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Chapter 0: Problem 2

Show that \(5-\sqrt{2}\) is an irrational number.

Short Answer

Expert verified
Assuming that \(5-\sqrt{2}\) is rational and can be expressed as \(\frac{p}{q}\), we derive a contradiction when trying to solve for \(p\), as we obtain \(p=5q - q\sqrt{2}\), in which \(q\sqrt{2}\) must not be an integer. This contradicts our assumption, therefore proving that \(5 - \sqrt{2}\) is an irrational number.

Step by step solution

01

Assume \(5-\sqrt{2}\) is rational

Let's assume, for the sake of contradiction, that \(5-\sqrt{2}\) is a rational number. This means that it can be expressed as a fraction. Let's express it in the simplest form as a fraction: \(5 - \sqrt{2} = \frac{p}{q}\) Where \(p\) and \(q\) are integers with no common factors other than 1 (i.e., they are co-prime), and \(q \neq 0\).
02

Express \(\sqrt{2}\) in terms of \(p\) and \(q\)

Next, express \(\sqrt{2}\) in terms of \(p\) and \(q\): \(\sqrt{2} = 5 - \frac{p}{q}\)
03

Rearrange the equation for \(p\)

Now rearrange the equation to solve for \(p\): \(p = 5q - q\sqrt{2}\)
04

Observe the properties of \(p\)

If \(q\) is an integer, then \(5q\) is clearly an integer. Also, if \(q\sqrt{2}\) is an integer, then \((5q - q\sqrt{2})\) must be an integer as well. However, we know that \(\sqrt{2}\) is irrational. Therefore, we can conclude that \(q\sqrt{2}\) is not an integer.
05

Derive a contradiction

Since we have assumed that \(5-\sqrt{2}\) is rational, it can be expressed as \(\frac{p}{q}\) where both \(p\) and \(q\) are integers. But now we have found that for the equation to hold true, \(p = 5q - q\sqrt{2}\), and \(q\sqrt{2}\) must not be an integer. This leads to a contradiction as \(p\) cannot be an integer.
06

Conclude that \(5-\sqrt{2}\) is irrational

Since our assumption that \(5-\sqrt{2}\) is rational leads to a contradiction, we can conclude that \(5-\sqrt{2}\) is an irrational number.

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