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(a) \( A_{3}=\left\{\frac{2 n^{2}-n-10}{2 n^{2}+3 n-2}: n \in \mathbb{Z}\right\} \) (b) \( A_{4}=\{x \in \mathbb{Q}: x \in(0,6]\} \)

Ask by Fitzgerald Garrett. in South Africa
Mar 08,2025

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**(a)** The set \( A_3 \) can be described as all numbers of the form \( 1 - \frac{4}{k} \), where \( k \) is any odd integer except -5. **(b)** The set \( A_4 \) includes all rational numbers between 0 and 6, including 6. \[ \boxed{A_3 = \left\{1 - \frac{4}{k} : k \text{ is an odd integer, } k \neq -5\right\}, \quad A_4 = \{x \in \mathbb{Q} : 0 < x \leq 6\}} \]

Solution

**(a) Finding an alternate description for** \[ A_{3}=\left\{\frac{2n^{2}-n-10}{2n^{2}+3n-2}: n\in\mathbb{Z}\right\}. \] 1. **Factor the denominator.** Notice that \[ 2n^2+3n-2 = (2n-1)(n+2). \] 2. **Factor the numerator.** We look at \[ 2n^2 - n -10. \] Compute the discriminant: \[ D = (-1)^2 - 4\cdot2\cdot(-10)= 1+80=81. \] Since \( \sqrt{81}=9 \) the roots are \[ n=\frac{1\pm 9}{4} \quad \Longrightarrow \quad n=\frac{10}{4}=\frac{5}{2} \quad \text{or} \quad n=\frac{-8}{4}=-2. \] Therefore, the numerator factors (with a suitable constant) as \[ 2n^2 - n -10 = 2\Bigl(n-\frac{5}{2}\Bigr)(n+2). \] (One may check by expanding that this factorization is correct.) 3. **Cancel the common factor (with a restriction).** For \( n\neq -2 \) the factor \( (n+2) \) may be cancelled. (Indeed, if \( n=-2 \) then the denominator is zero.) Thus, for \( n\neq -2 \) we have \[ \frac{2n^{2}-n-10}{2n^{2}+3n-2} = \frac{2\Bigl(n-\frac{5}{2}\Bigr)(n+2)}{(2n-1)(n+2)} = \frac{2\Bigl(n-\frac{5}{2}\Bigr)}{2n-1}. \] 4. **Rewrite in a simpler form.** Write \[ n-\frac{5}{2}=\frac{2n-5}{2}. \] Then \[ \frac{2\Bigl(n-\frac{5}{2}\Bigr)}{2n-1}=\frac{2\cdot\frac{2n-5}{2}}{2n-1}=\frac{2n-5}{2n-1}. \] Hence, for \( n\in\mathbb{Z}\setminus\{-2\} \) \[ \frac{2n^{2}-n-10}{2n^{2}+3n-2}=\frac{2n-5}{2n-1}. \] 5. **A convenient re‐parameterization.** Notice that we can write \[ \frac{2n-5}{2n-1} = \frac{(2n-1)-4}{2n-1}= 1-\frac{4}{2n-1}. \] Since when \( n \) runs over all integers (except \( n=-2 \) to avoid \(2n-1=-5\) which came from the cancellation) the quantity \( k=2n-1 \) runs over all odd integers except \( k=-5 \). Thus, one acceptable answer is: \[ \boxed{A_3=\left\{1-\frac{4}{k}: k\in\{\, \text{all odd integers}\,\}\setminus\{-5\}\right\}.} \] --- **(b) Description of** \[ A_{4}=\{x\in\mathbb{Q}: x\in(0,6]\}. \] This set is simply the collection of all rational numbers that lie in the interval \( (0,6] \). It is a dense subset (with respect to the usual topology on \(\mathbb{R}\)) of the interval \( (0,6] \); that is, between any two real numbers in \( (0,6] \) there is a rational number. In boxed form, one may write: \[ \boxed{A_{4}=\{x\in\mathbb{Q}: 0

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Mind Expander

The set \( A_{3} \) is a rather interesting expression! By simplifying the fraction, you might observe that certain values of \( n \) lead to specific results, potentially collapsing the set to individual points or intervals. Exploring integers can produce boundaries or even reveal repeating rational values as \( n \) varies. Now, let’s think about \( A_{4} \). This set includes all rational numbers within the interval (0, 6], so it's a delightful array of fractions and whole numbers that dance right up to and including 6, but don’t quite reach 0. It’s perfect for discovering ratios that can fit nicely in real-world situations, like measuring ingredients for a recipe or budgeting within a set limit.

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