
Let a relation $R:N\to N$ such that $aRb\Leftrightarrow a\le {{b}^{3}}\forall a,b\in N$
Which of the following statements are true? Justify.
\[\begin{align}
& \left( i \right)\left( a,a \right)\in R,aRa \\
& \left( ii \right)If\text{ }\left( a,b \right)\in R\text{ then }\left( b,a \right)\in R \\
& \left( iii \right)If\text{ }\left( a,b \right),\left( b,c \right)\in R\text{ then }\left( a,c \right)\in R \\
\end{align}\]
Answer
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Hint: In this question, we are given a relation defined from the set of natural numbers to the set of natural numbers given by $a\le {{b}^{3}}$. We need to state the given statement as true or false. For proving we will consider some examples from natural numbers and then find if the statements are true or false.
Complete step-by-step solution
Here, we are given the relation R as $aRb\Leftrightarrow a\le {{b}^{3}}$. The relation R is defined from the set of natural numbers to the set of natural numbers.
Let us pick each statement one by one and try to prove it.
(i) We are given the statement that $\left( a,a \right)\in R$ which means, we have to prove that for any 'a' belonging to the set of natural number $a\le {{a}^{3}}$.
Natural numbers start from 1. So, we can see $1\le {{1}^{3}}\Rightarrow 1\le 1$ which is true. Similarly, for all natural numbers such that $2\le {{2}^{3}}\Rightarrow 2\le 8,3\le {{3}^{3}}\Rightarrow 3\le 27$ the relation is true.
Therefore, $\forall a\in N,\left( a,a \right)\in N$ or we can say aRa.
(ii) We are given the statement that if \[\left( a,b \right)\in R\] then \[\left( b,a \right)\in R\] which means we have to prove that $b\le {{a}^{3}}$ if $a\le {{b}^{3}}$.
We know, this statement is false.
Let us take an example to understand this, let us consider that a = 2 and b = 9.
As we can see $2\le {{9}^{3}}$ i.e. $2\le 729$ but $9\nleq {{2}^{3}}$ i.e. $98$ because 9>8.
Hence for any two natural numbers a and b $aRb$ but not $bRa$.
So the given statement is false.
(iii) We are given the statement that, if \[\left( a,b \right),\left( b,c \right)\in R\] then \[\left( a,c \right)\in R\] which means that, if $a\le {{b}^{3}}\text{ and }b\le {{c}^{3}}$ then we have to prove that $a\le {{c}^{3}}$.
Since $a\le {{b}^{3}}\text{ and }b\le {{c}^{3}}$ so we can write $a\le {{\left( {{c}^{3}} \right)}^{3}}\Rightarrow a\le {{c}^{9}}$ but this does not means that $a\le {{c}^{3}}$.
Let us understand this with the help of an example.
Let us suppose that \[a=26\in N,b=3\in N,c=2\in N\].
As we can see, $26\le {{3}^{3}}$ i.e. $26\le 27$ which is true, also $3\le {{2}^{3}}$ i.e. $3\le {{8}^{3}}$ which is true.
But we can see that $26\nleq {{2}^{3}}$ i.e. $26 \nleq 8$ because $26> 8$.
Hence for any three numbers a, b, and c, if aRb and bRc then it does not imply that aRc. So the given statement is false.
Note: Students should note that, if any relation satisfies (i) statement, it is said to be a reflexive relation. If it satisfies (ii) statement, it is said to be symmetric relation and if it satisfies (iii) statement, it is said to be transitive relation. If it satisfies all three statements, it is said to be an equivalence relation. Students can make mistakes of proving cRa instead of aRc in the last statement.
Complete step-by-step solution
Here, we are given the relation R as $aRb\Leftrightarrow a\le {{b}^{3}}$. The relation R is defined from the set of natural numbers to the set of natural numbers.
Let us pick each statement one by one and try to prove it.
(i) We are given the statement that $\left( a,a \right)\in R$ which means, we have to prove that for any 'a' belonging to the set of natural number $a\le {{a}^{3}}$.
Natural numbers start from 1. So, we can see $1\le {{1}^{3}}\Rightarrow 1\le 1$ which is true. Similarly, for all natural numbers such that $2\le {{2}^{3}}\Rightarrow 2\le 8,3\le {{3}^{3}}\Rightarrow 3\le 27$ the relation is true.
Therefore, $\forall a\in N,\left( a,a \right)\in N$ or we can say aRa.
(ii) We are given the statement that if \[\left( a,b \right)\in R\] then \[\left( b,a \right)\in R\] which means we have to prove that $b\le {{a}^{3}}$ if $a\le {{b}^{3}}$.
We know, this statement is false.
Let us take an example to understand this, let us consider that a = 2 and b = 9.
As we can see $2\le {{9}^{3}}$ i.e. $2\le 729$ but $9\nleq {{2}^{3}}$ i.e. $98$ because 9>8.
Hence for any two natural numbers a and b $aRb$ but not $bRa$.
So the given statement is false.
(iii) We are given the statement that, if \[\left( a,b \right),\left( b,c \right)\in R\] then \[\left( a,c \right)\in R\] which means that, if $a\le {{b}^{3}}\text{ and }b\le {{c}^{3}}$ then we have to prove that $a\le {{c}^{3}}$.
Since $a\le {{b}^{3}}\text{ and }b\le {{c}^{3}}$ so we can write $a\le {{\left( {{c}^{3}} \right)}^{3}}\Rightarrow a\le {{c}^{9}}$ but this does not means that $a\le {{c}^{3}}$.
Let us understand this with the help of an example.
Let us suppose that \[a=26\in N,b=3\in N,c=2\in N\].
As we can see, $26\le {{3}^{3}}$ i.e. $26\le 27$ which is true, also $3\le {{2}^{3}}$ i.e. $3\le {{8}^{3}}$ which is true.
But we can see that $26\nleq {{2}^{3}}$ i.e. $26 \nleq 8$ because $26> 8$.
Hence for any three numbers a, b, and c, if aRb and bRc then it does not imply that aRc. So the given statement is false.
Note: Students should note that, if any relation satisfies (i) statement, it is said to be a reflexive relation. If it satisfies (ii) statement, it is said to be symmetric relation and if it satisfies (iii) statement, it is said to be transitive relation. If it satisfies all three statements, it is said to be an equivalence relation. Students can make mistakes of proving cRa instead of aRc in the last statement.
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