Math 360, Fall 2013, Assignment 3
We admit, in geometry, not only infinite magnitudes, that is to say, magnitudes greater than any assignable magnitude, but infinite magnitudes infinitely greater, the one than the other. This astonishes our dimension of brains, which is only about six inches long, five broad, and six in depth, in the largest heads.
- - Voltaire
Carefully define the following terms, then give one example and one non-example of each:
- Isomorphism.
- Isomorphic.
- Structural property.
- Identity element.
- Group.
- Inverse element.
- Abelian group.
Carefully state the following theorems (you need not prove them):
- Uniqueness of identity element.
- Left and right cancellation laws.
Solve the following problems:
- Section 3, problems 2, 3, 4, 8, 9, 10, and 17.
- Section 4, problems 3, 4, 5, 6, 10, 11, 12, and 13.
Questions:
- Directions from 2,3, and 4:
There is a binary operation \(*\) defined on the set \(S=\{a,b,c,d,e\}\), and computed with the table:
| * | a | b | c | d | e |
| a | a | b | c | b | d |
| b | b | c | a | e | c |
| c | c | a | b | b | a |
| d | b | e | b | e | d |
| e | d | b | a | d | c |
- 2.2
Compute \((a*b)*c\) and \(a*(b*c)\). Can you say on the basis of this computation whether \(*\) is associative?
- 2.3
Compute \((b*d)*c\) and \(b*(d*c)\). Can you say on the basis of this computation whether \(*\) is associative?
- 2.4
Is \(*\) commutative? Why?
- 2.8
Is \(*\) defined on \(\mathbb{Q}\) by letting \(a*b=ab+1\) commutative or associative? (Inclusive or).
- 2.9
Is \(*\) defined on \(\mathbb{Q}\) by letting \(a*b=ab/2\) commutative or associative? (Inclusive or).
- 2.10
Is \(*\) defined on \(\mathbb{Z}^+\) by letting \(a*b=2^{ab}\) commutative or associative? (Inclusive or).
- 2.17
Define \(*\) on \(\mathbb{Z}^+\) by \(a*b=a-b\) Is this a binary operation? If not, state whether condition 1, condition 2, or both are violated, where Condition 1 is:
Exactly one element is assigned to each possible ordered pair of elements of \(\mathbb{Z}^+\).
And Condition 2 is:
For each ordered pair of elements of \(\mathbb{Z}^+\), the element assigned to it is again in \(\mathbb{Z}^+\).
- Directions 3-10
Determine whether \(\phi\) is an isomorphism from the first binary structure to the second. If it isn't, say why.
- 3.3
From \((\mathbb{Z},+)\) to \((\mathbb{Z},+)\), where \(\phi(n) = 2n\)
- 3.4
From \((\mathbb{Z},+)\) to \((\mathbb{Z},+)\) where \(\phi(n) = n+1\)
- 3.5
From \((\mathbb{Q},+)\) to \((\mathbb{Q},+)\) where \(\phi(x) = x/2\)
- 3.6
From \((\mathbb{Q},\cdot)\) to \((\mathbb{Q},\cdot)\) where \(\phi(x)=x^2\).
- 3.10
From \((M_2(\mathbb{R}),\cdot)\) to \((\mathbb{R},\cdot)\) where \(\phi(A)\) is the determinant of \(A\).
- Directions 11-13
Let \(F\) be the set of all functions \(f:\mathbb{R}\rightarrow \mathbb{R}\) that have derivatives of all orders. Then follow the instructions for 3-10.
- 3.11
From \((F,+)\) to \((F,+)\) where \(\phi(f)=f^{\prime}\).
- 3.12
From \((F,+)\) to \((\mathbb{R},+)\) where \(\phi(f)=f^{\prime}(0)\).
- 3.13
From \((F,+)\) to \((F,+)\) where \(\phi(f)(x) = \int_0^xf(t)dt\).
Solutions:
Definitions:
- Isomorphism.
Definition:
An isomorphism from a binary structure \((A,*)\) to another binary structure \((B,\circ)\) is a function \(\varphi:A\rightarrow B\) such that \(\varphi\) is bijective, and:
\(\forall x,y \in A: \varphi(x*y) = \varphi(x)\circ \varphi(y)\)
Example:
Take the binary structures (from class) of \((\mathbb{Z}_n,+_n)\) and \((U_n,\cdot)\). The function \(\varphi:\mathbb{Z}_n \rightarrow U_n\) given by \(\varphi([x]) = \omega^x\), where \(x\in\mathbb{Z}\) and \(\omega = \cos\frac{2\pi}{n} + i\sin\frac{2\pi}{n}\), is an isomorphism.
Non-Example:
The function \(f\)from the binary structure \((\mathbb{Q},+)\) to \((\mathbb{R},+)\) given by \(f(x) = x\) is not an isomorphism, because it is not surjective. It does meet the other two requirements for an isomorphism.
- Isomorphic
Definition:
Two binary structures are isomorphic if there is an isomorphism between them.
