The Lumenary · Darasa la 6–8

Integers

Jina: ______________________

Tarehe: ______________________

Utajifunza

1.
Compute $-8 + 5$ .
2.
Compute $-4 - (-9)$ .
3.
Compute $(-3) \cdot (-4) \cdot (-1)$ .
4.
Compute $(-6) + (-7)$ .
5.
In the proof that $(-a)(-b) = ab$ , we wrote $(-a)(-b) + (-a)b = (-a)\big((-b) + b\big)$ . Which property justifies this step?
- A.The distributive law
- B.Commutativity of multiplication
- C.The sign rule itself
- D.Additive identity
6.
For which integers $a$ is there an integer $b$ with $a \cdot b = 1$ ? (Choose all that apply.)
- A. $a = 1$
- B. $a = -1$
- C. $a = 2$
- D. $a = 0$
7.
At dawn in Ulaanbaatar the temperature is $-23\,^\circ\mathrm{C}$ . By noon it has risen by $15$ degrees. What is the noon temperature, in $^\circ\mathrm{C}$ ?
8.
Write find_mult_partner(a, limit) that searches the integers $b$ with $-\text{limit} \le b \le \text{limit}$ and returns the first $b$ with a * b == 1, or None if there is none. Then think: why does searching further never help when $|a| \ge 2$ ?
Kikokotozi — Mazoezi

Majibu — Integers

1.
Starting at $-8$ and moving $5$ to the right lands on $-3$ . So $-8 + 5 = -3$ .
2.
$-4 - (-9) = -4 + 9 = 5$ .
3.
$(-3)(-4) = 12$ by the sign rule, and $12 \cdot (-1) = -12$ . An odd number of negative factors gives a negative product.
4.
$(-6) + (-7) = -13$ . The sign rule $(-a)(-b) = ab$ applies to multiplication, not addition.
5.
The step is the distributive law $c(x + y) = cx + cy$ , used with $c = -a$ , $x = -b$ , $y = b$ . Using the sign rule here would be circular reasoning.
6.
Only $a = 1$ and $a = -1$ . If $ab = 1$ then $|a|\,|b| = 1$ , and the only way two whole numbers multiply to $1$ is $1 \cdot 1$ . So $|a| = 1$ .
7.
$-23 + 15 = -8$ , so it is $-8\,^\circ\mathrm{C}$ at noon.
8.
```
def find_mult_partner(a, limit):
    for b in range(-limit, limit + 1):
        if a * b == 1:
            return b
    return None
```
Searching further never helps for $|a| \ge 2$ : if $b \ne 0$ then $|ab| = |a|\,|b| \ge 2$ , and if $b = 0$ then $ab = 0$ . Either way $ab \ne 1$ . The search is evidence; this short argument is the proof.