Question

If ab belongs to R if -(-a)=a prove ??
ii) -1(a)=-a
iii) -(a+b)=(-a)+(-b)
iv) -(a)(-b)=ab
Question: Find real number x such that |x+1|+|x-2|<7??

Answer 1

Answer to Question #82335, Math / Real Analysis

Question

(1) For $a, b$ belongs to $R$

(i) $-(-a) = a$

(ii) $-1(a) = -a$

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

(iv) $-(a)(-b) = ab$

(2) Find real number $x$ such that $|x + 1| + |x + 2| < 7$

Solution

(1) Given: $a, b \in R$

(i) Since $-a$ is additive inverse of $a$ it satisfies $-a + a = a + (-a) = 0$. That is $a$ is an additive inverse of $-a$.

Since $-(-a)$ is additive inverse of $-a$ it satisfies

$$-a + \left(-(-a)\right) = -(-a) + (-a) = 0.$$

As we know that additive inverse is unique, we get $a = -(-a)$

Hence, proved

(ii) To prove for every $a \in R$ the additive inverse of $a$ is the product of additive inverse of 1 and $a$, it is sufficient to prove that $(-1)a + a = a + (-1)a = 0$

We will show the second equality holds

$$\begin{array}{l} a + (-1)a = 1 \cdot a + (-1) \cdot a \quad \text{by multiplicative identity} \\ = (1 + (-1)) \cdot a \quad \text{by distributivity} \\ = 0 \cdot a \quad \text{by additive inverse} \\ = 0 \quad \text{for every } a \in R, 0 \cdot a = a \cdot 0 = 0 \\ \end{array}$$

So both equalities hold and as $(-1)a$ and $-a$ are both additive inverse of $a$, $-1(a) = -a$

Hence, proved

(iii) Since $-(a + b)$ is an additive inverse of $a + b$ . We have $-(a + b) + (a + b) = (a + b) + -(a + b)$

We also have

$(a + b) + ((-a) + (-b)) = ((-a) + (-b)) + (a + b)$ by commutativity of addition

$= (-a) + ((-b) + a) + b$ by associativity of addition

$= (-a) + (a + (-b)) + b$ by commutativity of addition

$= ((-a) + a) + ((-b) + b)$ by associativity of addition

$= 0 + 0$ by additive inverse of a and b

$= 0$ by additive identity

Hence, $(-a) + (-b)$ is also additive inverse of $a + b$ .

By uniqueness of additive inverse we get $-(a + b) = (-a) + (-b)$

Hence, proved

(iv)

$(-a)(-b) = (-1(a))(-1(b))$ -1(a)=-a

$= (-1)(a)(-1)(b)$ by commutativity of multiplication

$= (-1)(-1)(a)(b)$ by associativity of multiplication

$= (a)(b)$ $(-1)(-1) = 1$

Hence, proved

(2) $|x + 1| + |x - 2| < 7$

Here we have two check points -1 and 2.

Consider below three cases

Case (i): If $x < -1$ then $|x + 1| + |x - 2|$ expands as $-x - 1 - x + 2 < 7$ gives $x > -3$ . Hence, we get $-3 < x < -1$ .

Case (ii): If $-1 \leq x < 2$ then $|x + 1| + |x - 2|$ expands as $x + 1 - x + 2 < 7$ gives $3 < 7$ which is true for all $x \in R$ . Hence, for $-1 \leq x < 2$ given inequality holds true.

Case (iii): If $x \geq 2$ then $|x + 1| + |x - 2|$ expands as $x + 1 + x - 2 < 7$ gives $x < 4$ . Hence, we get $2 \leq x < 4$ .

combine all three ranges we get

$$(-3 < x < -1) \text{ or } (-1 \leq x < 2 \text{ and true } \forall x \in R) \text{ or } (2 \leq x < 4)$$

That is $-3 < x < 4$

Hence, $-3 < x < 4$ or $x \in (-3,4)$

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