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# System prepended metadata

title: Week 3&4

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# Week 3&4

## Problem Set 1

Suppose a, b, c are any three integers. We must show that $(a-b)(a-c)(b-c)$ is even.

We consider the cases that there are 0, 1, 2, 3 even integers in a, b, and c.

(i) a, b, and c are even integers (there are 3 even integers in a, b, and c)

By definition of even numbers, let $a=2k_1$, $b=2k_2$, $c=2k_3$. $k_1,k_2,k_3$ are some integers.

$$
\begin{align*}
(a-b)(a-c)(b-c) &= (2k_1-2k_2)(2k_1-2k_3)(2k_2-2k_3)\\
&= 2[(k_1-k_2)(2k_1-2k_3)(2k_2-2k_3)]\\
&= 2c_1
\end{align*}
$$

Where $c_1$ is some integer since integers are closed under multiplication and subtraction. By definition of even numbers, $(a-b)(a-c)(b-c)$ is even.

(ii) a, b, and c are odd integers  (there are 0 even integers in a, b, and c)

By definition of odd numbers, let $a=2k_1+1$, $b=2k_2+1$, $c=2k_3+1$. $k_1,k_2,k_3$ are some integers.

$$
\begin{align*}
(a-b)(a-c)(b-c) &= ((2k_1+1)-(2k_2+1))((2k_1+1)-(2k_3+1))((2k_2+1)-(2k_3+1))\\
&= (2k_1-2k_2)(2k_1-2k_3)(2k_2-2k_3)\\
&= 2[(k_1-k_2)(2k_1-2k_3)(2k_2-2k_3)]\\
&= 2c_2
\end{align*}
$$

Where $c_2$ is some integer since integers are closed under multiplication and subtraction. By definition of even numbers, $(a-b)(a-c)(b-c)$ is even.

(iii) a is even, b and c are odd integers (there are 1 even integers in a, b, and c)

By definition of even and odd numbers, let $a=2k_1$, $b=2k_2+1$, $c=2k_3+1$. $k_1,k_2,k_3$ are some integers.

$$
\begin{align*}
(a-b)(a-c)(b-c) &= (2k_1-(2k_2+1))(2k_1-(2k_3+1))((2k_2+1)-(2k_3+1))\\
&= (2k_1-2k_2-1)(2k_1-2k_3-1)(2k_2-2k_3)\\
&= 2[(2k_1-2k_2-1)(2k_1-2k_3-1)(k_2-k_3)]\\
&= 2c_3
\end{align*}
$$

Where $c_3$ is some integer since integers are closed under multiplication and subtraction. By definition of even numbers, $(a-b)(a-c)(b-c)$ is even.

(iv) a is odd, b and c are even integers (there are 2 even integers in a, b, and c)

By definition of even and odd numbers, let $a=2k_1+1$, $b=2k_2$, $c=2k_3$. $k_1,k_2,k_3$ are some integers.

$$
\begin{align*}
(a-b)(a-c)(b-c) &= ((2k_1+1)-2k_2)((2k_1+1)-2k_3)(2k_2-2k_3)\\
&= (2k_1+1-2k_2)(2k_1+1-2k_3)(2k_2-2k_3)\\
&= 2[(2k_1+1-2k_2)(2k_1+1-2k_3)(k_2-k_3)]\\
&= 2c_4
\end{align*}
$$

Where $c_4$ is some integer since integers are closed under multiplication and subtraction and addition. By definition of even numbers, $(a-b)(a-c)(b-c)$ is even.

Combine (i), (ii), (iii), and (iv), we have considered the cases that there are 0, 1, 2, 3 even integers in a, b, and c. Therefore no matter what a, b, and c is, $(a-b)(a-c)(b-c)$ is even.

## Problem Set 15

![](https://i.imgur.com/S3ajwpv.jpg)
![](https://i.imgur.com/VRlsfal.jpg)
