Tag: Fields

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

by RoRi

June 30, 2015

Prove that $x^2+1 = 0$ has no solution $x \in \mathbb{R}$ .

Proof. From Theorem I.20 we know that if $x \neq 0$ , then $x^2 > 0$ . From Theorem I.21 we know $1 > 0$ . Thus, if $x \neq 0$ , then $x^2 + 1 > 0$ . On the other hand, if $x = 0$ , then $x^2+1=1 > 0$ . Hence, there is no $x \in \mathbb{R}$ such that $x^2 + 1 = 0. \qquad \blacksquare$

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) + (b-c) = (a-c)

by RoRi

June 29, 2015

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

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

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

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

by RoRi

June 29, 2015

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

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

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