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Chapter 0: Problem 5

Show that \(4+9 \sqrt{2}\) is an irrational number.

Short Answer

Expert verified
Assuming \(4 + 9\sqrt{2}\) is a rational number, we find a contradiction by showing that both \(a\) and \(b\) share a common factor other than 1 (which breaks the assumption that they are coprime). Thus, \(4 + 9 \sqrt{2}\) must be an irrational number.

Step by step solution

01

Assume it's a rational number

Let's assume that \(4 + 9\sqrt{2}\) is a rational number. Then it can be represented as the ratio of two integers, say \(a\) and \(b\), where \(b \neq 0\) and gcd\((a, b) = 1\) (i.e., \(a\) and \(b\) are coprime). So, \[ 4 + 9\sqrt{2} = \frac{a}{b} \]
02

Rearrange and square both sides

We want to isolate the radical (\(\sqrt{2}\)) on one side of the equation. So, we can rewrite the equation as \[ 9\sqrt{2} = \frac{a}{b} - 4 \] Next, square both sides of the equation to get rid of the radical: \[ (9\sqrt{2})^2 = \left(\frac{a}{b} - 4\right)^2 \]
03

Simplify and solve for a

Simplify the equation and solve for \(a\): \[ 81 \cdot 2 = \frac{a^2}{b^2} - 8 \cdot \frac{a}{b} + 16 \] \[ 162 = \frac{a^2 - 8ab + 16b^2}{b^2} \] Since \(b^2\) must divide the right side of the equation, we can see that \(b^2\) must divide into \(162\). Thus, \(a^2 - 8ab + 16b^2 = 162b^2\), and we can solve for \(a\): \[ a^2 = 8ab + (162 - 16) b^2 = 8ab + 146b^2 \]
04

Show a contradiction

Now we know that \(a^2\) can be written as a sum of terms with \(b\) as a factor. Since both \(8ab\) and \(146b^2\) have \(b\) as a factor, their sum (\(a^2\)) must also have \(b\) as a factor. This contradicts our initial assumption that \(a\) and \(b\) are coprime (gcd\((a, b) = 1\)), as now both \(a\) and \(b\) share a common factor other than 1, which is \(b\). Therefore, our assumption that \(4 + 9\sqrt{2}\) is a rational number must be incorrect, and so \(4 + 9\sqrt{2}\) is an irrational number.

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