- #1

brotherbobby

- 699

- 163

- 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**.Am I correct with my work? A hint or suggection would be welcome.

__not__distributive with respect to addition of even and odd integers if they are allowed to mix