9c^2 = 3b^2 \] Divide both sides by 3:
\[ 3c^2 = b^2 \quad \text{--- (Equation 2)} \]

• From Equation 2, we see that $3$ divides $b^2$.
Therefore, by the same theorem, $3$ must also divide $b$.

• From our steps, we have shown that 3 divides $a$ and 3 also divides $b$.
This means that $a$ and $b$ have a common factor of 3.

• This contradicts our initial assumption that $a$ and $b$ are co-prime (having no common factors other than 1).
This contradiction has arisen because of our incorrect assumption that $\sqrt{3}$ is rational.


Step 4: Final Answer:
Hence, by contradiction, $\sqrt{3}$ is an irrational number.