Q.Let $R$ be a relation defined on the set of natural numbers $\mathbb{N}$ as follows: $R = \{(x, y) : x \in \mathbb{N}, \ y \in \mathbb{N}, \ 2x + y = 41\}$. Find the domain and range of the relation $R$. Also verify whether $R$ is reflexive, symmetric and transitive.

Key idea: The relation is defined by a linear equation $2x + y = 41$ over $\mathbb{N}$. Domain and range come from natural-number solutions; the relation is reflexive, symmetric, and transitive only if it satisfies the respective definitions.

Step 1 – Domain and range.

For $x, y \in \mathbb{N}$, $2x + y = 41 \implies y = 41 - 2x$. Since $y \ge 1$, we have $41 - 2x \ge 1 \implies x \le 20$. Also $x \ge 1$. So $x = 1, 2, \dots, 20$.

Corresponding $y$ values: $y = 39, 37, \dots, 1$ (odd numbers from 1 to 39).

Thus domain = $\{1, 2, \dots, 20\}$, range = $\{1, 3, 5, \dots, 39\}$.

Step 2 – Reflexive?

Need $(x, x) \in R$ for every $x$ in domain. That requires $2x + x = 3x = 41$, i.e. $x = 41/3$, not a natural number. So no $(x, x)$ exists. Not reflexive.

Step 3 – Symmetric?

If $(x, y) \in R$, then $2x + y = 41$. For symmetry we need $(y, x) \in R$, i.e. $2y + x = 41$. Adding the two equations gives $3(x+y) = 82$, so $x+y = 82/3$, impossible for integers. Not symmetric.

Step 4 – Transitive? …

We are working with natural numbers $\mathbb{N}$, which here means positive integers $\{1,2,3,\dots\}$. The condition $2x + y = 41$ ties $x$ and $y$ together: for each $x$, $y = 41 - 2x$. Since $y$ must also be a natural number, $41 - 2x \ge 1$, which gives $x \le 20$. So $x$ can be $1,2,\dots,20$, and $y$ takes the corresponding values $39,37,\dots,1$ — all odd numbers from 1 to 39.

Now let’s check the three properties.

1. Domain and range first.

   For $x = 1$, $y = 39$; for $x = 2$, $y = 37$; …; for $x = 20$, $y = 1$.

   So domain $= \{1,2,3,\dots,20\}$.

   Range $= \{39,37,35,\dots,1\}$, i.e., all odd numbers from 1 to 39.

2. Reflexive?

   A relation is reflexive if $(x,x) \in R$ for every $x$ in the domain. That would require $2x + x = 3x = 41$, i.e., $x = 41/3$, which is not an integer, let alone a natural number. So no pair $(x,x)$ exists at all.

   $R$ is not reflexive.

3. Symmetric?

   Symmetry means: if $(x,y) \in R$, then $(y,x) \in R$.

   Take $(1,39)$: $2(1) + 39 = 41$, so it’s in $R$. For symmetry we’d need $(39,1)$ in $R$, i.e., $2(39) + 1 = 79$, not $41$. So symmetry fails.

   In fact, if $(x,y) \in R$, then $2x + y = 41$. For $(y,x)$ to be in $R$, we’d need $2y + x = 41$. Subtracting the two equations gives $x - y = 0$, so $x = y$, which we already know is impossible. So no pair is symmetric.

   $R$ is not symmetric.

4. Transitive?

   Transitivity: if $(x,y) \in R$ and $(y,z) \in R$, then $(x,z) \in R$.

   But note: for $(y,z)$ to be in $R$, $y$ must be in the domain $\{1,\dots,20\}$. However, the $y$ values in $R$ are all odd numbers from 1 to 39 — many are larger than 20. For example, take $(1,39)$: $y=39$ is not in the domain, so there is no pair $(39,z)$ in $R$. The only way to have a chain is if $y$ itself is ≤ 20.

   Let’s test a possible chain: $(10,21)$? No, $2(10)+21=41$, so $(10,21) \in R$. But $21$ is not in the domain, so no $(21,z)$ exists.

   What about $(20,1)$? $2(20)+1=41$, so $(20,1) \in R$. Now $1$ is in the domain: is there a $(1,z)$? Yes, $(1,39)$. So we have $(20,1)$ and $(1,39)$. For transitivity we’d need $(20,39)$: $2(20)+39=79 \neq 41$, so it fails. …

Method: Analysing a relation defined by an equation

Use this when a relation on $\mathbb{N}$ (or $\mathbb{Z}$, $\mathbb{R}$) is given by an equation like $px + qy = k$, and you must find its domain/range and test reflexive/symmetric/transitive.

Steps

Step 1: Solve the defining equation for $y$ and impose the set constraints.

Write $y$ in terms of $x$, then use $x \ge 1$ and $y \ge 1$ (for $\mathbb{N}$) to bound $x$. The allowed $x$-values form the domain; the corresponding $y$-values form the range. This is where the honest range comes from — never assume $y$ ranges over the whole set.

Step 2: Test reflexive by substituting $y = x$.

A relation is reflexive only if $(x,x)$ satisfies the equation for every $x$ in the set. Put $y = x$; if the resulting equation has no natural-number solution (or holds only for one special $x$), it is not reflexive.

Step 3: Test symmetric by swapping the roles of $x$ and $y$. …

Common Mistakes

Mistake 1: Taking the range as 'all of $\mathbb{N}$' or all $y$ that appear, without the lower bound.

Why it's wrong: $y=41-2x$ must itself be a natural number, so $x\le 20$ and $y$ only takes the odd values $1,3,\dots,39$. Correct approach: impose $y\ge 1$ to bound $x$, then read off the honest domain $\{1,\dots,20\}$ and range $\{1,3,\dots,39\}$.

Mistake 2: Declaring $R$ reflexive without testing $(x,x)$.

Why it's wrong: reflexive needs $2x+x=41$, i.e. $x=41/3$, which is not natural, so no self-pair exists at all. Correct approach: substitute $y=x$ and check a solution actually exists in $\mathbb{N}$. …