Q.Show that the relation R in the set $\mathbf{R}$ of real numbers, defined as $R = \{(a, b) : a \le b^2\}$ is neither reflexive nor symmetric nor transitive.

Reflexive? Need $a>a$ for all $a$ — false. Not reflexive.

Symmetric? If $a>b$ then $b>a$ is false. Not symmetric.

Transitive? If $a>b$ and $b>c$ then $a>c$ — true. Transitive.

Concept and Intuition

We are checking whether a relation defined by a specific algebraic condition is reflexive, symmetric, transitive, or all three (equivalence). The key is to test each property using simple numbers, especially rational and irrational ones. The presence of $\sqrt{2}$ is a deliberate trap: it shifts the usual "difference is rational" idea into something that behaves differently under sign changes and addition.

Step-by-step reasoning

1. Reflexive property

   For reflexivity, we need $xRx$ for every real $x$.

   Compute $x - x + \sqrt{2} = \sqrt{2}$, which is irrational.

   So $xRx$ holds for all $x$.

   Result: R is reflexive.

2. Symmetric property

   For symmetry, if $xRy$ then we must have $yRx$.

   Suppose $xRy$ means $x - y + \sqrt{2}$ is irrational.

   Then $yRx$ requires $y - x + \sqrt{2}$ to be irrational.

   Notice that $y - x + \sqrt{2} = -(x - y) + \sqrt{2}$.

   If $x - y$ is rational, then $x - y + \sqrt{2}$ is irrational (since $\sqrt{2}$ is irrational), but $-(x - y) + \sqrt{2}$ is also irrational (same reasoning). So that case works.

   But what if $x - y$ is irrational? Then $x - y + \sqrt{2}$ could be rational. Example: let $x - y = 1 - \sqrt{2}$. Then $x - y + \sqrt{2} = 1$, which is rational — so $xRy$ fails. But we need a counterexample where $xRy$ holds but $yRx$ fails.

   Choose $x = \sqrt{2}$, $y = 0$. Then $x - y + \sqrt{2} = \sqrt{2} + \sqrt{2} = 2\sqrt{2}$, irrational → $xRy$ holds.

   Now check $yRx$: $y - x + \sqrt{2} = 0 - \sqrt{2} + \sqrt{2} = 0$, which is rational → $yRx$ fails.

   Result: R is not symmetric.

3. Transitive property

   For transitivity, if $xRy$ and $yRz$, we need $xRz$.

   Let $x = 0$, $y = \sqrt{2}$, $z = 2\sqrt{2}$.

   Check $xRy$: $0 - \sqrt{2} + \sqrt{2} = 0$, rational → $xRy$ fails. So this triple doesn't test transitivity.

   We need a triple where both $xRy$ and $yRz$ hold.

   Try $x = 0$, $y = 1$, $z = 2$.

   $xRy$: $0 - 1 + \sqrt{2} = \sqrt{2} - 1$, irrational → holds.

   $yRz$: $1 - 2 + \sqrt{2} = \sqrt{2} - 1$, irrational → holds.

   $xRz$: $0 - 2 + \sqrt{2} = \sqrt{2} - 2$, irrational → holds. That works, but we need a counterexample.

   Let’s try $x = 0$, $y = \sqrt{2}$, $z = 2\sqrt{2}$ again but adjust so $xRy$ holds.

   Actually, pick $x = 0$, $y = 1 - \sqrt{2}$, $z = 2 - 2\sqrt{2}$.

   $xRy$: $0 - (1 - \sqrt{2}) + \sqrt{2} = -1 + 2\sqrt{2}$, irrational → holds.

   $yRz$: $(1 - \sqrt{2}) - (2 - 2\sqrt{2}) + \sqrt{2} = -1 + 2\sqrt{2}$, irrational → holds.

   $xRz$: $0 - (2 - 2\sqrt{2}) + \sqrt{2} = -2 + 3\sqrt{2}$, irrational → holds. Still works.

   We need a case where the sum of two irrationals becomes rational.

   Let $x = 0$, $y = \sqrt{2}$, $z = 0$.

   $xRy$: $0 - \sqrt{2} + \sqrt{2} = 0$, rational → fails.

   Better: Let $x = 0$, $y = 1$, $z = 1 + \sqrt{2}$.

   $xRy$: $0 - 1 + \sqrt{2} = \sqrt{2} - 1$, irrational → holds.

   $yRz$: $1 - (1 + \sqrt{2}) + \sqrt{2} = 0$, rational → fails.

   Try $x = 0$, $y = \sqrt{2}$, $z = 2\sqrt{2}$.

