/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 144 Using the result that \(\sqrt{2}... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 144

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

01

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: \[2^{\frac{1}{6}} = \frac{p}{q}\]
02

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: \[2q^6 = p^6\]
03

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.
04

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: \[2q^6 = (2k)^6\]
05

Simplify the equation

Now, simplify the equation: \[2q^6 = 2^6k^6\] Divide both sides by 2: \[q^6 = 2^5k^6\]
06

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.
07

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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