Exercise 1.1 · Q4: $a \le b$ on $\R$

Basics you need

Solution

1. Translate the rule into a statement about a difference

   By Basic 1,

   $$(a,b)\in R \iff a \le b \iff b - a \ge 0.$$

   Test it on a few pairs before proving anything:

   Pair $(a,b)$	$b-a$	$b - a \ge 0$⁠?	In $R$⁠?
   $(1,2)$	$2-1 = 1$	yes	yes
   $(2,1)$	$1-2 = -1$	no	no
   $(3,3)$	$3-3 = 0$	yes	yes
   $(-5,-2)$	$-2-(-5) = 3$	yes	yes
   $(\tfrac12, -3)$	$-3-\tfrac12 = -\tfrac72$	no	no

   An inequality like $a \le b$ compares two numbers; a difference $b-a$ is a single number whose sign we can track through addition. Reducing every question to "is this difference $\ge 0$⁠?" lets the transitive step use the simple fact in Basic 5. The table already hints at the answer: $(1,2)$ is in $R$ but $(2,1)$ is not.
   $a \le b$ and $b - a \ge 0$ say the same thing: subtracting $a$ from both sides of an inequality keeps it true.

2. Reflexive: is every real number related to itself?

   We must show $(a,a)\in R$ for every $a \in \R$⁠. Take any real number $a$⁠. Then

   $$a - a = 0 \ge 0,$$

   so $a \le a$⁠, that is $(a,a)\in R$⁠. Since $a$ was arbitrary, $R$ is reflexive.

   This is where the "or equal to" in $\le$ matters. The pair $(a,a)$ needs $a \le a$⁠, and that is true only because equality is allowed. (For the strict relation $a < b$ the same step fails, since $a < a$ is false for every $a$⁠.) The argument used no special property of $a$⁠, so it covers every real number at once.
   Reflexive means $(a,a)\in R$ for every element of the set (Definition 3(i), Section 1.2).

3. Transitive: does $a \le b$ and $b \le c$ give $a \le c$⁠?

   Take arbitrary reals $a, b, c$ with $(a,b)\in R$ and $(b,c)\in R$⁠. By Step 1,

   $$b - a \ge 0 \qquad\text{and}\qquad c - b \ge 0.$$

   We want information about $c - a$⁠. Add the two differences; the middle number $b$ cancels:

   $$(b - a) + (c - b) = c - a.$$

   The left side is a sum of two numbers that are each $\ge 0$⁠, so it is $\ge 0$ (Basic 5). Hence $c - a \ge 0$⁠, that is $a \le c$⁠, and $(a,c)\in R$⁠. So $R$ is transitive.

   Transitivity is an "if … then" statement, so we start from the two given facts and derive the third. Adding is the right move because it removes $b$⁠: $-b + b = 0$⁠. The conclusion we need, $a \le c$⁠, does not mention $b$⁠, so $b$ has to disappear. On the number line this says: if $a$ is at or left of $b$⁠, and $b$ is at or left of $c$⁠, then $a$ is at or left of $c$⁠.
   Transitive: $(a,b)\in R$ and $(b,c)\in R \Rightarrow (a,c)\in R$ (Definition 3(iii)). The sum of two non-negative numbers is non-negative.

4. Not symmetric: one pair that breaks it

   Symmetric would mean: whenever $(a,b)\in R$⁠, also $(b,a)\in R$⁠. Try $a = 1$⁠, $b = 2$⁠:

   $$\begin{aligned} (1,2)&:\ 2 - 1 = 1 \ge 0, &&\text{so } 1 \le 2 \text{ and } (1,2)\in R;\\ (2,1)&:\ 1 - 2 = -1 < 0, &&\text{so } 2 \le 1 \text{ is false: } (2,1)\notin R. \end{aligned}$$

   So $(1,2)\in R$ but $(2,1)\notin R$⁠. The implication fails for this pair, and $R$ is not symmetric.

   Symmetry is a claim about all pairs in $R$⁠, so one pair where it fails disproves it completely. We chose two different numbers on purpose: for $a = b$ the pair and its reverse are the same pair, so they can never be a counterexample. In fact, if both $a \le b$ and $b \le a$ hold, then $a = b$⁠. So the reverse of a pair in $R$ is in $R$ only when the two entries are equal; every pair with $a < b$ is a counterexample.
   To disprove "for all", give one explicit counterexample and show the arithmetic. Basics: proving and disproving

5. Conclude

   $R$ is reflexive (Step 2) and transitive (Step 3), but not symmetric (Step 4). This is exactly what the question asked us to show. Since symmetry fails, $R$ is not an equivalence relation.

   An equivalence relation needs all three properties (Definition 4). Relations like $\le$ that are reflexive and transitive but only "one-way" are the standard model of an ordering, which is a different kind of relation from "being alike".

Every fact used, and where it comes from