Ring with unity - (-1)^2 = 1

Exercise 1 in section 7.1 of Dummit and Foote: Abstract Algebra is as follows:

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Let R be a ring with 1

Show that $\displaystyle {(-1)}^2 = 1$

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The web site Project Crazy Project gives the following proof:

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Let $\displaystyle x \in R$

Then $\displaystyle {(-1)}^2 x = (-1)(-x) = x$

Similarly $\displaystyle x {(-1)}^2 = (-x)(-1) = x$

By the uniqueness of 1, $\displaystyle {(-1)}^2 = 1$

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I am not sure why $\displaystyle {(-1)}^2 x = (-1)(-x)$

I can see that by associativity of multiplication $\displaystyle {(-1)}^2 x = (-1)[(-1)(x)]$

But then why is (-1)x = -x ... or is this just a formal way of expressing the same thing?

Further why is (-1)(-x) = x?

Peter

Re: Ring with unity - (-1)^2 = 1

Quote:

Originally Posted by

**Bernhard** But then why is (-1)x = -x?

$\displaystyle \begin{aligned}(-1)x + x &= (-1)x + (1)x\quad\text{(property of 1)} \\ &= (-1+1)x\quad\text{(distributive axiom)} \\ &= 0x\quad\text{(definition of –1)} \\ &= 0\quad\text{(property of 0)} \end{aligned}$

Thus (–1)x is the negative of x, or in other words (–1)x = –x.

Quote:

Originally Posted by

**Bernhard** Further why is (-1)(-x) = x?

From the above result with –x in place of x, it follows that (–1)(–x) = –(–x) = x.