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Find at least two different ways to fill in the ellipses in the set descriptions given. For example, \(\\{2,4, \ldots, 12\\}\) could be written either \(\mid 2 n: 1 \leq n \leq 6\) and \(n \in \mathbb{N})\) or \(\mid n+1: n \in\\{1,3,5,7,11\\}\\}\) (a) \([1,3, \ldots, 31)\) (b) \(\\{1,2, \ldots, 26 \mid\) (c) \(\\{2,5, \ldots, 32\\}\)

Short Answer

Expert verified
(a) Sequence: \(\{2n - 1 \mid 1 \leq n \leq 16\}\); \(\{n \mid n \text{ is odd, } 1 \leq n \leq 31\}\). (b) Sequence: \(\{n \mid 1 \leq n \leq 26\}\). (c) Sequence: \(\{3n - 1 \mid 1 \leq n \leq 11\}\).

Step by step solution

01

Step A1: Identify the Sequence Pattern for (a)

Observe the sequence given in (a): \([1, 3, \ldots, 31]\). This is an arithmetic sequence where each number increases by 2. The first term is 1 and the last term is 31.
02

Step A2: First Method for (a)

Using the arithmetic sequence notation, the sequence can be represented as \(1 + (n-1)\times 2\). For the sequence, this leads to \(2n - 1\). With the last term being 31, solve for \(n\): \(2n - 1 = 31\), giving \(n = 16\). Thus, it can be written as \(\{2n - 1 \mid 1 \leq n \leq 16, n \in \mathbb{N}\}\).
03

Step A3: Second Method for (a)

Alternatively, consider the sequence as odd numbers \(\{1, 3, 5, \ldots, 31\}\) which can be represented more directly as \(\{n \mid n \text{ is odd and } 1 \leq n \leq 31\}\).
04

Step B1: Identify the Sequence Pattern for (b)

The sequence in (b) is \(\{1, 2, \ldots, 26 \mid\). It is a straightforward arithmetic sequence where the first term is 1, the common difference is 1, and the last term is 26.
05

Step B2: First Method for (b)

This sequence includes all natural numbers from 1 to 26. It can be written as \(\{n \mid 1 \leq n \leq 26, n \in \mathbb{N}\}\).
06

Step B3: Second Method for (b)

Another method is to use the formula for an arithmetic sequence where each term is represented as \(n\). Hence, the sequence can also be expressed as \(\{n \mid n \text{ is an integer and } n = n + 0, 1 \leq n \leq 26\}\).
07

Step C1: Identify the Sequence Pattern for (c)

The sequence (c) \(\{2, 5, \ldots, 32\}\) is an arithmetic sequence where each term increases by 3. The first term is 2, and the last term is 32.
08

Step C2: First Method for (c)

Using the formula for arithmetic sequences, start with the first term: \(2 + (n-1)\times 3 = 2 + 3(n-1)\). Setting the last term 32, solve for \(n\): \(2 + 3(n-1) = 32\), resulting in \(3n - 3 = 30\), \(3n = 33\), giving \(n = 11\). Thus, \(\{2 + 3(n-1) \mid n = 1, 2, \ldots, 11, n \in \mathbb{N}\}\).
09

Step C3: Second Method for (c)

Alternatively, represent the sequence as: each term follows the pattern of adding 3 to the previous odd number starting from 2 (even): \(\{3n - 1 \mid n \text{ such that } 1 \leq n \leq 11\}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

sequence patterns
In the world of mathematics, identifying sequence patterns is crucial, especially when working with series or sequences. A sequence is simply an ordered list of numbers following a specific rule.
In an arithmetic sequence, each term after the first is obtained by adding a constant difference to the previous term. This pattern is known as the common difference. For example, the sequence \([1, 3, 5, \ldots, 31]\) follows a pattern where each number increases by 2.
The constant increase (common difference) helps to form an expression for the arithmetic sequence. If the first term is \(a\) and the common difference is \(d\), then the \(n\)-th term is expressed as \(a + (n-1)\times d\). By recognizing these patterns, you can understand how sequence elements are formed and predict upcoming numbers in the sequence easily.
natural numbers
Natural numbers play a fundamental role in understanding arithmetic sequences. They are the basic counting numbers like 1, 2, 3, and so on.
When dealing with sequences, it's crucial to specify that each position (or term) in the sequence uses a natural number as its base.
For example, in the sequence \(\{2n - 1 \mid 1 \leq n \leq 16, n \in \mathbb{N}\}\), the natural numbers \(\mathbb{N}\) specify that \(n\) can take any integer value from 1 to 16 inclusive.
This makes sure that all terms in the sequence are valid and adhere to the rules of natural numbers, ensuring no fractions or negative numbers are included in sequences intended to be arithmetic with whole numbers.
set notation
Set notation is a mathematical language used to define collections of objects or numbers. When describing sequences, set notation is particularly useful because it offers a clear and concise way to represent the properties of the sequence.
For instance, the expression \(\{n \mid n \text{ is odd and } 1 \leq n \leq 31\}\) uses set notation to describe all the odd numbers between 1 and 31. In this context, \(\mid\) means "such that" and helps establish the rule or condition that members of the set must satisfy.
Set notation provides a universal language that mathematicians can use to effectively communicate ideas without ambiguity. This helps in ensuring any set or pattern described is easily understandable, making it a powerful tool in mathematical discourse.

