Problem 85 Using the result that \(\sqrt{2}... [FREE SOLUTION]

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Precalculus A Prelude to Calculus

Sheldon Axler

$Math Studyset 91影视 Explanations$ Math

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Chapter 3: Problem 85

Using the result that $\sqrt{2}$ is irrational, explain why $2^{1 / 6}$ is irrational.

Short Answer

Expert verified

Assuming $2^{\frac{1}{6}}$ is rational, we can find integers $p$ and $q$ such that $2^{\frac{1}{6}} = \frac{p}{q}$. By raising both sides to the power of 6 and simplifying the equation, we find that both $p$ and $q$ are even, which contradicts our initial assumption that they have no common factors other than 1. This contradiction proves that $2^{\frac{1}{6}}$ is irrational.

Step by step solution

Assume $2^{\frac{1}{6}}$ is rational

Let's assume $2^{\frac{1}{6}}$ is rational. It means there exist integers $p$ and $q$ with no common factors other than 1 (they are coprime) such that: \[2^{\frac{1}{6}} = \frac{p}{q}\]

Raise both sides to the power of 6

Let's raise both sides to the power of 6 to eliminate the fraction: \[\left(2^{\frac{1}{6}}\right)^6 = \left(\frac{p}{q}\right)^6\] This simplifies to: \[2^1 = \frac{p^6}{q^6}\] This can be also written as: \[2q^6 = p^6\]

Examine the parity of $p^6$

The above equation states that $2q^6$ is equal to $p^6$. Since 2 is a factor of the left-hand side, it implies that $p^6$ must be even. This means that $p$ must also be even as an even number raised to any power is always even.

Rewrite $p$ and substitute

Since $p$ is even, we can express it as $p = 2k$ for some integer k. Now, substitute this expression for $p$ in the equation: \[2q^6 = (2k)^6\]

Simplify the equation

Now, simplify the equation: \[2q^6 = 2^6k^6\] Divide both sides by 2: \[q^6 = 2^5k^6\]

Examine the parity of $q^6$

The new equation determines that $q^6$ has 2 as a factor since it is equal to $2^5k^6$. This implies that $q^6$ is even, which means $q$ must also be even, because an even number raised to any power remains even.

Find the contradiction

In Step 3, we determined that $p$ is even, and now in Step 6, we found that $q$ is even as well. This contradicts our original assumption that $p$ and $q$ have no common factors other than 1, as they both have at least a common factor of 2. The contradiction indicates that our initial assumption that $2^{\frac{1}{6}}$ is rational was incorrect. Therefore, we can conclude that $2^{\frac{1}{6}}$ is irrational.

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91影视

Using the result that \(\sqrt{2}\) is irrational, explain why \(2^{1 / 6}\) is irrational.

Short Answer

Step by step solution

Assume \(2^{\frac{1}{6}}\) is rational

Raise both sides to the power of 6

Examine the parity of \(p^6\)

Rewrite \(p\) and substitute

Simplify the equation

Examine the parity of \(q^6\)

Find the contradiction

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