in Set Theory & Algebra retagged by
2,309 views
14 votes

Let $A$ be a set of $n(>0)$ elements. Let $N_r$ be the number of binary relations on $A$ and let $N_f$ be the number of functions from $A$ to $A$

  1. Give the expression for $N_r,$ in terms of $n.$
  2. Give the expression for $N_f,$ terms of $n.$
  3. Which is larger for all possible $n,N_r$ or $N_f$
in Set Theory & Algebra retagged by
2.3k views

1 comment

A relation on a set S={1,2,3} → {(1,2),(1,3)} this is a relation but not a function.

Every function is a relation but converse is false.
0

Subscribe to GO Classes for GATE CSE 2022

4 Answers

28 votes
 
Best answer

A. Consider a $2D$ matrix of $n\times n$ size where each dimension represent the $n$ elements of $A.$ So, we get $n^2$ elements where each one can be viewed as an ordered pair $(i,j), 1\leq i,j \leq n.$ Now if we consider a set with these $n^2$ elements, any subset of it will be a binary relation from $A$ to $A$.

So, no. of binary relations, $N_r = 2^{n^2}$

For a function from a set of $n$ elements to a set of $m$ elements we have $m$ choice for each of the $n$ elements of the domain. Thus, we get $m^n$ possibilities. Here, $m = n.$ So,

No. of functions, $N_f = n^n$

Now, number of functions must be smaller than the number of relations because every function is also a relation. We can prove this formally as follows:

$ \displaystyle \lim_{n\to \infty} \dfrac{2^{n^2}}{n^n} $

$\dfrac{\infty}{\infty} \text{ form}.$ So, applying L'Hopital's rule (Limit remains same after taking derivative of both numerator and denominator)

$= \displaystyle \lim_{n \to \infty} \dfrac{2n.\log 2. 2^{n^2}}{nn^{n-1}}$

We can continue like this applying L'Hopital's rule and we should eventually get $1$ in the denominator (as after each derivation power of $n$ decreases by $1$) which gives the limit value as $\infty.$ So, $2^{n^2}$ is faster growing than $n^n.$

So, for sufficiently large $n,$ $N_r > N_f$

edited by

6 Comments

edited by
Here, no need to do further calculation which one is greater. A function is a subset of relation.

Every function is a relation but every relation is not a function.So, clearly relation size is greater than that of number of function.
27

(as after each derivation power of n decreases by 1)

Where is this happening?

0

@Divy Kala-No need to do derivation to tell you which one is greater.

2
Yup.. functions are a subset of relations. But I think the explanation given in the answer is wrong, the denominator does not decrease at all.
0

@Ayush Upadhyaya functions is a subset of relations because one to many type of relations are excluded from being functions right?

2

$(2^{x^{2}})/(x^{x})$ has limit infinity.

 

but the derivation is wrong.

derivative  of $x^{x}$  is $x^{x} (1+\log x)$.

 

 

so while in numerator with each derivative we are getting “x” in denominator, with each derivative, we are getting(1+logn) extra.  so limit is infinity

2
3 votes

A.  Number of binary relations on set A = Nr = 2^(n^2)

B.  Number of functions on set A = Nf = n^n

C.  No of relation > No of function( Nr >Nf) ,because function is restricted version of relation so it is obvious.By the way, we can check by taking log at both sides.

3 Comments

edited by

Set > Cartesian product > relation > function

1
How cartesian product is a subset of set????
3

The cartesian product consists of sets as it the cross product of a set with a set resulting in a set consisting of many sets. That's is why it comes under the sets.

0
1 vote

$1.N_{r}=2^{n^2}$

$2.N_{f}=n^n$

$3. N_{r}>N_{f}$

$f:R\rightarrow R,$ $f(x)=\sqrt{x}$

Is this a function ? NO !

$f(4)=\sqrt {4}=\pm 2$

But it is certainly a relation.

$\therefore$ Every function is a relation, but not every relation is a function.

0 votes
$N_r = 2^{n^{2}}$ as total elements in the Cartesian product is $n \times n$ out of which each entry may or may not be included in the relation.

$N_f = n^{n}$ Each element can map to any of the n elements.

Take log of both the $N_f = n^{n}$ and $N_r = 2^{n^{2}}$

we get $n* log n  $ and $n^2 log2$ we know that $n^2 log2$ > $n* log n  $ so $N_r$ > $N_f$

Related questions

Ask
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