Question

The whole numbers a and b are divisible by c. Is a+b divisible by c? Is b−a divisible by c? Explain your reasoning.

Answer 1

Answer:

$a+b$ is divisible by $c$ since $d$ and $e$ are also whole numbers.

$b-a$ is divisible by $c$ since $d$ and $e$ are also whole numbers.

Step-by-step explanation:

Let $a$, $b$ whole number that are divisible by $c$. Then, we have that $d = \frac{a}{c}$ and $e = \frac{b}{c}$, which are also whole numbers. We proceed to prove that $a + b$ is divisible by $c$:

1) $d = \frac{a}{c}$, $e = \frac{b}{c}$, $a$, $b$ , $c$, $d$, $e$ $\in \mathbb{Z}$ Given

2) $d + e = \frac{a}{c}+\frac{b}{c}$ Compatibility with addition

3) $d+e = a\cdot c^{-1}+b\cdot c^{-1}$ Definition of division

4) $d + e = c^{-1}\cdot (a+b)$ Commutative and distributive properties

5) $d+e = (a+b) \cdot c^{-1}$ Commutative properties

6) $d+e = \frac{a+b}{c}$ Definition of division/Result

Therefore, $a+b$ is divisible by $c$ since $d$ and $e$ are also whole numbers.

We proceed to prove that $b-a$ is divisible by $c$:

1) $d = \frac{a}{c}$, $e = \frac{b}{c}$, $a$, $b$ , $c$, $d$, $e$ $\in \mathbb{Z}$ Given

2) $d\cdot (-1) = \frac{a}{c}\cdot (-1)$ Compatibility with multiplication

3) $e + d\cdot (-1) = \frac{b}{c}+\frac{a}{c}\cdot (-1)$ Compatibility with addition

4) $e + d\cdot (-1) = b\cdot c^{-1}+(a\cdot c^{-1})\cdot (-1)$ Definition of division

5) $e + d\cdot (-1) = b\cdot c^{-1} + [a\cdot (-1)]\cdot c^{-1}$ Associative and commutative properties

6) $e + d\cdot (-1) = c^{-1}\cdot [b+a\cdot (-1)]$ Commutative and distributive properties.

7) $e+d\cdot (-1) = [b+a\cdot (-1)]\cdot c^{-1}$ Commutative properties.

8) $e + d\cdot (-1) = \frac{b+a\cdot (-1)}{c}$ Definition of division

9) $e-d = \frac{b-a}{c}$ $(-1)\cdot a = -a$/Definition of subtraction

Therefore, $b-a$ is divisible by $c$ since $d$ and $e$ are also whole numbers.