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,

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Stumbling Robot

A Fraction of a Dot
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Tag: Inverses

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

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

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Taking reciprocals is order-reversing

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

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,

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,

Prove that if and are positive reals with , then .

So, by Theorem I.19, we have

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 .

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 ,