search
Log In
Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true
1 vote
187 views

Let $X$ be the set $\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10 \}$. Define the set $\mathcal{R}$ by $\mathcal{R} = \{(x,y) \in X \times X : x$ and $y$ have the same remainder when divided by $3\}$.

Then the number of elements in $\mathcal{R}$ is

  1. $40$
  2. $36$
  3. $34$
  4. $33$
in Set Theory & Algebra
recategorized by
187 views

2 Answers

3 votes

The given relation R is equivalence relation(since it satisfies reflexive,symmetric,transitive properties).

The  given equivalence relation R partitions the domain into 3 non empty,disjoint,exhaustive subsets {1,4,7,10}, {2,5,8}, {3,6,9}.

Together these 3 subsets form a partition.

we have to form the ordered pairs in equivalence relation from the partition.

The ordered pairs from {1,3,7,10}={(1,1),(1,4),(1,7),(1,10),(4,1),(4,4),(4,7),(4,10),(7,1)(7,4),(7,7),(7,10),(10,1),(10,4),(10,7),(10,10)}

or for the ordered pair(x,y) from the subset {1,3,7,10}, x has 4 choices and y also has 4 choices.

Therefore 16 ordered pairs are formed from subset {1,3,7,10}.

The ordered pairs from {2,5,8}={(2,2),(2,5),(2,8),(5,2),(5,5),(5,8),(8,2),(8,5),(8,8).

or for the ordered pair (x,y) from the subset {2,5,8} ,x has 3 choices and y also has 3 choices.

Therefore 9 ordered pairs are formed from subset {2,5,8}.

similarly 9 order pairs are formed from subset  {3,6,9}.

Therefore there are 16+9+9=34 ordered pairs possible.

 

2 votes
We  can have 10 pairs that is $\{(1,1), (2,2),  \ldots ,(10,10)\}$ and each of the pair have same remainder.

For $1 \rightarrow \{4, 7, 10\} \text{ will have same remiander and they can form pair in 12 ways}$

For $2 \rightarrow \{5, 8\} \text{ will have same remiander and they can form pair in 6 ways}$

For $3 \rightarrow \{6, 9\} \text{ will have same remiander and they can form pair in 6 ways}$ with each other

$(3,6), (6,3), (3,9), (9,3), (6,9), (9,6)$

So total number of ways = $10 + 12 + 6 + 6 = 34$

Hence option $C$ is the answer
0
correct.

Related questions

1 vote
2 answers
1
199 views
A set contains $2n+1$ elements. The number of subsets of the set which contain at most $n$ elements is $2^n$ $2^{n+1}$ $2^{n-1}$ $2^{2n}$
asked Sep 23, 2019 in Set Theory & Algebra Arjun 199 views
0 votes
0 answers
2
145 views
Let $X$ be a nonempty set and let $\mathcal{P}(X)$ denote the collection of all subsets of $X$. Define $f: X \times \mathcal{P}(X) \to \mathbb{R}$ by $f(x,A)=\begin{cases} 1 & \text{ if } x \in A \\ 0 & \text{ if } x \notin A \end{cases}$ Then $f(x, A \cup B)$ equals $f(x,A)+f(x,B)$ $f(x,A)+f(x,B)\: – 1$ $f(x,A)+f(x,B)\: – f(x,A) \cdot f(x,B)$ $f(x,A)\:+ \mid f(x,A)\: – f(x,B) \mid $
asked Sep 23, 2019 in Set Theory & Algebra Arjun 145 views
0 votes
0 answers
3
139 views
Consider the sets defined by the real solutions of the inequalities $A = \{(x,y):x^2+y^4 \leq 1 \} \:\:\:\:\:\:\:\: B = \{ (x,y):x^4+y^6 \leq 1\}$ Then $B \subseteq A$ $A \subseteq B$ Each of the sets $A – B, \: B – A$ and $A \cap B$ is non-empty none of the above
asked Sep 23, 2019 in Set Theory & Algebra Arjun 139 views
1 vote
2 answers
4
222 views
Let $A$ be a set of $n$ elements. The number of ways, we can choose an ordered pair $(B,C)$, where $B,C$ are disjoint subsets of $A$, equals $n^2$ $n^3$ $2^n$ $3^n$
asked Sep 23, 2019 in Combinatory Arjun 222 views
...