Tag: Inverses

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

by RoRi

September 5, 2015

Let $f(x) = 2x+5$ . Show that $f$ is strictly monotonic on $\mathbb{R}$ . Find the domain of the inverse of $f$ , denoted by $g$ . Find a formula for computing $g(y)$ for each $y$ in the domain of $g$ .

First, to show $f$ is monotonic let $x_1, x_2 \in \mathbb{R}$ with $x_1 < x_2$ . Then

$\begin{align*} x_1 < x_2 && \implies && 2x_1 < 2x_2 \\ && \implies && 2x_1 + 5 < 2x_2 + 5 \\ && \implies && f(x_1) < f(x_2). \end{align*}$

Hence, $f$ is strictly increasing on $\mathbb{R}$ .

Next,

$\begin{align*} y = 2x + 5 && \implies && x &= \frac{y-5}{2} \\ && \implies && g(y) &= \frac{y-5}{2} \qquad \text{for all } y \in \mathbb{R}. \end{align*}$

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

by RoRi

September 5, 2015

Let $f(x) = x + 1$ . Show that $f$ is strictly monotonic on $\mathbb{R}$ . Find the domain of the inverse of $f$ , denoted by $g$ . Find a formula for computing $g(y)$ for each $y$ in the domain of $g$ .

First, to show $f$ is monotonic let $x_1, x_2 \in \mathbb{R}$ with $x_1 < x_2$ . Then

$x_1 + 1 < x_2 + 1 \quad \implies \quad f(x_1) < f(x_2).$

Hence, $f$ is strictly increasing on $\mathbb{R}$ .

Next,

$y = x+1 \quad \implies \quad x = y-1 \quad \implies \quad g(y) = y-1 \quad \text{for all } y \in \mathbb{R}.$

Taking reciprocals is order-reversing

by RoRi

June 30, 2015

Prove that if $a$ and $b$ are positive reals with $a < b$ , then $0 < b^{-1} < a^{-1}$ .

Proof. Since $a$ and $b$ are positive we have $a \neq 0$ and $b \neq 0$ , so $a^{-1}$ and $b^{-1}$ exist. Then, $a > 0$ and $b > 0$ implies $a^{-1} > 0$ and $b^{-1} > 0$ (by I.3.5, Exercise #4). Hence,

$a^{-1} b^{-1} > 0$

So, by Theorem I.19, we have

$\begin{align*} &&a(a^{-1} b^{-1}) &< b(a^{-1}b^{-1}) \\ \implies&& b^{-1} &< a^{-1}. \qquad \blacksquare \end{align*}$

Prove equivalence of different forms for additive inverses of fractions

by RoRi

June 29, 2015

Prove that if $b \neq 0$ then

$- \left( \frac{a}{b} \right) = \frac{-a}{b} = \frac{a}{-b}.$

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

$\begin{align*} - \left(\frac{a}{b} \right) &= -(ab^{-1}) &(\text{Def of } b^{-1}) \\ &= -(b^{-1}a) \\ &= \left( (-b)^{-1} a \right) &(\text{Thm I.12}) \\ &= a(-b^{-1}) &(\text{Thm I.12}) \\ &= \frac{a}{-b} &(\text{Def of } (-b)^{-1}). \end{align*}$

Then for the other equality, similarly, we have

$\begin{align*} -\left( \frac{a}{b} \right) &= (-a)b^{-1} \\ &= \frac{-a}{b}. \qquad \blacksquare \end{align*}$

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

by RoRi

June 29, 2015

Prove that $-\left( \frac{a}{b} \right) = \frac{-a}{b} = \frac{a}{-b}$ if $b \neq 0$ .

Proof. Since $b \neq 0$ by Theorem I.9 (I.3.3, Exercise #1, part (c) we have $\frac{a}{b} = ab^{-1}$ . So,

$\begin{align*} -\left( \frac{a}{b} \right) &= -(ab^{-1}) \\ &= -(b^{-1} a) & (\text{commutativity}) \\ &= ((-b^{-1})a) & (\text{Theorem I.12}) \\ &= a(-b^{-1}) & (\text{commutativity}) \\ &= \frac{a}{-b} & (\text{Theorem I.9, again}) \end{align*}$

Similarly,

$-(ab^{-1}) = (-a)b^{-1} = \frac{-a}{b}. \qquad \blacksquare$

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

by RoRi

June 29, 2015

Prove that $(ab)^{-1} = a^{-1} b^{-1}$ if $a, b \neq 0$ .

Proof. Since $a, b \neq 0$ we know there exist elements $a^{-1}, b^{-1} \in \mathbb{R}$ such that $aa^{-1} = 1$ and $bb^{-1} = 1$ . So,

$\begin{align*} (ab)^{-1} &= (ab)^{-1} \cdot (aa^{-1}) \cdot (bb^{-1}) \\ &= (ab)^{-1} \cdot (ab) \cdot (a^{-1} b^{-1}) & (\text{Using assoc. and comm.}) \\ &= a^{-1} b^{-1}. \qquad \blacksquare \end{align*}$

Prove -(a-b) = -a+b

by RoRi

June 29, 2015

Prove that $-(a-b) = -a+b$ .

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

$-(a+(-b)) = -a + -(-b).$

By Theorem I.4, $-(-b) =b$ , so

$-a+ -(-b) = -a+b. \qquad \blacksquare$

Prove -(a+b) = -a-b

by RoRi

June 29, 2015

Prove that $-(a+b) = -a-b$ .

Proof. By Theorem I.3, we have $-a-b = -a+(-b)$ , so

$(a+b) + (-a-b) = (a+b) + (-a) + (-b) = (a+(-a))+(b+(-b)) = 0+0 = 0.$

Thus, indeed $-a-b$ is the additive inverse of $(a+b)$ ; hence, by definition

$-(a+b) = -a-b. \qquad \blacksquare$

Zero has no reciprocal

by RoRi

June 29, 2015

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 $a \in \mathbb{R}$ such that $a \cdot 0 = 1$ (this is the definition of multiplicative inverse). Then, by part (b) of this exercise, we know that $a \cdot 0 = 0$ for any $a$ . Hence,

$a \cdot 0 = 1 \quad \text{and} \quad a \cdot 0 = 0 \qquad \implies \qquad 1 = 0$