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# Show a function is monotonic and find a formula for its inverse

Let . Show that is strictly monotonic on . Find the domain of the inverse of , denoted by . Find a formula for computing for each in the domain of .

First, to show is monotonic let with . Then Hence, is strictly increasing on .

Next, # Show a function is monotonic and find a formula for its inverse

Let . Show that is strictly monotonic on . Find the domain of the inverse of , denoted by . Find a formula for computing for each in the domain of .

First, to show is monotonic let with . Then Hence, is strictly increasing on .

Next, # Taking reciprocals is order-reversing

Prove that if and are positive reals with , then .

Proof. Since and are positive we have and , so and exist. Then, and implies and (by I.3.5, Exercise #4). Hence, So, by Theorem I.19, we have # 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) = -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 , 