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$A=R-\{-1\}$ and $a*b = a+b+ab $

  1. Show that * is a binary operation on A
  2. Show that * is associative
  3. Show that there is an identity element in A for *
  4. Show that every element in A has an inverse with respect to *

I'm new to this lesson. All I know to prove is associativity,

let $a,b,c\in A$

Consider:$$(a*b)*c=(a+b+ab)*(c)$$ $$(a*b)*c=[a+b+ab+c+(a+b+ab)c]$$ $$(a*b)*c=a+b+c+ab+ac+bc+abc$$

Now consider: $$a*(b*c)=a*(b+c+bc)$$ $$a*(b*c)=[a+(b+c+bc)+a(b+c+bc)]$$ $$a*(b*c)=a+b+c+ab+ab+ac+abc$$ Since we get the same result it is associative.

How to prove part 1? and How do I prove that identity exist? Thanks in advance!

P.S: also I'm curious to know why -1 is omitted in the set?

emil
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1 Answers1

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For the binary operation, you need to prove that $a*b\ne-1$ iff $a$, $b\ne-1$, that is $$a*b+1=a+b+ab+1\ne0.$$

For identity, you want an $e$ with $a*e=e*a=a$. As $*$ is commutative, all one needs is that $a*e=a$, that is $$a+e+ae=a.$$ Can you solve that for $e$ in terms of $a$? And is the result independent of $a$?

Once you have done that, do inverses. You then need to solve $a*b=e$ for $b$ in terms of $a$, that is $$a+b+ab=e$$ where you now know $e$.

Angina Seng
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  • It is very clear after your hint. so identity would be $e=0$. Thanks a lot. – emil Dec 15 '18 at 06:11
  • For the binary proof operation how can we prove that $a*b\neq-1$ There can be a lot of possibilities to get -1 as answer. Isn't it? – emil Dec 15 '18 at 06:44
  • @emil Why don't you try to solve the equation $a+b+ab=-1$ for $b$ in terms of $a$? After all, it's only a linear equation. – bof Dec 15 '18 at 06:55
  • What the theory says is any two elements which are inside the given set should also yield an element inside the given set. Since -1 does not belong to set, what we need to prove is when a and b are equal to -1 we should not get -1 as the result. Am I correct? So it is about proving not equal to -1. I find it difficult to prove this. – emil Dec 15 '18 at 07:06
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    @emil Assuming that $a+b+ab=-1$ and solving for $b$ I get $$(1+a)b=-1-a$$. Assuming that $a\ne-1$ I can divide by $1+a$ and get $$b=\frac{-1-a}{1+a}=\frac{-(1+a)}{1+a}=-1,$$ so it appears that the equation $a+b+ab=-1$ can only hold if $a=-1$ or $b=-1$. – bof Dec 15 '18 at 07:14
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    @emil Maybe everything will be easier if you note that the definition of $ab$ can be rewritten as $$ab=(a+1)(b+1)-1$$ or $$a*b+1=(a+1)(b+1).$$ For instance, if the left side of the last equation is equal to $0$, then the right side is also equal to $0$. – bof Dec 15 '18 at 07:18
  • This clarifies my doubts. Thank you very much! – emil Dec 15 '18 at 07:21