Example:
\((\mathbb{Z}_n,+_n)\) is isomorphic to \((U_n,\cdot)\).
Non-Example:
The structures \((\mathbb{Q},+)\) and \((\mathbb{R},+)\) are not isomorphic.
- Structural Property
Definition:
A structural property is a property that is preserved under isomorphism. Let's say we have two binary structures \((A,*)\) and \((B,\circ)\). If we also have a property of binary structures, i.e. a statement about the structures that is either true or false, then this property is structural if it is preserved by the isomorphism. That is to say, if \((A,*)\) has the property, and there is an isomorphism from \((A,*)\) to \((B,\circ)\), then \((B,\circ)\) also has the property. I think second-order logic might be required to give a proper mathematical definition of this, so I'm going to leave it like this.
Example:
Commutativity of a binary structure's operation is structural.
Non-Example:
The type of the elements contained in a binary structure (real numbers, complex numbers, etc.) is not a structural property. Meaning the answer to the question "Are the elements of this binary structure elements of the set 'X'?" is not a structural property.
- Identity Element
Definition:
An identity element is an element of a binary structure that acts like 1 does for multiplication or 0 does for addition. More formally, given an algebraic structure \((A,*)\), an identity element \(i\) is one such that:
\(\forall a \in A: a*i = a = i*a\)
Example:
1 is an identity element for \((\mathbb{R},*)\).
Non-Example:
1 is not an identity element for \((\mathbb{R},+)\).
- Group
Definition:
A group is a binary structure \((G,*)\) with the properties:
\(\forall x,y, z \in G: x*(y*z) = (x*y)*z\)
\(\exists i \in G: \forall x \in G: x*i = i*x = x\)
\( \forall x\in G: \exists y\in G: x*y = 1 = y*x\)
So it's a binary structure that is associative, has an identity, and has inverse elements. Intuitively, it's a binary structure that looks like multiplication of invertible matrices, or (kind of) like addition of integers. (Addition of integers is also commutative. Groups do not have to be commutative).
Example:
Non-Example:
- Inverse Element
Definition:
Negative numbers are inverse elements of the positive numbers (for addition) and reciprocals are inverse elements for everything else (in multiplication). More formally, given a binary structure \((A,*)\) that has an identity element \(i\), the inverse of an element \(a \in A\) is an element </math>a^{-1}\in A</math> such that
\(a*a^{-1} = i = a^{-1}*a\)
Example:
-5 is the inverse of 5, over addition of integers.
Non-Example:
-5 is not the inverse of 5 over multiplication of rational numbers. 1/5 is.
- Abelian Group
Definition:
An abelian group is a group that is also commutative.
Example:
\((\mathbb{Z},+)\) is an abelian group (because addition of integers is associative, it commutes, there is an identity (0), and every element has an inverse (the negative of that element).
Non-Example:
Multiplication of invertible matrices is not an abelian group, because matrix multiplication does not commute.
Theorems:
- Uniqueness of Identity Element
If a binary structure \((A,*)\) has an identity element \(i\), then that element is unique. Meaning there are no other elements that are also identities. Formally, if there are two elements \(i,j\in A\), and \((a*i=i*a=a)\) and \( (a*j = j*a = j)\), then \( i=j\).
- Left and Right Cancellation Laws
Given elements \(x,y,z\) in a group \((G,*)\), \(x*z = y*z \rightarrow x=y\), and \(z*x = z*y \rightarrow x=y\). You can cancel equal elements from both sides of an equation, if they're being used on the same side of an operation. (\(x*z=z*y\) does not imply that \(x=y\) unless the group is commutative/abelian.
Questions:
- 2.2
Answer:
\((a*b)*c = a\) and \(a*(b*c)=a\) This is not enough to say whether the operation is associative.
- 2.3
Answer:
\((b*d)*c = a\) and \(b*(d*c) = c\). The operation is not associative.
- 2.4
Answer:
The operation is not commutative, because \(e*b = b\) while \(b*e=c\).
- 2.8
Answer:
The operation is commutative, since \(ab+1 = ba+1\). The operation is not associative, because \(a*(b*c) = a(bc+1)+1 = abc+a+q \neq (a*b)*c = (ab+1)c+1 = abc + c +1\).
- 2.9
Answer:
The operation is commutative, since \(ab/2 = ba/2\) It is associative, since \(a(bc/2)/2 = abc/4 = (ab/2)c/2 = abc/4\).
- 2.10
Answer:
The operation is commutative, since \(2^{ab} = 2^{ba}\) It is not associative, since \(a*(b*c) = 2^{a2^{bc}} = 2^a\cdot2^{2^{bc}} = 2^a\cdot4^{bc} \neq (a*b)*c = 2^{c2^{ab}} = 2^c\cdot2^{2^{ab}}=2^a\cdot4^{ab}\).
- 2.17
Answer:
- 3.3
Answer:
- 3.4
Answer:
- 3.5
Answer:
- 3.6
Answer:
- 3.10
Answer:
- 3.11
Answer:
- 3.12
Answer:
- 3.13
Answer:
