# Precursor to Algebra: Herstein shows a way!

Reference: Abstract Algebra, 3rd edition, I. N. Herstein, Prentice Hall International Edition.

Problems:

1. Let S be a set having an operation * which assigns an element a*b of S for any $a,b \in S$. Let us assume that the following two rules hold:

i) If a, b are any objects in S, then $a*b = a$

ii) If a, b are any objects in S, then $a*b=b*a$

Show that S can have at most one object.

II. Let S be the set of all integers $0, \pm 1, \pm 2, \pm 3, \ldots, \pm n, \ldots$. For a, b in S, define * by $a*b=a-b$. Verify the following:

(i) $a*b \neq b*a$ unless $a=b$.

(ii) $(a*b)*c \neq a*(b*c)$ in general. Under what conditions on a, b, c is $(a*b)*c=a*(b*c)$?

(iii) the integer 0 has the property that $a*0=a$ for every a in S.

(iv) For a in S, $a*a=0$.

III) Let S consist oif f the two objects $\square$ and $\triangle$. We define the operation * on S by subjecting $\square$ and $\triangle$ to the following conditions:

i) $\square * \triangle = \triangle = \triangle * \square$i

ii) $\square * \square =\square$

iii) $\triangle * \triangle=\square$

of verify by explicit calculations that if a, b, c are any elements of S (that is, a, b and c can be any of $\square$ or $\triangle$ then

i) $a*b$ is in S

ii) $(a*b)*c=a*(b*c)$

iii) $a*b=b*a$

iv) There is a particular a in S such that $la=latex a*b=b*a=b$ for all b in S.

,v) Given b in S, then $b*b=a$ where a is the parituclar element in part “iv” above.

Cheers,

Nalin Pithwa

This site uses Akismet to reduce spam. Learn how your comment data is processed.