Prove that sqrt(6) is irrational.
Understand the Problem
The question is asking to prove that the square root of 6 is an irrational number. This involves demonstrating that it cannot be expressed as a fraction of two integers.
Answer
$\sqrt{6}$ is irrational.
Answer for screen readers
$\sqrt{6}$ is an irrational number.
Steps to Solve
- Assume the opposite for contradiction
Assume that $\sqrt{6}$ is a rational number. This means we can express it as a fraction of two integers:
$$ \sqrt{6} = \frac{a}{b} $$
where $a$ and $b$ are integers, and $b \neq 0$. We can also assume that $a$ and $b$ have no common factors (i.e., the fraction is in its simplest form).
- Square both sides
To eliminate the square root, we square both sides of the equation:
$$ 6 = \frac{a^2}{b^2} $$
This means:
$$ a^2 = 6b^2 $$
- Analyze divisibility
From the equation $a^2 = 6b^2$, we can deduce that $a^2$ is divisible by 6. Since 6 factors into $2 \cdot 3$, it follows that $a^2$ must be divisible by both 2 and 3. This implies that $a$ itself must be divisible by both 2 and 3 (because if a prime divides a square, it must also divide the number itself).
- Set $a$ in terms of $k$
Letâ€™s denote $a$ as follows:
$$ a = 6k $$
for some integer $k$.
- Substitute for $a$ and simplify
Now substitute $a$ back into the equation $a^2 = 6b^2$:
$$(6k)^2 = 6b^2$$
This simplifies to:
$$ 36k^2 = 6b^2 $$
Dividing both sides by 6:
$$ 6k^2 = b^2 $$
- Analyze $b^2$
From the equation $b^2 = 6k^2$, we see that $b^2$ is also divisible by 6, which means $b$ must also be divisible by both 2 and 3.
- Reach a contradiction
Since both $a$ and $b$ are divisible by 2 and 3, this contradicts our initial assumption that $a$ and $b$ had no common factors. Thus, our assumption that $\sqrt{6}$ is rational must be false.
$\sqrt{6}$ is an irrational number.
More Information
An irrational number cannot be expressed as a fraction of two integers, and $\sqrt{6}$ fits this description. Similar proofs exist for other square roots of non-perfect squares, reinforcing the concept of irrationality in mathematics.
Tips
- A common mistake is not recognizing that both $a$ and $b$ being divisible by 6 contradicts the original assumption about the fraction being in simplest form. To avoid this, always check if your assumptions lead to contradictions.