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Chapter 0: Problem 3

Show that \(3 \sqrt{2}\) is an irrational number.

Short Answer

Expert verified
Assuming \(3\sqrt{2}\) is rational and can be expressed as the simplified fraction \(\frac{a}{b}\), where \(a\) and \(b\) are coprime integers, we derive an equation \(18b^2 = a^2\). From this, we conclude that both \(a\) and \(b\) must be even numbers. However, this contradicts the initial assumption that \(a\) and \(b\) are coprime integers, as they share a common factor of \(2\). Thus, \(3\sqrt{2}\) must be an irrational number.

Step by step solution

01

Assume \(3\sqrt{2}\) is Rational

Let's assume that \(3\sqrt{2}\) is a rational number. That means it can be expressed as a simplified fraction \(\frac{a}{b}\), where \(a\) and \(b\) are coprime integers: \(3\sqrt{2} = \frac{a}{b}\) Now, let's manipulate this equation to eliminate the square root symbol.
02

Square Both Sides

To eliminate the square root symbol, we can square both sides of the equation: \((3\sqrt{2})^2 = (\frac{a}{b})^2\) Which simplifies to: \(18 = \frac{a^2}{b^2}\)
03

Re-arrange The Equation

To further simplify the equation, let's multiply both sides by \(b^2\): \(18b^2 = a^2\)
04

Analyze Both Sides Of The Equation

The left side of the equation, \(18b^2\), is obviously divisible by \(2\) since \(18\) is an even number and multiplying by \(b^2\) won't change the fact that it is even. This means that \(a^2\) on the right side of the equation is also divisible by \(2\). Since \(a^2\) is divisible by \(2\), it implies that \(a\) must also be an even number. So, let \(a = 2m\) for some integer \(m\).
05

Substitute \(a\) and Simplify

Now, let's substitute \(a = 2m\) back into the equation \(18b^2 = a^2\): \(18b^2 = (2m)^2\) Which simplifies to: \(18b^2 = 4m^2\) Divide both sides by \(2\): \(9b^2 = 2m^2\)
06

Analyze Both Sides Of The Equation Again

Now the right side of the equation, \(2m^2\), is divisible by \(2\). Since \(9b^2\) is equal to \(2m^2\), it implies that \(9b^2\) is also divisible by \(2\), and thus \(b^2\) must be divisible by \(2\). Since \(b^2\) is divisible by \(2\), it implies that \(b\) must also be an even number. So, \(a\) and \(b\) are both even numbers.
07

Identify the Contradiction

We initially assumed that \(a\) and \(b\) are coprime integers. However, since both \(a\) and \(b\) are even numbers, they both have a common factor of \(2\), which contradicts our initial assumption that they are coprime integers. Therefore, our initial assumption that \(3\sqrt{2}\) is a rational number must be incorrect, and hence, \(3\sqrt{2}\) is an irrational number.

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Most popular questions from this chapter

A copying machine works with paper that is 8.5 inches wide, provided that the error in the paper width is less than 0.06 inch. (a) Write an inequality using absolute values and the width \(w\) of the paper that gives the condition that the paper's width fails the requirements of the copying machine. (b) Write the set of numbers satisfying the inequality in part (a) as a union of two intervals.

Show that if \(a

(a) Show that if \(a \geq 0\) and \(b \geq 0\), then \(|a+b|=|a|+|b|\) (b) Show that if \(a \geq 0\) and \(b<0\), then \(|a+b| \leq|a|+|b|\) (c) Show that if \(a<0\) and \(b \geq 0\), then \(|a+b| \leq|a|+|b|\) (d) Show that if \(a<0\) and \(b<0\), then \(|a+b|=|a|+|b|\) (e) Explain why the previous four items imply that $$|a+b| \leq|a|+|b|$$ for all real numbers \(a\) and \(b\).

Write each union as a single interval. $$(-3, \infty) \cup[-5, \infty)$$

Write each union as a single interval. $$(-9,-2) \cup[-7,-5]$$
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