Exercise 1.1 · Q5: $a \le b^3$ on $\R$

Basics you need

Solution

1. Get a feel for the rule

   $(a,b)\in R$ means $a \le b^3$⁠. A few pairs:

   Pair $(a,b)$	$b^3$	Is $a \le b^3$⁠?	In $R$⁠?
   $(1,2)$	$8$	$1 \le 8$	yes
   $(2,1)$	$1$	$2 \le 1$ is false	no
   $(\tfrac12,\tfrac12)$	$\tfrac18$	$\tfrac12 \le \tfrac18$ is false	no
   $(2,2)$	$8$	$2 \le 8$	yes
   $(-2,-2)$	$-8$	$-2 \le -8$ is false	no

   Computing a few cases first shows where the trouble is. Already the table contains a pair whose reverse is not in $R$⁠, and two numbers not related to themselves. The next steps turn these observations into proper disproofs.
   For each pair, cube the second entry, then compare with the first.

2. Reflexive? No: $\tfrac12$ is not related to itself

   Reflexive would need $a \le a^3$ for every real $a$⁠. Try $a = \tfrac12$⁠:

   $$\left(\tfrac12\right)^3 = \tfrac12\cdot\tfrac12\cdot\tfrac12 = \tfrac18, \qquad \tfrac12 = \tfrac48 > \tfrac18.$$

   So $\tfrac12 \le \left(\tfrac12\right)^3$ is false and $\left(\tfrac12, \tfrac12\right) \notin R$⁠. Hence $R$ is not reflexive.

   It is worth seeing exactly which numbers fail. Factor:

   $$a^3 - a = a(a^2 - 1) = a(a-1)(a+1).$$

   $(a,a)\in R$ means $a^3 - a \ge 0$⁠. The three factors change sign at $-1$⁠, $0$ and $1$⁠, and checking the sign on each stretch gives

   $$a \le a^3 \iff -1 \le a \le 0 \ \text{ or } \ a \ge 1.$$

   So the numbers with $(a,a)\notin R$ are exactly those with $a < -1$ or $0 < a < 1$⁠. For example $a=-2$⁠: $(-2)^3 = -8$ and $-2 > -8$⁠.

   Reflexivity is a "for every $a$" statement, so one number that fails settles it. We picked $\tfrac12$ because numbers between 0 and 1 shrink when cubed (Basic 1). The factorisation is not needed for the disproof, but it explains the answer: $R$ is reflexive on parts of $\R$⁠, just not on all of it, and "reflexive" requires all of it.
   A product of three factors is $\ge 0$ when an even number of them is negative (or one is zero). Sign check, $a = -2$⁠: $(-)(-)(-) < 0$⁠; $a = -\tfrac12$⁠: $(-)(-)(+) > 0$⁠; $a = \tfrac12$⁠: $(+)(-)(+) < 0$⁠; $a = 2$⁠: $(+)(+)(+) > 0$⁠.

3. Symmetric? No: $(1,2)$ is in $R$⁠, $(2,1)$ is not

   Take $a = 1$⁠, $b = 2$⁠.

   $$\begin{aligned} (1,2)&:\ \text{is } 1 \le 2^3 = 8\,? &&\text{yes, so } (1,2)\in R;\\ (2,1)&:\ \text{is } 2 \le 1^3 = 1\,? &&\text{no, so } (2,1)\notin R. \end{aligned}$$

   The implication $(a,b)\in R \Rightarrow (b,a)\in R$ fails for this pair, so $R$ is not symmetric.

   Reversing a pair changes which number gets cubed. $(1,2)$ compares 1 with $2^3 = 8$⁠, which is easy to satisfy; $(2,1)$ compares 2 with $1^3 = 1$⁠, which fails. One such pair disproves symmetry for the whole relation.
   Symmetric: $(a,b)\in R \Rightarrow (b,a)\in R$ for all $a, b$ (Definition 3(ii)). One failure is enough.

4. Transitive? No: build a chain that breaks

   We need $a, b, c$ with

   $$a \le b^3 \ \text{ and } \ b \le c^3, \qquad\text{but}\qquad a > c^3.$$

   How to find one. The first link lets $a$ be as large as $b^3$⁠; the conclusion only allows $a$ up to $c^3$⁠. So we want $b^3$ to be much bigger than $c^3$⁠, while $b \le c^3$ still holds. Take $b = 3$⁠: then $b^3 = 27$⁠. Choose $c$ so that $c^3$ is just above 3: $c = \tfrac32$ gives $c^3 = \tfrac{27}{8} = 3.375$⁠. Finally pick any $a$ between $3.375$ and $27$⁠, say $a = 10$⁠.

   Check every part.

   $$\begin{aligned} (10, 3)&:\ \text{is } 10 \le 3^3 = 27\,? &&\text{yes;}\\ (3, \tfrac32)&:\ \text{is } 3 \le \left(\tfrac32\right)^3 = \tfrac{27}{8} = 3.375\,? &&\text{yes;}\\ (10, \tfrac32)&:\ \text{is } 10 \le \left(\tfrac32\right)^3 = 3.375\,? &&\text{no.} \end{aligned}$$

   So $(10,3)\in R$ and $(3,\tfrac32)\in R$⁠, but $(10,\tfrac32)\notin R$⁠. Both links are in $R$ but the direct pair is not, so $R$ is not transitive.

   Transitivity says "two links always give a direct link". To disprove it we need one chain where they do not. The search idea above is the useful part: each link "allows" the first number to be up to the cube of the second, and cubing a number above 1 makes it much larger. So the allowance of two links together ($a$ up to $27$⁠) is far more generous than a single link from $a$ to $c$ ($a$ up to $3.375$⁠). Any $a$ with $3.375 < a \le 27$ works; $a = 4$ would do as well as $a = 10$⁠.
   Transitive: $(a,b)\in R$ and $(b,c)\in R \Rightarrow (a,c)\in R$ for all $a,b,c$ (Definition 3(iii)). A counterexample must show both links in $R$ and the direct pair not in $R$⁠.

5. Conclude

   $R$ is not reflexive (Step 2), not symmetric (Step 3) and not transitive (Step 4). So $R$ is neither reflexive nor symmetric nor transitive.

   The question says "check whether", so the answer must state each property separately, with its reason. Here all three fail, each by its own counterexample.

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