Exercise 1.1 · Q2: $a \le b^2$ on $\R$

Basics you need

Solution

1. Picture the relation

   Draw $b$ across and $a$ up. The pair $(a,b)$ is in $R$ exactly when the point lies on or below the curve $a = b^2$⁠.

   A picture shows at once why the properties fail. The curve is not symmetric about the line $a = b$⁠, so swapping $a$ and $b$ can move a point from inside to outside. And near $b$ between 0 and 1 the curve dips below the line $a=b$⁠, so points $(t,t)$ there lie outside the region. We use the picture only to find counterexamples; each one is then checked with exact arithmetic.

2. Not reflexive: the pair $\left(\tfrac12, \tfrac12\right)$

   Reflexive would need $a \le a^2$ for every real $a$⁠. Take $a = \tfrac12$⁠:

   $$a^2 = \left(\tfrac12\right)^2 = \tfrac14.$$

   So the condition $a \le a^2$ reads $\tfrac12 \le \tfrac14$⁠, which is false, since $\tfrac12 = \tfrac24 > \tfrac14$⁠. Hence $\left(\tfrac12,\tfrac12\right)\notin R$⁠, and $R$ is not reflexive.

   Exactly which $a$ fail? $a \le a^2$ means $a^2 - a \ge 0$⁠, that is $a(a-1)\ge 0$⁠. A product is $\ge 0$ when both factors have the same sign or one is 0, which happens for $a \le 0$ or $a \ge 1$⁠. So $(a,a)\in R$ for those $a$⁠, and $(a,a)\notin R$ exactly for $0 < a < 1$⁠.

   Most numbers we first think of, like 2 or −3, do satisfy $a \le a^2$ ($2 \le 4$⁠, $-3 \le 9$⁠). Testing only those would wrongly suggest $R$ is reflexive. The search has to go where squaring shrinks numbers: between 0 and 1 (Basic 4). One such number is all a disproof needs; the factorisation simply shows the whole set of failures, the open interval $(0,1)$⁠.
   A product $xy \ge 0$ exactly when $x$ and $y$ are both $\ge 0$ or both $\le 0$⁠. Here $x = a$⁠, $y = a-1$⁠.

3. Not symmetric: the pair $(1, 2)$

   We need $(a,b)\in R$ with $(b,a)\notin R$⁠. Take $a = 1$⁠, $b = 2$⁠:

   $$\begin{aligned} (1,2)&:\ 1 \le 2^2 = 4, \ \text{ true}; \\ (2,1)&:\ 2 \le 1^2 = 1, \ \text{ false}. \end{aligned}$$

   So $(1,2)\in R$ but $(2,1)\notin R$⁠.

   So $R$ is not symmetric. (The pair $(1,4)$ works just as well: $1 \le 16$ is true but $4 \le 1$ is false. Any such pair works.)

   For the reverse pair $(b,a)$⁠, the roles swap: now $b$ is compared with $a^2$⁠. We want the first inequality true and the second false, that is $a \le b^2$ and $b > a^2$⁠. With $a = 1$ both say the same thing in practice: $a^2 = 1$⁠, so any $b > 1$ gives $b > a^2$⁠, and then $b^2 > 1 = a$ as well. The simplest choice is $b = 2$⁠.
   A symmetry counterexample makes "$(a,b)\in R$" true and "$(b,a)\in R$" false (Basic 3). Both halves are checked above.

4. Not transitive: the pairs $(3,2)$ and $\left(2,\tfrac32\right)$

   We need $(a,b)\in R$ and $(b,c)\in R$ but $(a,c)\notin R$⁠. Take $a = 3$⁠, $b = 2$⁠, $c = \tfrac32$⁠:

   $$\begin{aligned} (3,2)&:\ 3 \le 2^2 = 4, &&\text{true;}\\ \left(2,\tfrac32\right)&:\ 2 \le \left(\tfrac32\right)^2 = \tfrac94 = 2.25, &&\text{true;}\\ \left(3,\tfrac32\right)&:\ 3 \le \tfrac94 = 2.25, &&\text{false.} \end{aligned}$$

   Both links are in $R$⁠, but the shortcut is not. So $R$ is not transitive.

   How were these numbers found? Written out, we need $$b \le c^2 \ <\ a \le b^2.$$ So $c^2$ must fit in the gap between $b$ and $b^2$⁠, and $a$ must sit above $c^2$ but not above $b^2$⁠. That gap only exists when $b < b^2$⁠, for example $b = 2$ with gap from 2 to 4. Pick $c$ with $c^2$ in that gap: $c = \tfrac32$ gives $c^2 = 2.25$⁠. Then pick $a$ between $2.25$ and $4$⁠: $a = 3$⁠. On the figure, $(3,2)$ and $\left(2,\tfrac32\right)$ are green, while $\left(3,\tfrac32\right)$ is rose, sitting just above the curve.
   A transitivity counterexample needs two linked pairs (the second starts where the first ends, here at $b = 2$⁠) and a missing shortcut $(a,c)$ (Basic 2, Basic 3).

5. Conclude

   Each property has failed at a specific, checked example: $\left(\tfrac12,\tfrac12\right)\notin R$⁠; $(1,2)\in R$ but $(2,1)\notin R$⁠; $(3,2),\left(2,\tfrac32\right)\in R$ but $\left(3,\tfrac32\right)\notin R$⁠. Hence $R$ is neither reflexive nor symmetric nor transitive.

   The question says "show", so the answer is a proof, and three verified counterexamples are a complete proof of three "not" statements. No general argument is needed for a property that fails.

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