114 Examples

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Here are some examples of the definitions and theorems given on the 114 page.


Contents

Mappings and Functions

Definitions and Properties

  • The following are mappings:
  1. f : \mathbb{R}\rightarrow \mathbb{R}, f(x) = x^2 for all x\in\mathbb{R}.
  2. g : \mathbb{Z}\rightarrow \mathbb{R}, g(x) = e^{x+1} for all x\in\mathbb{Z}.
  3. h : \mathbb{C}\rightarrow \mathbb{R}, h(a+bi) = \sqrt{a^2+b^2} for all a,b\in\mathbb{R}.
  • The mappings f : \mathbb{R}\rightarrow \mathbb{R} and g : \mathbb{R}\rightarrow \mathbb{R} given by f(x) = x+1 and g(x) = \frac{x^2+x+1}{x+1} + \frac{x}{x+1} are not equal. This is because, although if we simplify g we will get f, the mapping g cannot be defined at x = -1, and therefore the mappings do not have the same domain. If we define g(-1) = 0, then the mappings are equal.
  • The following mappings are injective:
  1. f(x) = x
  2. g(x) = 2x+1
  3. h(x) = \frac{e^x}{4}

The following mappings are not injective:

  1. f(x) = x^2, as f(-1) = f(1), but 1\neq -1
  2. g(x) = (\sin x)^2, as g(0) = g(2\pi), but 0 \neq 2\pi
  • The following mappings are surjective:
  1. f(x) = x, by definition.
  2. g(x) = 2x+1, as g\left(\frac{x-1}{2}\right) = x.
  3. h(x) = \frac{e^x}{4}, as h(\ln 4x) = x.

The following mappings are not surjective:

  1. f : \mathbb{R} \rightarrow \mathbb{R}, f(x) = \frac{1}{x}, as the does not exist a real number such that \frac{1}{x} = 0
  • Note that the above three functions that were both one to one and onto are bijective.
  • Let f : \mathbb{Z}\rightarrow\mathbb{Q} and g : \mathbb{Q}\rightarrow \mathbb{R} be defined by f(n) = \frac{n}{n+1} where f(-1) = 0, and g(x) = 2e^x. Thus, if we have h = g\circ f, then h(n) = g(f(n)) = 2e^{\frac{n}{n+1}}, where h(-1) = 2.

Operations

  • Given the set \mathbb{N}, the relation n*m = n^m is an operation on that set.
  • We have seen many examples of abridged forms of Cayley tables before. Consider for example:
× 1 2 3 4 5
1 1 2 3 4 5
2 2 4 6 8 10
3 3 6 9 12 15
4 4 8 12 16 20
5 5 10 15 20 25

The above is a section of the Cayley table for the natural numbers under the operation of multiplication.

  • We know that a+(b+c) = (a+b)+c for all real numbers, thus addition is associative on \mathbb{R}. However, the operation * defined by n*m = n^m is not associative, as 2*(3*2) = 2^9 = 512 but (2*3)*2 = 8^2 = 64.
  • We know that a+b = b+a for all real numbers, thus addition is commutative on \mathbb{R}. However, the operation * defined by n*m = n^m is not commutative, as 2*3 = 8 but 3*2 = 9.
  • Given the set \mathbb{Z} under addition, we know that for all a\in\mathbb{Z}, 0+a = a+0 = a. Thus, 0 is the identity for addition. Furthermore, given any a\in\mathbb{Z}, a+(-a) = (-a)+a = 0, and thus -a is the inverse of a.

Groups and Permutations

Introduction to Groups

  • The following are groups:
  1. (\mathbb{Z}, +)
  2. (\mathbb{Q}, \cdot)
  3. The set G = \left\lbrace 1, -1 \right\rbrace under multiplication.
  4. The set of all matrices under matrix addition.
  5. The set of formal pairs E-E^{\prime} of vector bundles over a compact Hausdorff space X where E_1 - E_1^{\prime} = E_2 - E_2^{\prime} if E_1 \oplus E_2^{\prime} is stably isomorphic to E_2 \oplus E_1^{\prime} (where two bundles E and E^{\prime} are stably isomorphic if they become isomorphic upon the addition of a trivial bundle, E\oplus\varepsilon^n \approx E^{\prime} \oplus \varepsilon^n).

The following are not:

  1. (\mathbb{N}, +), because inverses do not exist.
  2. (\mathbb{Z}, \cdot) because inverses do not exist.
  3. (\mathbb{Q}, *) where n*m = n^m, as the operation is not associative.
  4. The set of all matrices under matrix multiplication.
  • Note that all groups are also semi-groups. Also, (\mathbb{N}, +) and (\mathbb{N}, \cdot) are also semi-groups.
  • Also note that as addition and multiplication are commutative, we know that all the abvoe groups whose operations are eithe addition or multiplication are Abelian groups.
  • The order of all of the groups involving \mathbb{Z} or \mathbb{Q} are infinite. However the third group above has order 2.

Permutations

The elements of S_3 are given by:

\begin{pmatrix} 1 & 2 & 3 \\ 1 & 2 & 3 \end{pmatrix}

\begin{pmatrix} 1 & 2 & 3 \\ 2 & 1 & 3 \end{pmatrix}

\begin{pmatrix} 1 & 2 & 3 \\ 3 & 2 & 1 \end{pmatrix}

\begin{pmatrix} 1 & 2 & 3 \\ 1 & 3 & 2 \end{pmatrix}

\begin{pmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \end{pmatrix}

\begin{pmatrix} 1 & 2 & 3 \\ 3 & 1 & 2 \end{pmatrix}

Writen more compactly, we have S_3 = \left\lbrace \begin{pmatrix} 1 \end{pmatrix}, \begin{pmatrix} 1 & 2 \end{pmatrix}, \begin{pmatrix} 1 & 3 \end{pmatrix}, \begin{pmatrix} 2 & 3 \end{pmatrix}, \begin{pmatrix} 1 & 2 & 3 \end{pmatrix}, \begin{pmatrix} 1 & 3 & 2 \end{pmatrix},  \right\rbrace