Prove that √3 is irrational.

Understand the Problem

The question is asking for a proof that the square root of 3 is an irrational number. This means demonstrating that it cannot be expressed as a fraction p/q, where p and q are integers and q is not zero. A common method is proof by contradiction.

Answer

$\sqrt{3}$ is irrational.

Steps to Solve

Assume the opposite

Assume that $\sqrt{3}$ is rational. This means we can write it as a fraction $\frac{p}{q}$, where $p$ and $q$ are integers, and $q \neq 0$. We also assume that this fraction is in its simplest form, meaning $p$ and $q$ have no common factors (other than 1).

Write the equation and square both sides

If $\sqrt{3} = \frac{p}{q}$, then squaring both sides gives us:

$$3 = \frac{p^2}{q^2}$$

Rearrange the equation

Multiply both sides by $q^2$ to get:

$$3q^2 = p^2$$

Deduce that $p^2$ is a multiple of 3

From the equation $3q^2 = p^2$, we can see that $p^2$ is a multiple of 3, since it equals 3 times $q^2$.

Deduce that $p$ is a multiple of 3

If $p^2$ is a multiple of 3, then $p$ must also be a multiple of 3. This is because if $p$ were not a multiple of 3, it could be written as $3k+1$ or $3k+2$. Squaring these gives $(3k+1)^2 = 9k^2 + 6k + 1 = 3(3k^2 + 2k) + 1$ or $(3k+2)^2 = 9k^2 + 12k + 4 = 3(3k^2 + 4k + 1) + 1$. In both cases, the result is not a multiple of 3. Therefore, $p$ must be a multiple of 3.

Express $p$ as a multiple of 3

Since $p$ is a multiple of 3, we can write $p = 3k$ for some integer $k$.

Substitute $p = 3k$ back into the equation

Substitute $p = 3k$ into the equation $3q^2 = p^2$:

$$3q^2 = (3k)^2$$ $$3q^2 = 9k^2$$

Simplify the equation

Divide both sides by 3:

$$q^2 = 3k^2$$

Deduce that $q^2$ is a multiple of 3

From the equation $q^2 = 3k^2$, we can see that $q^2$ is a multiple of 3.

Deduce that $q$ is a multiple of 3

If $q^2$ is a multiple of 3, then $q$ must also be a multiple of 3 (using the same logic as in step 5).

State the contradiction

We have now shown that both $p$ and $q$ are multiples of 3. This contradicts our initial assumption that $\frac{p}{q}$ was in its simplest form and that $p$ and $q$ have no common factors (other than 1).

Conclude that $\sqrt{3}$ is irrational

Since our assumption that $\sqrt{3}$ is rational leads to a contradiction, our initial assumption must be false. Therefore, $\sqrt{3}$ is irrational.

$\sqrt{3}$ is irrational.

More Information

This proof follows the classic proof by contradiction method, similar to the proof that $\sqrt{2}$ is irrational.

Tips

A common mistake is failing to explicitly state the initial assumption that the fraction $p/q$ is in its simplest form. Without this assumption, the contradiction is not as clear. Another mistake is not explicitly explaining why if $p^2$ is divisible by 3, then $p$ must also be divisible by 3.