   $xRy$: $0 - \sqrt{2} + \sqrt{2} = 0$, rational → fails.

   Let’s construct: We want $x - y + \sqrt{2}$ irrational and $y - z + \sqrt{2}$ irrational, but $x - z + \sqrt{2}$ rational.

   Let $x - y = a$, $y - z = b$, then $x - z = a + b$.

   We need $a + \sqrt{2}$ irrational, $b + \sqrt{2}$ irrational, but $(a+b) + \sqrt{2}$ rational.

   Choose $a = 1$, $b = -1 + \sqrt{2}$. Then $a + \sqrt{2} = 1 + \sqrt{2}$ (irrational), $b + \sqrt{2} = -1 + 2\sqrt{2}$ (irrational), but $a+b = \sqrt{2}$, so $(a+b)+\sqrt{2} = 2\sqrt{2}$ (irrational) — not rational.

   Choose $a = \sqrt{2}$, $b = -\sqrt{2}$. Then $a + \sqrt{2} = 2\sqrt{2}$ (irrational), $b + \sqrt{2} = 0$ (rational) — fails.

   Choose $a = 1 - \sqrt{2}$, $b = \sqrt{2}$. Then $a + \sqrt{2} = 1$ (rational) — fails.

   Choose $a = \sqrt{2}$, $b = 1 - \sqrt{2}$. Then $a + \sqrt{2} = 2\sqrt{2}$ (irrational), $b + \sqrt{2} = 1$ (rational) — fails.

   The trick: Let $a = \sqrt{2}$, $b = 1$. Then $a + \sqrt{2} = 2\sqrt{2}$ (irrational), $b + \sqrt{2} = 1 + \sqrt{2}$ (irrational), $a+b = 1 + \sqrt{2}$, so $(a+b)+\sqrt{2} = 1 + 2\sqrt{2}$ (irrational) — not rational.

   We need $a+b = k - \sqrt{2}$ for some rational $k$ so that $(a+b)+\sqrt{2}=k$ is rational.

   Let $a = \sqrt{2}$, $b = 1 - 2\sqrt{2}$. Then $a + \sqrt{2} = 2\sqrt{2}$ (irrational), $b + \sqrt{2} = 1 - \sqrt{2}$ (irrational), $a+b = 1 - \sqrt{2}$, so $(a+b)+\sqrt{2} = 1$ (rational). Perfect!

   Now pick concrete numbers: Let $x = 0$, then $y = -a = -\sqrt{2}$, and $z = y - b = -\sqrt{2} - (1 - 2\sqrt{2}) = -\sqrt{2} - 1 + 2\sqrt{2} = \sqrt{2} - 1$.

   Check: $xRy$: $0 - (-\sqrt{2}) + \sqrt{2} = 2\sqrt{2}$ (irrational) → holds.

   $yRz$: $(-\sqrt{2}) - (\sqrt{2} - 1) + \sqrt{2} = -\sqrt{2} - \sqrt{2} + 1 + \sqrt{2} = 1 - \sqrt{2}$ (irrational) → holds.

   $xRz$: $0 - (\sqrt{2} - 1) + \sqrt{2} = 1$ (rational) → fails.

   So transitivity fails.

4. Equivalence property

   Since the relation is not symmetric and not transitive, it cannot be an equivalence relation.

The key here is to check each property — reflexivity, symmetry, transitivity — against the given set. Many students mistakenly think a relation must have more pairs to be symmetric or transitive, but the definitions are about conditions on the pairs that are present, not about having extra ones.

1. Reflexivity requires every element to be related to itself. The set is $\{1,2,3\}$, and $R$ contains $(1,1)$, $(2,2)$, and $(3,3)$. Every element appears, so $R$ is reflexive.

2. Symmetry requires that whenever $(a,b)$ is in $R$, $(b,a)$ must also be in $R$. Here, every pair is of the form $(x,x)$. For such a pair, the reverse is the same pair, so it is automatically present. Thus $R$ is symmetric.

3. Transitivity requires that whenever $(a,b)$ and $(b,c)$ are in $R$, then $(a,c)$ must also be in $R$. In $R$, the only pairs are $(1,1)$, $(2,2)$, $(3,3)$. So the only possible "chain" is something like $(1,1)$ and $(1,1)$ — which would require $(1,1)$, and it is there. No other chains exist, so the condition holds vacuously. Hence $R$ is transitive.

Since $R$ is reflexive, symmetric, and transitive, it is an equivalence relation. The options list "symmetric only", "transitive only", and "reflexive only" — but $R$ is all three, so only option (B) fits.

