Prove that has no solution .

*Proof.*From Theorem I.20 we know that if , then . From Theorem I.21 we know . Thus, if , then . On the other hand, if , then . Hence, there is no such that

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Stumbling Robot

A Fraction of a Dot
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Tag: Fields

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There is no real number that is the square root of -1

* Proof. * From Theorem I.20 we know that if , then . From Theorem I.21 we know . Thus, if , then . On the other hand, if , then . Hence, there is no such that
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Prove equivalence of different forms for additive inverses of fractions

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Prove -(a/b) = (-a)/b = a/(-b) if b is not zero

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Prove that the inverse of a product is the product of the inverses

* Proof. * Since we know there exist elements such that and . So,
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Prove (a-b) + (b-c) = (a-c)

* Proof. * From Theorem I.3 we know , and . So,
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Prove -(a-b) = -a+b

* Proof. * By Theorem I.3 we have . Then, using I.3.3, Exercise #5 we have,
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Prove -(a+b) = -a-b

* Proof. * By Theorem I.3, we have , so
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Zero has no reciprocal

* Proof. * The proof is by contradiction. Suppose otherwise, that there is some such that (this is the definition of multiplicative inverse). Then, by part (b) of this exercise, we know that for any . Hence,
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Multiplicative identity is its own multiplicative inverse

* Proof. * On the one hand since 1 is the multiplicative identity we have,
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Additive identity is its own inverse

* Proof. * On the one hand, by Axiom 4, we have,

Prove that has no solution .

Prove that if then

*Proof.* We can use the first exercise of this section (Section I.3.3, Exercise #1) and the previous exercise (Section I.3.3, Exercise #8) to compute

Then for the other equality, similarly, we have

Prove that if .

Prove that .

Prove that .

By Theorem I.4, , so

Prove that .

Thus, indeed is the additive inverse of ; hence, by definition

Prove from the field axioms that the additive identity, 0, has no multiplicative inverse.

since equality is transitive. However, this contradicts field Axiom 4 that and must be distinct elements

Prove that .

On the other hand, from Theorem I.10 (Exercise I.3.3, #1 part (d), we have . Hence,

Therefore, since ,

Prove that .

On the other hand, since is the additive inverse of 0, we have,

Therefore,