Prove that sqrt(6) is irrational.
Understand the Problem
The question is asking to demonstrate that the square root of 6 cannot be expressed as a fraction of two integers, which is the definition of an irrational number. To prove this, we can use a proof by contradiction. We assume that sqrt(6) is rational and can be expressed as a/b where a and b are integers with no common factors, then show that this assumption leads to a contradiction.
Answer
$\sqrt{6}$ is irrational.
Answer for screen readers
The square root of 6 is irrational.
Steps to Solve
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Assume the opposite
Let's 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 with no common factors (in simplest form). -
Square both sides
To eliminate the square root, we square both sides of the equation:
$$ 6 = \frac{a^2}{b^2} $$ -
Rearrange the equation
Next, multiply both sides by $b^2$ to eliminate the fraction:
$$ 6b^2 = a^2 $$ -
Analyze the equation
This equation tells us that $a^2$ is equal to $6b^2$. Thus, $a^2$ must be even (since it is 6 times a whole number). If $a^2$ is even, then $a$ must also be even (as the square of an odd number is odd). -
Express $a$ as an even integer
Since $a$ is even, we can write it as:
$$ a = 2k $$
for some integer $k$. -
Substitute back into the equation
Now substitute $a$ into the equation $6b^2 = a^2$:
$$ 6b^2 = (2k)^2 $$
This simplifies to:
$$ 6b^2 = 4k^2 $$ -
Rearrange and simplify
Dividing both sides by 2 gives us:
$$ 3b^2 = 2k^2 $$
This tells us that $2k^2$ is divisible by 3, which means that $k^2$ is also divisible by 3. Therefore, $k$ must also be divisible by 3. -
Conclusion
Since both $a$ and $b$ are even (both have a factor of 2 and 3), there must be a common factor between $a$ and $b$, which contradicts our assumption that $a$ and $b$ are in simplest form (with no common factors).
Hence, our assumption that $\sqrt{6}$ is rational must be false, proving that $\sqrt{6}$ is irrational.
The square root of 6 is irrational.
More Information
This proof demonstrates that the square root of 6 cannot be expressed as a simple fraction, which is a key example in real number theory of irrationality.
Tips
- Assuming that a number is rational without proving that it can be simplified to integers with no common factors.
- Not recognizing that if $a^2$ is even, then $a$ itself must be even, which is crucial for the subsequent steps.
- Mismanaging the algebra when isolating $a$ and $b$, particularly in the final steps of the proof.
