In what structures does $ (-1)^2 = 1$?

Question

Does $ (-1)^2 = 1$ anywhere you have associativity and an inverse element?

Thanks!

See this answer on the Law of Signs. There we see that it also is true for odd functions under composition. — Bill Dubuque, Jul 01 '15 at 19:11
Thank you @BillDubuque. So it's also true for these Near-Rings ? — Gil-Mor, Jul 02 '15 at 08:01

score 6 · Accepted Answer · answered Jul 01 '15 at 17:30

6

I assume you mean $(-1)^{2} = 1$ for this answer. Furthermore, I assume you mean rings, not groups as in the tag, because I assume the square is the multiplicative operation, yet the minus sign refers to additive inverse. These things should be clarified.

The answer is yes. In general, we can show that $(-1)(x) = -x$ in the following way:

$x + (-1)(x) = (1+-1)(x) = 0x = 0$.

Note that the only requirement here is distributivity, guaranteed in any ring.

If we let $x=-1$, the desired result is there.

answered Jul 01 '15 at 17:30

Jonathan Hebert

5,272

oh so it's true where you have an inverse and distributivity (not associativity..) and that's rings.. So, are there special groups where this is also true? – Gil-Mor Jul 01 '15 at 17:33
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@Gil-Mor Repeat: Not groups – Hagen von Eitzen Jul 01 '15 at 17:35
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Your question doesn't make sense in the context of groups. What do you want $(-1)^{2}$ to mean in a group where you are using additive notation to designate the inverse as $-1$? A group only has one binary operation. – Jonathan Hebert Jul 01 '15 at 17:36
Yes I see.. Thank you very much.. have a good day.. – Gil-Mor Jul 01 '15 at 17:37

In what structures does $ (-1)^2 = 1$?

1 Answers1