Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $\ast$ be defined in $\mathbb Z_8$ as follows: $$\begin{aligned} a \ast b = a +b+2ab\end{aligned}$$

Determine all the invertible elements in $(\mathbb Z_8, \ast)$ and determine, if possibile, the inverse of the class $[4]$ in $(\mathbb Z_8, \ast)$.

Identity element

We shall say that $(\mathbb Z_8, \ast)$ has an identity element if: $$\begin{aligned} (\forall a \in \mathbb Z_8) \text { } (\exists \varepsilon \in \mathbb Z_8 : a \ast \varepsilon = \varepsilon \ast a = a)\end{aligned}$$

$$\begin{aligned} a+\varepsilon+2a\varepsilon = a \Rightarrow \varepsilon +2a\varepsilon = 0 \Rightarrow \varepsilon(1+2a) = 0 \Rightarrow \varepsilon = 0 \end{aligned}$$ As $\ast$ is commutative, similarly we can prove for $\varepsilon \ast a$.

$$\begin{aligned} a \ast 0 = a+0+2a0 = a \end{aligned}$$ $$\begin{aligned} 0\ast a = 0+a+20a = a\end{aligned}$$

Invertible elements and $[4]$ inverse

We shall state that in $(\mathbb Z_8, \ast)$ there is the inverse element relative to a fixed $a$ if and only if exists $\alpha \in (\mathbb Z_8, \ast)$ so that:

$$\begin{aligned} a\ast \alpha = \alpha \ast a = \varepsilon \end{aligned}$$ $$\begin{aligned} a+\alpha +2a\alpha = 0 \end{aligned}$$ $$\begin{aligned} \alpha(2a+1) \equiv_8 -a \Rightarrow \alpha \equiv_8 -\frac{a}{(2a+1)} \end{aligned}$$

In particular looking at $[4]$ class, it follows: $$\begin{aligned} \alpha \equiv_8 -\frac{4}{(2\cdot 4+1)}=-\frac{4}{9} \end{aligned}$$

therefore: $$\begin{aligned} 9x \equiv_8 -4 \Leftrightarrow 1x \equiv_8 4 \Rightarrow x=4 \end{aligned}$$

which seems to be the right value as $$\begin{aligned} 4 \ast \alpha = 4 \ast 4 = 4+4+2\cdot 4\cdot 4 = 8 + 8\cdot 4 = 0+0\cdot 4 = 0 \end{aligned}$$

Does everything hold? Have I done anything wrong, anything I failed to prove?

share|improve this question
1  
The third line under Identity element should be $a+\varepsilon+2a\varepsilon=a$. I would have edited it for you, but the edit, while important, was deemed too short to be accepted. –  Rick Decker Jul 26 '12 at 15:32
    
@RickDecker oops... you're right, a typo while copying. Corrected, though. Thank you. –  haunted85 Jul 26 '12 at 15:38
    
Your definition of "identity" is wrong. –  Arturo Magidin Jul 26 '12 at 18:34

3 Answers 3

up vote 3 down vote accepted

Let me mention that your definition of identity is wrong. You write:

We shall say that $(\mathbb{Z}_8,\ast)$ has an identity if $$(\forall a \in \mathbb Z_8) \text { } (\exists \varepsilon \in \mathbb Z_8 : a \ast \varepsilon = \varepsilon \ast a = a)$$

Under this formulation, the element $\varepsilon$ is allowed to change depending on $a$. This is not the usual meaning of "identity"! The identity is usually required to be the same for every element.

For example, consider the operation $\odot$ on $\mathbb{Z}$ given by $$a\odot b = 2a-b.$$ Then this operation satisfies the definition you give: given an integer $a$, let $\varepsilon=a$. Then $a\odot \varepsilon = 2a-a = a$, and $\varepsilon\odot a = 2a-a = a$. However, there is no element $\varepsilon$ that "works" for every integer: for instance, if $1\odot\varepsilon = 1$ then this means that $2-\varepsilon = 1$, so $\varepsilon = 1$. But then $2\odot\varepsilon\neq 2$.

The order of the quantifiers matters! By putting the existential quantifier after the universal quantifier, you allow the value of $\varepsilon$ to depend on the value of $a$. The correct formula for the existence of an identity is: $$\exists\varepsilon\in\mathbb{Z}_8(\forall a\in \mathbb{Z}_8\colon a\ast \varepsilon = \varepsilon\ast a = a).$$

Of course, if there is a $\varepsilon$ that works independent of $a$, then then there is one that works if we allow it to depend on $a$. In other words, $\exists a(\forall b P(a,b))$ always implies $\forall a(\exists b P(a,b))$. The converse does not hold, however, as the example above shows. (Or, for another example: it is true that for every person $a$ there is a person $b$ such that $b$ is the mother of $a$; i.e., everyone has someone who is their mother. It is not true, however, that there is a person $b$ such that for every person $a$, $b$ is the mother of $a$; i.e., there isn't someone who is everyone's mother).

There is also, in my opinion, a gap in the argument for inverses. You are correct that $\varepsilon = 0$ is an identity; and also that if $a$ has an inverse, then $$a+\alpha + 2a\alpha \equiv 0 \pmod{8}.$$ Hence, $\alpha(1+2a)\equiv -a\pmod{8}$ holds.

Now, you need to justify that you can "divide by $1+2a$". That is, that $1+2a$ has a multiplicative inverse modulo $8$. This can be done by noting that $1+2a$ is necessarily odd, and hence relatively prime to $8$. Thus, it can be solved in all cases, so every element has a $\ast$-inverse.

I will note, also, that you should have "$a\in\mathbb{Z}_8$" and not "$a\in(\mathbb{Z}_8,\ast)$". You want an element of the underlying set of your algebraic object, rather than an element of the ordered pair.

share|improve this answer

All looks correct, but in a couple cases you should explicitly mention why $\rm\:1+2a\:$ is cancellable or invertible mod $8$ (being odd it is coprime to $2$ so coprime to $8$ hence $\ldots$). In fact, it is idempotent $\rm\: (1\!+\!2a)^2 = 1 + 4a(a\!+\!1)\equiv 1\pmod 8,\:$ by $\rm\:2\:|\:a(a+1),\:$ so $\rm\:1/(1\!+\!2a) = 1\!+\!2a,\:$ which enables you to give a simpler formula for inverses in your group.

$\rm\ odd^2\equiv 1\pmod 8\:$ is frequently useful in elementary number theory, so it's worth remembering.

share|improve this answer

From your post:

Determine all the invertible elements in $(\mathbb Z_8, \ast)$

You have shown that the inverse of $a$ is $\alpha \equiv_8 -\frac{a}{(2a+1)}$. I would go a step further and list out each element which has an inverse since you only have eight elements to check.

share|improve this answer
    
isn't enough showing $\alpha \equiv_8 -\frac{a}{(1+2a}$ to answer that part of the question? –  haunted85 Jul 26 '12 at 15:00
    
thank you for your advice, I'm wondering what if I ever have to deal with $\mathbb Z_n$ with very large $n$, would it be ok to just leave the general answer? Would there be further steps that I better do in order to support my proof? –  haunted85 Jul 26 '12 at 15:09
1  
@haunted85 I think if $n$ were large, then a general answer such as what you gave here would likely be sufficient, unless it suggests some simple pattern, such as all odd integers less than $n$. –  Code-Guru Jul 26 '12 at 15:13
    
This should be a comment in my opinion –  Belgi Jul 26 '12 at 15:40
    
@haunted85 I have updated my answer after our conversation in the comments. I will delete the comments and suggest you do likewise. –  Code-Guru Jul 26 '12 at 15:44

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.