Question:
Let $S = \{(x, y) \mid x, y \in \mathbb{N}, 1 \le x \le 15, 1 \le y \le 20\}$ be a set. Let $\mathcal{R}$ be the equivalence relation on $S$ defined by $(x, y) \mathcal{R} (x', y')$ if and only if $x + y = x' + y'$. Then the number of equivalence classes of $\mathcal{R}$ on $S$ is
Let $S = \{(x, y) \mid x, y \in \mathbb{N}, 1 \le x \le 15, 1 \le y \le 20\}$ be a set. Let $\mathcal{R}$ be the equivalence relation on $S$ defined by $(x, y) \mathcal{R} (x', y')$ if and only if $x + y = x' + y'$. Then the number of equivalence classes of $\mathcal{R}$ on $S$ is
Show Hint
An equivalence class for $(x, y) \mathcal{R} (x', y')$ is determined by the output of the function $f(x, y) = x + y$.
The number of equivalence classes is equal to the size of the range of this function over the domain $S$.
The number of equivalence classes is equal to the size of the range of this function over the domain $S$.
Updated On: Jun 16, 2026
- 34
- 35
- 15
- 20
Show Solution
Verified By Collegedunia
The Correct Option is A
Solution and Explanation
Step 1: Understanding the Question:
The problem defines a set $S$ consisting of ordered pairs of natural numbers within specified ranges.
An equivalence relation $\mathcal{R}$ partitions the set into distinct equivalence classes where two elements belong to the same class if their coordinates sum to the same value.
We need to find the total number of distinct possible sums.
Step 2: Key Formula or Approach:
Let $k = x + y$ be the sum of the coordinates.
An equivalence class is formed for each unique value of $k$ that can be achieved by at least one pair $(x, y) \in S$.
The number of equivalence classes is simply the number of distinct values of $k$ that can be formed.
Step 3: Detailed Explanation:
• The coordinates of any pair $(x, y) \in S$ must satisfy:
\[ 1 \le x \le 15 \quad (x \in \mathbb{N}) \]
\[ 1 \le y \le 20 \quad (y \in \mathbb{N}) \]
• Let us find the minimum possible value of $k = x + y$:
\[ k_{\min} = x_{\min} + y_{\min} = 1 + 1 = 2 \]
• Let us find the maximum possible value of $k = x + y$:
\[ k_{\max} = x_{\max} + y_{\max} = 15 + 20 = 35 \]
• Since $x$ and $y$ are natural numbers, the sum $k$ must be an integer.
Can every integer value from 2 to 35 be achieved?
- For any $k \in [2, 16]$, we can choose $x = 1 \in [1, 15]$ and $y = k - 1 \in [1, 20]$.
- For any $k \in [17, 35]$, we can choose $y = 20 \in [1, 20]$ and $x = k - 20$. Since $17 \le k \le 35$, we have $1 \le k - 20 \le 15$, which is a valid value for $x$.
- Thus, every integer $k \in [2, 35]$ is achievable by at least one pair in $S$.
• The number of distinct values of $k$ is:
\[ N = 35 - 2 + 1 = 34 \]
• Each of these values corresponds to exactly one equivalence class.
Step 4: Final Answer:
The number of equivalence classes of $\mathcal{R}$ on $S$ is 34.
The problem defines a set $S$ consisting of ordered pairs of natural numbers within specified ranges.
An equivalence relation $\mathcal{R}$ partitions the set into distinct equivalence classes where two elements belong to the same class if their coordinates sum to the same value.
We need to find the total number of distinct possible sums.
Step 2: Key Formula or Approach:
Let $k = x + y$ be the sum of the coordinates.
An equivalence class is formed for each unique value of $k$ that can be achieved by at least one pair $(x, y) \in S$.
The number of equivalence classes is simply the number of distinct values of $k$ that can be formed.
Step 3: Detailed Explanation:
• The coordinates of any pair $(x, y) \in S$ must satisfy:
\[ 1 \le x \le 15 \quad (x \in \mathbb{N}) \]
\[ 1 \le y \le 20 \quad (y \in \mathbb{N}) \]
• Let us find the minimum possible value of $k = x + y$:
\[ k_{\min} = x_{\min} + y_{\min} = 1 + 1 = 2 \]
• Let us find the maximum possible value of $k = x + y$:
\[ k_{\max} = x_{\max} + y_{\max} = 15 + 20 = 35 \]
• Since $x$ and $y$ are natural numbers, the sum $k$ must be an integer.
Can every integer value from 2 to 35 be achieved?
- For any $k \in [2, 16]$, we can choose $x = 1 \in [1, 15]$ and $y = k - 1 \in [1, 20]$.
