Exercise 1.1 · Q8: $|a-b|$ even on {1, …, 5}

Basics you need

Solution

1. Remove the absolute value

   For any $a, b$⁠, $|a-b|$ is either $a-b$ or $-(a-b)$ (Basic 2), and a number is even exactly when its negative is even (Basic 1). So

   $$\begin{aligned} (a,b)\in R &\iff |a-b| \text{ is even}\\ &\iff a - b \text{ is even}\\ &\iff a - b = 2k \text{ for some integer } k. \end{aligned}$$
   Absolute values are awkward in algebra because $|x| + |y|$ need not equal $|x+y|$⁠. The transitive step needs to add differences, so we first replace "$|a-b|$ is even" by the equivalent "$a-b$ is even". This is valid because removing the absolute value changes at most the sign, and the sign does not affect evenness.
   Example: $|2-5| = |-3| = 3$ is odd and $2 - 5 = -3$ is odd; $|1-5| = |-4| = 4$ is even and $1 - 5 = -4 = 2\cdot(-2)$ is even.

2. Reflexive

   Take any $a\in A$⁠. Then $|a - a| = |0| = 0 = 2\cdot 0$⁠, which is even. So $(a,a)\in R$ for every $a$⁠, and $R$ is reflexive.

   Every element differs from itself by 0, and 0 counts as even. The argument does not depend on which $a$ we take.
   0 is even, because $0 = 2\cdot 0$ with $0$ an integer (Basic 1).

3. Symmetric

   Suppose $(a,b)\in R$⁠, so $|a-b|$ is even. Since $b - a = -(a-b)$⁠, we have

   $$|b - a| = |-(a-b)| = |a-b|,$$

   which is even. So $(b,a)\in R$⁠, and $R$ is symmetric.

   The absolute value makes symmetry immediate: the distance from $a$ to $b$ is the same as from $b$ to $a$⁠. We started from $(a,b)\in R$ and derived $(b,a)\in R$⁠, as the definition requires.
   $|-x| = |x|$ for every real $x$ (Basic 2).

4. Transitive

   Suppose $(a,b)\in R$ and $(b,c)\in R$⁠. By Step 1 there are integers $k$ and $m$ with

   $$a - b = 2k \qquad\text{and}\qquad b - c = 2m.$$

   Adding makes $b$ cancel:

   $$a - c = (a-b) + (b-c) = 2k + 2m = 2(k+m).$$

   $k + m$ is an integer, so $a - c$ is even, and by Step 1 $|a-c|$ is even. So $(a,c)\in R$⁠, and $R$ is transitive.

   Being reflexive, symmetric and transitive, $R$ is an equivalence relation.

   We add the two equations because that removes the middle element $b$⁠, and the conclusion $(a,c)\in R$ does not mention $b$⁠. Using two letters $k$ and $m$ is important: the two differences need not be equal (for example $5 - 1 = 4$ and $1 - 3 = -2$⁠). This is why Step 1 removed the absolute values: the signed differences add up exactly, $4 + (-2) = 2 = 5 - 3$⁠.
   Even + even = even: $2k + 2m = 2(k+m)$⁠. Integers are closed under addition.

5. All of $\{1,3,5\}$ are related, and all of $\{2,4\}$ are related

   Check every pair inside each set (9 ordered pairs from $\{1,3,5\}$⁠, 4 from $\{2,4\}$⁠). The differences that occur:

   Set	Pairs $(a,b)$	$|a-b|$	Even?
   $\{1,3,5\}$	$(1,1),(3,3),(5,5)$	$0$	yes
   	$(1,3),(3,1),(3,5),(5,3)$	$2$	yes
   	$(1,5),(5,1)$	$4$	yes
   $\{2,4\}$	$(2,2),(4,4)$	$0$	yes
   	$(2,4),(4,2)$	$2$	yes

   All 13 pairs have an even difference, so every element of $\{1,3,5\}$ is related to every element of $\{1,3,5\}$⁠, and likewise for $\{2,4\}$.

   In general: if $a = 2i+1$ and $b = 2j+1$ are both odd, $a-b = 2(i-j)$ is even; if $a = 2i$⁠, $b = 2j$ are both even, $a - b = 2(i-j)$ is even.

   The sets are small, so a complete table is a full proof. The general line underneath explains why it works and would work for any odd or even numbers.
   Same parity $\Rightarrow$ even difference (Basic 4).

6. No element of $\{1,3,5\}$ is related to any element of $\{2,4\}$

   There are $3\times 2 = 6$ pairs with $a\in\{1,3,5\}$ and $b\in\{2,4\}$⁠:

   $$\begin{aligned} |1-2| &= 1, & |1-4| &= 3,\\ |3-2| &= 1, & |3-4| &= 1,\\ |5-2| &= 3, & |5-4| &= 1. \end{aligned}$$

   Every one is odd, so none of these pairs is in $R$⁠. By symmetry of $|\cdot|$⁠, the 6 reversed pairs $(b,a)$ have the same odd values and are not in $R$ either. So no element of $\{1,3,5\}$ is related to any element of $\{2,4\}$.

   "No element of one set is related to any element of the other" is a statement about all 6 cross pairs (12 counting both orders), so we check them all. In general an odd number minus an even number is odd (Basic 4), which is why every cross difference came out as 1 or 3.
   So $R$ has exactly two equivalence classes: $[1] = \{1,3,5\}$ and $[2] = \{2,4\}$⁠. They are disjoint and their union is $A$⁠. Basics: equivalence classes

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