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Most popular questions from this chapter

(a) Find the resolvant of \((p \vee q)\) and \((\neg p \vee r)\) on \(p\). (b) Find the resolvant of \((p \vee q \vee r \vee s)\) and \((\neg p \vee \neg q \vee t)\) on \(p\). (c) Find the resolvant of \((p \vee q)\) and \(\neg p\) on \(p\). (d) Find the resolvant of \((p)\) and \((\neg p)\) on \(p\). (e) Which resolvant above from parts (a) through (d) is a tautology? Which is tautologically false?

Find a formula in negation normal form equivalent to the negation of $$ \forall x \exists y((P(x, y) \wedge Q(x, y)) \rightarrow R(x, y)) $$.

For any two integers \(m\) and \(n,\) we say \(m\) divides \(n\) if there is an integer \(k\) such that \(n=\) \(m k\). (Many programming languages give easy ways to say that, such as \(n \% m=0\) or \(n\) div \(m=0 .\) ) Define \(D i v(m, n)\) to be \(m\) divides \(n\). Translate each of the following propositions and quantified formulas into a clear English sentence. Label each as being true or false, with the universe as the set \(\mathbb{Z}\). (a) \(\operatorname{Div}(5,7)\) (b) \(\operatorname{Div}(4,16)\) (c) \(\operatorname{Div}(16,4)\) (d) \(\operatorname{Div}(-8,0)\) (e) \(\forall m(\forall n(\operatorname{Div}(m, n)))\) (f) \(\forall n(\operatorname{Div}(1, n))\) (g) \(\forall m(\operatorname{Div}(m, 0))\) (h) \(\forall m(\forall n(\operatorname{Div}(m, n) \rightarrow \operatorname{Div}(n, m)))\) (i) \(\forall m(\forall n(\forall p((D i v(m, n) \wedge \operatorname{Div}(n, p)) \rightarrow \operatorname{Div}(m, p))))\) (j) \(\forall m(\forall n((D i v(m, n) \wedge \operatorname{Div}(n, m)) \rightarrow m=n))\)

Let \(\phi=(p \vee q) \rightarrow(r \wedge \neg s)\). For each of the following interpretations of \(p, q, r,\) and \(s,\) compute \(I(\phi)\) using the truth tables for \(\neg, \vee, \wedge, \rightarrow,\) and \(\leftrightarrow\) (a) \(I(p)=T, I(q)=T, I(r)=T,\) and \(I(s)=F\) (b) \(I(p)=T, I(q)=T, I(r)=F,\) and \(I(s)=F\) (c) \(I(p)=F, I(q)=T, I(r)=T,\) and \(I(s)=T\) (d) \(I(p)=F_{,} I(q)=F_{,} I(r)=T,\) and \(I(s)=T\)

(a) The conjunction of \(n\) formulas \(p_{1}, p_{2}, \ldots, p_{n}\) is defined to be the formula \(\left(\ldots\left(\left(p_{1} \wedge p_{2}\right) \wedge p_{3}\right) \wedge \ldots\right) \wedge p_{n} .\) For \(n=0,\) there is a special case: The conjunction of zero formulas is defined to be \(T\). For \(n=1\), that conjunction simplifies to \(p_{1}\). Let \(\phi\) be the conjunction of \(p_{1}, p_{2}, \ldots, p_{n} .\) Prove that for any interpretation \(I, I(\phi)=T\) if and only if \(I\left(p_{i}\right)=T\) for each \(i\) such that \(1 \leq i \leq n .\) (Hint: Use induction.) (b) Let \(\phi\) be the formula $$ \left(\ldots\left(\left(p_{1} \leftrightarrow p_{2}\right) \leftrightarrow p_{3}\right) \leftrightarrow \ldots\right) \leftrightarrow p_{n} $$ for \(n \geq 1 .\) For what interpretations \(I\) is \(I(\phi)=T ?\) (Hint: The answer involves counting how many of the \(p_{i}\) 's are true in \(I\). Prove the result by induction on \(n\).)

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