- For any $k \in [17, 35]$, we can choose $y = 20 \in [1, 20]$ and $x = k - 20$. Since $17 \le k \le 35$, we have $1 \le k - 20 \le 15$, which is a valid value for $x$.
- Thus, every integer $k \in [2, 35]$ is achievable by at least one pair in $S$.
• The number of distinct values of $k$ is:
\[ N = 35 - 2 + 1 = 34 \]
• Each of these values corresponds to exactly one equivalence class.
Step 4: Final Answer:
The number of equivalence classes of $\mathcal{R}$ on $S$ is 34.
Was this answer helpful?
0
0
Top NEST Mathematics Questions
- Let \(\theta \neq \dfrac{(2n+1)\pi}{2}\), for \(n\in\mathbb{Z}\). Consider the system of equations \[ x\cos\theta+y\sec\theta=0 \] \[ x\sin\theta+y\tan\theta=0 \] This system does not have a unique solution if and only if \(\theta\) belongs to:
- Let \(S\) be the set of all rational numbers \(r\) such that if \[ r=\frac{p}{q} \] with \(p,q\in\mathbb{Z}\), \(q\neq0\), then the two roots of the quadratic equation \[ x^{2}+2px+q^{2}=0 \] are equal. Then the number of elements in \(S\) is:
- Evaluate \[ \int_{0}^{\sqrt{\pi}} x\sin^{2}(x^{2})\,dx \]
- For \(x \in \mathbb{R}\), define \[ \langle x \rangle = x-[x] \] where \([x]\) denotes the greatest integer less than or equal to \(x\). Then for any arbitrary pair of real numbers \(x,y \in \mathbb{R}\), which of the following is always true?
- A possible solution of the system of equations \[ x^2 - 8xy + 16y^2 = 0 \] \[ (\log_{10} x)^2 + 2(\log_{10} x)(\log_{10} y) + (\log_{10} y)^2 = 4 \] is
View More Questions
Top NEST Number System Questions
- Let \(S\) be the set of all rational numbers \(r\) such that if \[ r=\frac{p}{q} \] with \(p,q\in\mathbb{Z}\), \(q\neq0\), then the two roots of the quadratic equation \[ x^{2}+2px+q^{2}=0 \] are equal. Then the number of elements in \(S\) is:
- A possible solution of the system of equations \[ x^2 - 8xy + 16y^2 = 0 \] \[ (\log_{10} x)^2 + 2(\log_{10} x)(\log_{10} y) + (\log_{10} y)^2 = 4 \] is
- Let $m$ and $p$ be real numbers such that the polynomial $f(x) = x^2 + mx + p$ has two distinct negative rational roots. Then the polynomial $g(x) = x^2 - (m^2 - 2p)x + p^2$ has distinct
- Amar, Akbar, and Anthony write a test independently. The probability that Amar passes the test and Akbar fails the test is $\frac{3}{20}$. The probability that Akbar passes the test and Anthony fails the test is $\frac{1}{4}$. The probability that Amar and Anthony both pass the test is $\frac{2}{5}$. Then the probability that at least one of Amar, Akbar and Anthony fails the test is
- The probability that the sum of two integers $m$ and $n$, where $m, n \in \{1, 2, \dots, 50\}$, chosen randomly and independently, being divisible by 3 is
View More Questions
Top NEST Questions
- In the 2024 Olympics, two runners, Noah Lyles and Kishane Thompson, finished the 100.00 m race in 9.784 s and 9.789 s respectively. Assuming that they move with constant speed throughout the race, the distance between the two runners when Noah Lyles touched the finishing line would be closest to:
- The Balmer series of hydrogenic spectral lines refers to an electron transitioning from \(n\geq 3\) to \(n=2\). A hydrogenic atom with atomic number \(Z=24\) undergoes a Balmer transition of the largest possible wavelength. The emitted photon has energy \(E_0\). Another hydrogenic atom with atomic number \(Z=25\) undergoes a similar Balmer transition with emitted photon energy \(E_1\). Then \(|E_1-E_0|\), in eV, is closest to:
- Three positive and two negative charges each of magnitude \(q\) are placed at five of the six vertices of a regular hexagon of side \(R\). The work done to bring a negative charge \(-q\) from infinity to the centre of the hexagon is:
- Consider a metal cylinder of heat capacity \(C\), thermal conductivity \(K\), mass \(m\) and radius \(r\). The combination that will have the dimension of time is:
- Let \(\theta \neq \dfrac{(2n+1)\pi}{2}\), for \(n\in\mathbb{Z}\). Consider the system of equations \[ x\cos\theta+y\sec\theta=0 \] \[ x\sin\theta+y\tan\theta=0 \] This system does not have a unique solution if and only if \(\theta\) belongs to:
View More Questions