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

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 # Prove equivalence of different forms for additive inverses of fractions

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

Prove that if .

Proof. Since by Theorem I.9 (I.3.3, Exercise #1, part (c) we have . So, Similarly, # Prove that the inverse of a product is the product of the inverses

Prove that if .

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

Prove that .

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

Prove that .

Proof. By Theorem I.3 we have . Then, using I.3.3, Exercise #5 we have, By Theorem I.4, , so # Prove -(a+b) = -a-b

Prove that .

Proof. By Theorem I.3, we have , so Thus, indeed is the additive inverse of ; hence, by definition # Zero has no reciprocal

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

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, since equality is transitive. However, this contradicts field Axiom 4 that and must be distinct elements # Multiplicative identity is its own multiplicative inverse

Prove that .

Proof. On the one hand since 1 is the multiplicative identity we have, On the other hand, from Theorem I.10 (Exercise I.3.3, #1 part (d), we have . Hence, Therefore, since , # Additive identity is its own inverse

Prove that .

Proof. On the one hand, by Axiom 4, we have, On the other hand, since is the additive inverse of 0, we have, Therefore, 