The key idea is that the domain of a relation is the set of all first coordinates (here $x$) that actually appear in some ordered pair satisfying the given condition. Since $y$ must be a natural number, we can't just pick any $x$ — we need $x = 8 - 2y$ to be positive (natural numbers are usually $1,2,3,\dots$) and $y$ itself must be natural.

Let's work through it:

1. Rewrite the equation for $x$ in terms of $y$:

   From $x + 2y = 8$, we get $x = 8 - 2y$.

   For $x$ to be a natural number, $8 - 2y$ must be a positive integer (usually $\ge 1$).

2. Find the possible natural values of $y$:

   Since $y \in \mathbb{N}$, the smallest $y$ is $1$.

   • If $y = 1$, then $x = 8 - 2(1) = 6$.
   • If $y = 2$, then $x = 8 - 4 = 4$.
   • If $y = 3$, then $x = 8 - 6 = 2$.
   • If $y = 4$, then $x = 8 - 8 = 0$, but $0$ is not a natural number (in most conventions).
   • For $y \ge 4$, $x$ becomes $0$ or negative, which are not natural.

3. Collect the $x$ values:

   The valid $(x,y)$ pairs are $(6,1)$, $(4,2)$, and $(2,3)$.

   So the domain (set of all $x$ that appear) is $\{2,4,6\}$.

Step 1 — Understand the relation.

Since all elements of $A$ are positive, we may divide by $bd$:

So $R$ says: the two ordered pairs have the same ratio. (This is precisely why $R$ is an equivalence relation — equality of ratios is reflexive, symmetric and transitive.)

Step 2 — Write the class.

The equivalence class of $(3,2)$ is

In lowest terms $\gcd(3,2)=1$, so every such pair is a common multiple:

Step 3 — Impose the constraint that BOTH entries lie in $A$.

$A=\{2,3,4,\dots,18\}$, so we need $2\le 3k\le 18$ and $2\le 2k\le 18$, i.e. $k\le 6$ (from $3k\le18$) and $k\ge 1$ (from $2k\ge2$).

Step 4 — List them.

Each passes the check $ad=bc$: e.g. $(3,2)R(18,12)$ since $3\cdot12=36=2\cdot18$ ✓.

Step 5 — Count.

There are exactly 6 ordered pairs in the class.

ANSWER: C

Concept & Intuition

Two numbers are co-prime (or relatively prime) if their greatest common divisor (GCD) is 1. Here, we have a relation from set A = {2, 3, 4} to set B = {3, 6, 7, 10}. For each x in A, we find all y in B such that gcd(x, y) = 1. The range of R is the set of all y that are paired with at least one x. So we simply check each y: does it have any co-prime partner in A? If yes, it belongs to the range.

Step-by-step reasoning

1. Check y = 3

   • gcd(2, 3) = 1 → co-prime.
   • So 3 is in the range (paired with 2, also with 4 since gcd(4,3)=1).

2. Check y = 6

   • gcd(2, 6) = 2 → not co-prime.
   • gcd(3, 6) = 3 → not co-prime.
   • gcd(4, 6) = 2 → not co-prime.
   • No x in A is co-prime with 6, so 6 is not in the range.

3. Check y = 7

   • gcd(2, 7) = 1 → co-prime.
   • So 7 is in the range.

4. Check y = 10

   • gcd(3, 10) = 1 → co-prime.
   • So 10 is in the range (also gcd(7,10)=1 but 7 is not in A; the point is at least one x works).

Thus the range = {3, 7, 10}.

Step 1 — State the conditions

$R = \{(a,b) : a = b-2,\ b > 6\}$, so a pair $(a,b)$ belongs to $R$ only if both $a=b-2$ and $b>6$ hold.

Step 2 — Test each option

• (A) $(2,4)$: $a=b-2 \Rightarrow 2 = 4-2 = 2$ ✓, but $b>6 \Rightarrow 4>6$ ✗. Rejected.
• (B) $(3,8)$: $a=b-2 \Rightarrow 3 \stackrel{?}{=} 8-2=6$ ✗. Rejected.
• (C) $(6,8)$: $a=b-2 \Rightarrow 6 = 8-2 = 6$ ✓, and $b>6 \Rightarrow 8>6$ ✓. Both hold.
• (D) $(8,7)$: $a=b-2 \Rightarrow 8 \stackrel{?}{=} 7-2=5$ ✗. Rejected.

Step 3 — Conclusion

Only $(6,8)$ satisfies both conditions.