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This is the second exercise in the second set of chapter 2 in Pinter's A Book of Abstract Algebra.

It is asked to check for commutativity, associativity, identity element in $\mathbb{R}$ and inverse elemente in $\mathbb{R}$ for each described operation, in this case i have a question about $x*y=|x+y|$.

My first question lies in the identity check, this is what i did:
$ x*e = |x + e| = \sqrt {(x+e)^2} = x$ so $e=0$, checking for $e*x $ gives same thing.

Alright, but if x is negative then the operation will not give back a negative x, but a positive x, thus this operation is not injective. The book explicity says that for something to be an operation it needs to be uniquely defined for every $x\, \in\, \mathbb{G}$ I thought about using $ \pm $ but it doesn't make sense to me in that case. Also, i don't have access to the book's solution.

My second question is about the associativity, i did: $ (x*y)*z = |x+y|*z = ||x+y|+z| = \sqrt{(\sqrt{(x+y)^2} + z)^2} = \sqrt{(x+y)^2 + 2z\sqrt{(x+y)^2} + z^2}$
$ x*(y*z) = x*|y+z| = |x + |y+z|| = \sqrt{(\sqrt{(y+z)^2} + x)^2} = \sqrt{(y+z)^2 + 2x\sqrt{(y+z)^2} + x^2}$

Using computational resources i get that $(x*y)*z = x*(y*z)$ will be true under certain conditions, but i just can't visualize it.
Thanks.

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  • $\begingroup$ So, you have managed to show in the first part that there is no identity element and in the second part that associativity does not hold. $\endgroup$ – Ben Sheller Sep 24 '15 at 1:38
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How can $0$ be an identity?

$-1\ast 0 = |-1 + 0| = |-1| = 1 \neq -1$.

It is not hard to see that there is NO real number $e$ that will yield:

$x\ast e = x$ for any $x < 0$.

Also:

$[(-1)\ast(-1)]\ast 1 = |(-1) + (-1)| \ast 1 = |-2|\ast 1 = 2\ast 1 = |2+1| = |3| = 3$, but:

$-1 \ast[(-1)\ast 1] = -1 \ast |-1 + 1| = -1 \ast |0| = -1 \ast 0 = |-1 + 0| = |-1| = 1$.

All is takes is ONE counter-example for disproof.

Without an identity, how can you have inverses?

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  • $\begingroup$ Yes, that was what i was looking for. I thought there could be something more to it. $\endgroup$ – Bernardo L.K. Sep 24 '15 at 2:06

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