question_answer

Let * be the binary operation on N given by a * b = LCM of a and b. Find

(1) 5 * 7, 20 * 16.

(ii) Is * commutative?

(iii) Is * associative?

(iv) Find the identity of * in N.

(v) Which elements of N are invertible for the operation *?

OR

Let N denote the set of all natural numbers and R be the relation on \[N\times N\] defined by d(a, b) R (c, d), if \[ad(b+c)\text{ }=\text{ }bc(a+b).\] Show that R is an   equivalence relation.

Answer:

The binary operation * on N is defined as

a * b = LCM of a and b.

(i) Consider, 5 * 7= LCM of 5 and 7 = 35 and 20 * 16= LCM of 20 and 16 = 80

(ii) It is known that,

LCM of a and b = LCM of b and \[a,\,\,\forall \,\,a,\,\,b\in N.\]

Thus, the operation '*? is commutative.

(iii) For \[a,\,\,b,\,\,c\in N,\] we have

(a * b)* c = (LCM of a and b)* c = LCM of a b and c

a * (b * c) = a * (LCM of b and c) = LCM of a b and c

Therefore, \[(a*b)*c=a\,\,*(b\,\,*c).\]

\[\therefore \]      \[a*b=b\,\,*a.\]

Thus the operation \[*\]is associative.

(iv) We know that,

LCM of a and 1 = a = LCM of 1 and \[a,\,\,\forall \,\,a\in N,\]

i.e. \[a*1=a=1*a,\] \[\forall a\in N\]

Thus, 1 is the identity of * in N.

(v) An element a in N is invertible with respect to the operation *, if there exists an element b in N such that a * b = e   = b * a.

Here, e = 1. This means that,

LCM of a and b = 1 = LCM of b and a

This case is possible only when a and b are equal to 1.

Thus, 1 is the only invertible element of N with respect to the operation*.

OR

We observe the following properties of relation R.

Reflexivity Let (a, b) be an arbitrary element of \[N\times N\]

Then,\[(a,b)\in N\times N\Rightarrow a,\,\,b\in N\Rightarrow ab(b+a)=ba(a+b)\]

[By commutativity of addition and multiplication on N]

\[\Rightarrow \]   \[(a,\,\,b)R(a,\,\,b)\]

Thus, (a, b)R(a, b)for all \[(a,\,\,b)\in N\,\times N.\] So, R is reflexive on \[N\times N.\]

Symmetry Let \[(a,\,\,b),(c,\,\,d)\in N\,\,\times N\] be such that

\[(a,\,\,b)R(c,\,\,d).\] Then, (a, b)R(c, d)

\[\Rightarrow \]   \[ad(b+c)=bc(a+b)\]

\[\Rightarrow \]   \[cb(d+a)=da(c+b)\]

[By commutativity of addition and multiplication on N]

\[\Rightarrow \]   (c, d)R(a, b)

Thus, (a, b)R(c, d)

\[\Rightarrow \] (c, d)R(a, b) for all(a, b) \[(c,\,\,d)\in N\times N.\]

So, R is Symmetry on \[N\,\times N.\]

Transitivity Let(a, b) (c, d)\[(e,\,\,f)\in N\times N\] such that (a, b)R(c, d)and (c, d)R(e, f). Then,

\[(a,\,\,b)R(c,\,\,d)\Rightarrow ad(b+c)=bc(a+d)\]

\[\Rightarrow \frac{b+c}{bc}=\frac{a+d}{ad}\]

\[\frac{1}{b}+\frac{1}{c}=\frac{1}{a}+\frac{1}{d}\]                                ?(i)

and \[(c,\,\,d)R(e,\,\,f)\Rightarrow cf(d+e)=de(c+f)\]

\[\Rightarrow \]   \[\frac{d+e}{de}=\frac{c+f}{cf}\] \[\Rightarrow \] \[\frac{1}{d}+\frac{1}{e}=\frac{1}{c}+\frac{1}{f}\]        ?(ii)

On addition Eqs. (i) and (ii), we get

\[\left( \frac{1}{b}+\frac{1}{c} \right)+\left( \frac{1}{d}+\frac{1}{e} \right)=\left( \frac{1}{a}+\frac{1}{d} \right)+\left( \frac{1}{c}+\frac{1}{f} \right)\]

\[\Rightarrow \]   \[\frac{1}{b}+\frac{1}{e}=\frac{1}{a}+\frac{1}{f}\] \[\Rightarrow \] \[\frac{b+e}{be}=\frac{a+f}{af}\]

\[\Rightarrow \]   \[af(b+e)=be(a+f)\] \[\Rightarrow \] \[(a,\,\,b)R(e,\,\,f)\]

Thus,    (a, b)R(c, d) and (c, d)R(e, f)

\[\Rightarrow \]   \[(a,\,\,b)R(e,\,\,f)\in N\,\,\times N\]

So, R is transitive on \[N\,\,\times N.\]

Hence, R is an equivalence relation \[N\,\,\times N.\]

The binary operation * on N is defined as

a * b = LCM of a and b.

