Set properties of even (##E##) and odd (##I##) integers

[brotherbobby]

Homework Statement

I copy and paste the question as it appears in the text down below. It is too much to type.

Relevant Equations

The usual rules for commutativity, associativity and distributivity for addition and multiplication of elements of a set along with definitions of identities and inverses.

Problem Statement : I copy and paste the problem as it appears in the text (Lang, Basic Mathematics, 1971).

Attempt : There are several questions in both a) and b) above. I type out the question and my answer each time.

a) (i) Show that addition for $E$ and $I$ is associative and commutative : All elements of $E$ and $I$ are (positive) integers. These integers satisfy the commutative and associative rules : (1) $a+b=b+a\;\; \textbf{and}\;\; a+(b+c) = (a+b) +c$, where $a,b\in E,I$.

(ii) Show that $E$ plays the role of a zero element for addition : We remember that the zero element in integers has a property that $a+0 =a=0+a$, for $a\in \mathbb{Z_+}$. When we look at even and odd (positive) integers, we find that $E+E = E$ and $I+E = I$. Thus $E$ is the "zero element" in this group for addition.

(iii) What are the additive inverses for $E$ and $I$? The additive inverse for an element $a\in \mathbb{Z}$ is a number $b\in \mathbb{Z}$ such that $a+b=0$ (commutativity is implied). Here the additive identity is $E$. Since $E+E = E$, $E$ is the additive inverse for $E$. Likewise, since $I+I = E$, $I$ is the additive inverse for $I$.

(It is from now on that doubts abound, with decreasing confidence in my answers.)

b) (i) Show that multiplication for $E$ and $I$ is commutative and associative : Same answer as in a) (i) above. Since all numbers of $E$ and $I$ are positive integers, which satisfy properties of commutativity of associativity for multiplication, so will the even and odd (positive) integers.

(ii) Which of $E$ or $I$ behaves like 1? We remember that the multipicative identity for an element $a$ is 1, such that $a\cdot 1 = a$, commutativity implied also, where $a\in Z$. But here we have to ask $E\;\;\cdot\;\; ?\;\; = E$. Clearly $E$. However $I\cdot E=E (\ne I)$. Hence $E$ is not the multipicative inverse (or 1) for this set. However, while $I\cdot I = I\;\;, \;\; E\cdot I=E$. Hence $I$ is the multiplicative inverse (1) for this group.

(iii) Which of them behaves like 0 for multiplication? Only one number, $0\in E$, behaves as the 0 for multiplication; not the other $E$'s nor any of the $I$'s.

(iv) Show that multiplcation is distributive with respect to addition : Let's take three even integers, $e_1, e_2, e_3$. It is trivial to show that upon operating $e_a \cdot (e_b+e_c)$ where $a,b,c$ denote the various combinations, the same answer will result since $E+E = E$ and $E\cdot E = E$. We remember that distributivity holds for all integers. If we take three odd integers, $i_1, i_2, i_3$, since we have $I+I = E$ and $I\cdot E = E$, we will get the same results if we attempt distributivity. Let's try a combination of $e_1, i_2, i_3$. In that case we have first : $e_1\cdot (i_2+i_3) = e_1\cdot e_4 = e_{14}$, while $(e_1+i_2)\cdot i_3= i_4\cdot i_3 = i_{34}\ne e_{14}$. Thus I have to conclude that, contrary to what is stated, multiplication is not distributive with respect to addition of even and odd integers if they are allowed to mix.Am I correct with my work? A hint or suggection would be welcome.

 

[Office_Shredder]

For associativity they are asking you to prove things like $(E+I)+E=E+(I+E)$

 

[pasmith]

You have binary operations defined on \{E, I\} as given in the question: I + I = E, E + E = E, E + I = I + E = I, etc. The question is asking you about these binary operations, not about addition or multiplication of integers.

 

[fresh_42]

b) (iv) is wrong, and you brought up your own argument for it. But from the beginning:

It is correct except for the last point b) (iv).
I think you are expected to show associativity and commutativity elementwise, same for distributivity.
However, your method is actually better if you had written it consistently.
By consistently I mean a) better description and b) using it for all tasks, not just the first.

a) You basically defined a function $\pi \, : \,\mathbb{Z}\longrightarrow \{E,I\}$ that has the properties
$$
\pi(a+b)=\pi(a)+\pi(b)\, \wedge \,\pi(a\cdot b)=\pi(a)\cdot \pi(b)
$$
In order to transport associativity and commutativity from $\mathbb{Z}$ to $\{E,I\}$ you need to show these two properties. The easiest way to do so is to interpret $\{E,I\}$ as the remainders of the division of integers by $2,$ i.e. $E=0$ and $I=1.$

If you had done this, which by the way is also necessary to complete your proof since it is not clear from what you wrote why associativity and commutativity are transported from the integers onto $\{E,I\},$ then you automatically have the properties of neutral and inverse elements, and the distributive law (b) (iv)) as a bonus.

 

[brotherbobby]

Thank you all. I will respond to you but in particular to @pasmith's comment below.

You have binary operations defined on {E,I} as given in the question: I+I=E, E+E=E, E+I=I+E=I, etc. The question is asking you about these binary operations, not about addition or multiplication of integers.

I understand that the operation of "+" and "$\times$" are arbitrary ones that satisfy the given rules for even and odd integers but in this case, they are clearly addition and multiplication. Let's have a look at the author's comments at the start of the exercise :

If he is saying "we know that", surely he means something we are already aware of, and not new operations we need to define.
However, for the sake of generality, I will assume that the operations are some arbitrary addition, say $\bigoplus$, and some arbitrary multiplication, say $\bigodot .$ If there's no confusion, I will simply use $E+I$ and $EI$ to represent addition and multiplication, respectively, between the elements.

I will carry out the proofs by hand, using Microsoft Word$^{\circledR}$. I hope I am not violating anything.

 

[FactChecker]

Thank you all. I will respond to you but in particular to @pasmith's comment below.

I understand that the operation of "+" and "$\times$" are arbitrary ones that satisfy the given rules for even and odd integers but in this case, they are clearly addition and multiplication.

In abstract algebra, they would be called addition and multiplication, but that does not mean that every mathematical property of the integers would hold for E and I.

Let's have a look at the author's comments at the start of the exercise :
View attachment 315271
If he is saying "we know that", surely he means something we are already aware of, and not new operations we need to define.
However, for the sake of generality, I will assume that the operations are some arbitrary addition, say $\bigoplus$, and some arbitrary multiplication, say $\bigodot .$ If there's no confusion, I will simply use $E+I$ and $EI$ to represent addition and multiplication, respectively, between the elements.

I will carry out the proofs by hand, using Microsoft Word$^{\circledR}$. I hope I am not violating anything.

View attachment 315276
View attachment 315277

No. You have identified E as the multiplicative zero of this abstract algebra. EI = E (=0) and EE = E (=0), so it behaves as a zero for multiplication.

View attachment 315278

The first part of last line should be: I(I+E) = I(I) = I and II+IE = I+E = I, so I(I+E)=II+IE.
The second part of the last line should be: (I+I)E = (E)E = E and IE+IE = E+E = E, so (I+I)E = IE+IE,