10
$\begingroup$

(Apologies if this one sounds like I have not done much research, or I did not aware already have an answer, but I have been searching everywhere and all of these structures presented here, even including the highly exotic division by zero proposals such as Wheels and Meadows, they all seemed to either retain the distributive law (both side) or at least retain the right distributive law

So my question then becomes

Q1 Any famous or widely used example of an algebraic structure where BOTH right and left distributive law fails?

Q2. Suppose I have an algebraic structure on the set $S$ with the following

$$S=\left\{a,b,c,d\right\}$$

with the axiom

$$a^2=a$$

with right addition defined as

$$+:(a_1,a_2)\rightarrow(a_1+a_2),a_1,a_2 \in S$$

and some left operation

$$\circ : (k,a)\rightarrow (k^2 a), k,a \in S$$

I am then interested in computing this entry in the Cayley table for $\circ$

$$a \circ(b+c)$$

How should I approach it since I don't have distributive laws (BOTH left and right) that allow me to simplify this expression to this and apply the axioms I know about $S$

$$a \circ b+a \circ c$$

$\endgroup$
4
  • 2
    $\begingroup$ To answer Q1; yes, check out lattices. As you correctly guessed, the absence of a distributivity law makes things much harder; normal forms are more complicated, etc. $\endgroup$
    – goblin GONE
    Apr 1, 2015 at 12:58
  • $\begingroup$ en.wikipedia.org/wiki/Normal_form_(abstract_rewriting) Is this the normal form you are referring to? $\endgroup$
    – Secret
    Apr 1, 2015 at 13:03
  • $\begingroup$ More or less, but you're overcomplicating it. Look at it this way. If you've got two laws saying that multiplication distributes over addition on all sides, then you can write all your expressions as sums of products. For example, you can write $(xxx+y)x+xy$ as $xxxx+yx+xy,$ or in other words $x^4+yx+xy$. See? Its a sum of products; that's a normal form into which every such expression can be converted. Without both distributive laws, however, this doesn't work, and things can get much hairier. $\endgroup$
    – goblin GONE
    Apr 1, 2015 at 13:07
  • $\begingroup$ The current answer is wrong. $\endgroup$
    – mr_e_man
    Apr 20 at 20:36

1 Answer 1

Reset to default
5
$\begingroup$

Consider linear functions over $\Bbb{R}$ of the form $f(x)=mx+b.$

This is visibly a vector space, and we may go further to obtain an associative algebra-like-object by giving this vector space a "multiplicative" binary operation defined by composition. This defines a left and right binary operation on this vector space.

That is $f_1 =m_1 x + b_1$ and $f_2 =m_2 x + b_2,$ then define

$$f_1\circ f_2 =m_1 m_2 x + m_1b_2+b_1.$$

Further, given $f_3=m_3 x + b_3,$ then

$$f_1\circ f_3 =m_1 m_3 x + m_1b_3+b_1,$$

and $$f_1\circ f_2+f_1\circ f_3 = m_1(m_2 +m_3)x+ m_1(b_2 +b_3)+2b_1.$$

Where $$f_1\circ (f_2+ f_3)= f_1\circ[(m_2+m_3 )x+b_2+b_3] =m_1(m_2+m_3)x+m_1(b_2+b_3)+b_1 $$

so $$[f_1\circ f_2+f_1\circ f_3]-[f_1\circ (f_2+ f_3)]=b_1.$$

Choosing any $b_1\ne0,$ yields an element $f_1$ which will not possess the distributive property.

So we have all the desirable properties of an algebra, without the left or right distributive property.

$\endgroup$
1
  • $\begingroup$ But it does have the right distributive property! $(f_1+f_2)\circ f_3=f_1\circ f_3+f_2\circ f_3$ $\endgroup$
    – mr_e_man
    Apr 20 at 20:14

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .