- Evaluate expressions: \(b=2\sqrt{35}\), \(d=2\), \(e=-2+\sqrt{3}\), \(f=5+2\sqrt{6}\), \(g=\sqrt{7}-\sqrt{3}\), \(h=16-4\sqrt{15}\).
- Determine truth values: (1) sometimes true, (2) sometimes true, (3) always true, (4) never true.
- Recognize that the set of irrational numbers is not closed under multiplication or addition as these operations can yield rational numbers.
- Final Answers: \(b. 2\sqrt{35}, d. 2, e. -2+\sqrt{3}, f. 5+2\sqrt{6}, g. \sqrt{7}-\sqrt{3}, h. 16-4\sqrt{15}\); 1. Sometimes true; 2. Sometimes true; 3. Always true; 4. Never true.
Explanation
- Problem Analysis and Methodology
We will evaluate the given expressions involving radicals and analyze the properties of irrational numbers. The rules for radicals are: \(\sqrt{a} \cdot \sqrt{b} = \sqrt{a \cdot b}\), \(\sqrt{a} \cdot \sqrt{a} = a\), and the binomial expansions \((a \pm b)^2 = a^2 \pm 2ab + b^2\). For the true/false statements, we will provide counterexamples or justifications for each.
- Calculation of Expressions
The expressions are as follows:
- (b) \(2 \sqrt{5} \cdot \sqrt{7} = 2 \sqrt{5 \cdot 7} = 2 \sqrt{35}\).
- (d) \(\frac{1}{\sqrt{5}} \cdot \frac{10}{\sqrt{5}} = \frac{10}{(\sqrt{5})^2} = \frac{10}{5} = 2\).
- (e) \(-2 + \sqrt{3}\) is already in simplest form.
- (f) \((\sqrt{3} + \sqrt{2})^2 = (\sqrt{3})^2 + 2(\sqrt{3})(\sqrt{2}) + (\sqrt{2})^2 = 3 + 2\sqrt{6} + 2 = 5 + 2\sqrt{6}\).
- (g) \(\sqrt{7} - \sqrt{3}\) is already in simplest form.
- (h) \((\sqrt{6} - \sqrt{10})^2 = (\sqrt{6})^2 - 2(\sqrt{6})(\sqrt{10}) + (\sqrt{10})^2 = 6 - 2\sqrt{60} + 10 = 16 - 2(2\sqrt{15}) = 16 - 4\sqrt{15}\).
- Analyzing Properties of Irrational Numbers
We evaluate the truth of each statement regarding irrational numbers:
- "The sum of two irrational numbers is rational." This is sometimes true. For example, \(\sqrt{2} + (-\sqrt{2}) = 0\) (rational), while \(\sqrt{2} + \sqrt{2} = 2\sqrt{2}\) (irrational).
- "The product of two irrational numbers is irrational." This is sometimes true. For example, \(\sqrt{2} \cdot \sqrt{2} = 2\) (rational), while \(\sqrt{2} \cdot \sqrt{3} = \sqrt{6}\) (irrational).
- "An integer can be written as a product of two irrational numbers." This is always true. For example, \(\sqrt{2} \cdot \sqrt{8} = \sqrt{16} = 4\).
- "Irrational numbers are closed with respect to multiplication." This is never true (or not true as a general property), since the product of two irrationals can be rational (as shown in item 2). Therefore, the set is not closed.
- Final Conclusion and Results Summary
In summary, the calculated values are \(2\sqrt{35}\), \(2\), \(-2 + \sqrt{3}\), \(5 + 2\sqrt{6}\), \(\sqrt{7} - \sqrt{3}\), and \(16 - 4\sqrt{15}\). The properties of irrational numbers demonstrate that closure does not hold, and sums or products can result in either rational or irrational values. The final answer is provided by the calculated steps above.
Examples
Understanding how to manipulate radicals and knowing the properties of irrational numbers is vital in fields like civil engineering. For instance, when designing structural supports involving diagonal bracing, the length of a beam is often calculated using the Pythagorean theorem, resulting in expressions like \(\sqrt{a^2 + b^2}\). Knowing whether these resulting values are rational or irrational helps engineers determine if they can be measured precisely using standard rulers or if they require more specialized tools. This foundational knowledge ensures structural calculations are robust and accurate.