(i) Consider, 5 * 7= LCM of 5 and 7 = 35 and 20 * 16= LCM of 20 and 16 = 80

(ii) It is known that,

LCM of a and b = LCM of b and \[a,\,\,\forall \,\,a,\,\,b\in N.\]

Thus, the operation '*? is commutative.

(iii) For \[a,\,\,b,\,\,c\in N,\] we have

(a * b)* c = (LCM of a and b)* c = LCM of a b and c

a * (b * c) = a * (LCM of b and c) = LCM of a b and c

Therefore, \[(a*b)*c=a\,\,*(b\,\,*c).\]

\[\therefore \]      \[a*b=b\,\,*a.\]

Thus the operation \[*\]is associative.

(iv) We know that,

LCM of a and 1 = a = LCM of 1 and \[a,\,\,\forall \,\,a\in N,\]

i.e. \[a*1=a=1*a,\] \[\forall a\in N\]

Thus, 1 is the identity of * in N.

(v) An element a in N is invertible with respect to the operation *, if there exists an element b in N such that a * b = e   = b * a.

Here, e = 1. This means that,

LCM of a and b = 1 = LCM of b and a

This case is possible only when a and b are equal to 1.

Thus, 1 is the only invertible element of N with respect to the operation*.

OR

We observe the following properties of relation R.

Reflexivity Let (a, b) be an arbitrary element of \[N\times N\]

Then,\[(a,b)\in N\times N\Rightarrow a,\,\,b\in N\Rightarrow ab(b+a)=ba(a+b)\]

[By commutativity of addition and multiplication on N]

\[\Rightarrow \]   \[(a,\,\,b)R(a,\,\,b)\]

Thus, (a, b)R(a, b)for all \[(a,\,\,b)\in N\,\times N.\] So, R is reflexive on \[N\times N.\]

Symmetry Let \[(a,\,\,b),(c,\,\,d)\in N\,\,\times N\] be such that

\[(a,\,\,b)R(c,\,\,d).\] Then, (a, b)R(c, d)

\[\Rightarrow \]   \[ad(b+c)=bc(a+b)\]

\[\Rightarrow \]   \[cb(d+a)=da(c+b)\]

[By commutativity of addition and multiplication on N]

\[\Rightarrow \]   (c, d)R(a, b)

Thus, (a, b)R(c, d)

\[\Rightarrow \] (c, d)R(a, b) for all(a, b) \[(c,\,\,d)\in N\times N.\]

So, R is Symmetry on \[N\,\times N.\]

Transitivity Let(a, b) (c, d)\[(e,\,\,f)\in N\times N\] such that (a, b)R(c, d)and (c, d)R(e, f). Then,

\[(a,\,\,b)R(c,\,\,d)\Rightarrow ad(b+c)=bc(a+d)\]

\[\Rightarrow \frac{b+c}{bc}=\frac{a+d}{ad}\]

x\[\frac{1}{b}+\frac{1}{c}=\frac{1}{a}+\frac{1}{d}\]                              ?(i)

and \[(c,\,\,d)R(e,\,\,f)\Rightarrow cf(d+e)=de(c+f)\]

\[\Rightarrow \]   \[\frac{d+e}{de}=\frac{c+f}{cf}\] \[\Rightarrow \] \[\frac{1}{d}+\frac{1}{e}=\frac{1}{c}+\frac{1}{f}\]        ?(ii)

On addition Eqs. (i) and (ii), we get

\[\left( \frac{1}{b}+\frac{1}{c} \right)+\left( \frac{1}{d}+\frac{1}{e} \right)=\left( \frac{1}{a}+\frac{1}{d} \right)+\left( \frac{1}{c}+\frac{1}{f} \right)\]

\[\Rightarrow \]   \[\frac{1}{b}+\frac{1}{e}=\frac{1}{a}+\frac{1}{f}\] \[\Rightarrow \] \[\frac{b+e}{be}=\frac{a+f}{af}\]

\[\Rightarrow \]   \[af(b+e)=be(a+f)\] \[\Rightarrow \] \[(a,\,\,b)R(e,\,\,f)\]

Thus,    (a, b)R(c, d) and (c, d)R(e, f)

\[\Rightarrow \]   \[(a,\,\,b)R(e,\,\,f)\in N\,\,\times N\]

So, R is transitive on \[N\,\,\times N.\]

Hence, R is an equivalence relation \[N\,\,\times N.\]