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Let $*$ denote the binary operation on a set $\mathbb{R}^3$ of ordered triples of real numbers defined such that: $$(A_0,A_1,A_2) * (B_0,B_1,B_2)= (A_0B_0,A_0B_1 + A_1B_0 ,A_0B_2 + A_1B_1 + A_2B_0)$$

for all $(A_0,A_1,A_2),(B_0,B_1,B_2)\in\mathbb{R}^3$.

I have proved that the set above is a monoid and I have worked out its identity element to be $(1,0,0)$.

I am trying to determine which of its elements are invertible and that's what I just can't figure out yet. This is what I have done

$$(A_0,A_1,A_2) * (B_0,B_1,B_2)= (1,0,0)$$

where $(B_0,B_1,B_2)$ is the inverse of $(A_0,A_1,A_2)$, so:

$$(A_0B_0,A_0B_1 + A_1B_0 ,A_0B_2 + A_1B_1 + A_2B_0) = (1,0,0)$$

Then:

$$A_0B_0=1 \implies B_0=\frac{1}{A_0}$$

$$A_0B_1 + A_1B_0=0 \implies B_1 =-\frac{A_1B_0}{A_0}$$

$$A_0B_2 + A_1B_1 + A_2B_0=0 \implies B_2= \frac{-A_2B_0 -A_1B_1}{A_0}$$

I do not know what to do next. I want to find which of the elements are invertible and what their inverses are.

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    You've actually already done all the real work that's needed. All that's left is to notice two things: (1) If $A_0=0$ then the $B_i$'s that you computed don't exist (because you can't divide by 0), so there is no inverse for $(A_0,A_1,A_2)$. (2) If $A_0\neq0$ then your formulas for the $B_i$'s give you an inverse for $(A_0,A_1,A_2)$.2012-12-30

2 Answers 2

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Suppose that $\langle a_0,a_1,a_2\rangle*\langle b_0,b_1,b_2\rangle=\langle 1,0,0\rangle$; then

$$\left\{\begin{align*} &a_0b_0=1\\ &a_0b_1+a_1b_0=0\\ &a_0b_2+a_1b_1+a_2b_0=0\;. \end{align*}\right.$$

This clearly implies that $a_0\ne 0$ and $b_0=\dfrac1{a_0}$, so the system becomes

$$\left\{\begin{align*} &a_0b_1+\frac{a_1}{a_0}=0\\ &a_0b_2+a_1b_1+\frac{a_2}{a_0}=0\;. \end{align*}\right.$$

The first of these equations implies that $b_1=-\dfrac{a_1}{a_0^2}$ and imposes no restriction on $a_1$. The second then reduces to

$$a_0b_2-\frac{a_1^2}{a_0^2}+\frac{a_2}{a_0}=0\;,$$

which can evidently be solved for $b_2$ without imposing any conditions on $a_2$. Thus, $\langle a_0,a_1,a_2\rangle$ is invertible precisely when $a_0\ne 0$, and I’ve computed most of the inverse above.

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    thank you ijust cant seem to make the connection which implies that a0!=02012-12-29
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    @Jack: You have to have $a_0b_0=1$. If $a_0=0$, this is impossible. If $a_0\ne 0$, it simply requires that we set $b_0=\dfrac1{a_0}$.2012-12-29
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As you can see, $(A_0,A_1,A_2)$ has an inverse iff $A_0\not=0$. (Otherwise you wont be able to solve for an inverse)

Actually this monoid is ismorphic to $R[x]/$ under multiplication

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    how do i list all the inverses for all invertible elements for this set2012-12-29
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    You want to find real numbers $(B_0,B_1,B_2)$ such that it is an inverse for $(A_0,A_1,A_2)$. The system of equations that you wrote will have a solution iff $A_0\not=0$2012-12-29
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    This would mean, of course, that $B_0 \neq 0$ as well, yes?2012-12-29
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    Sure. This is because $\frac{1}{A_0}\not=0$2012-12-29
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    So it would seem any element $X = \langle X_0, X_1, X_2\rangle$ in this "monoid/ring" is invertible provided $X_0 \ne 0$; that is, those elements $Y = \langle 0, Y_1, Y_2\rangle$ are not invertible?2012-12-29
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    Yes. I agree with you. I think that you feel that there is something incorrect2012-12-29
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    Amr: No...just elaborating for the sake of the OP: who is wanting to know how to determine which elements are not invertible, or which are...so I'm trying to comment instead of answer. Nothing at all is wrong with your answer! Just trying to help the asker make connections. Perhaps I should have commented below the actual question...it's just that the generalization comes precisely from your, and Brian's, answers.2012-12-29
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    @amWhy . Its OK!. One of the reasons I come to this site is to learn. This is why I asked you if there is something incorrect in my reasoning.2012-12-29